# NCERT Solutions for Class 11 chemistry chapter 2 – Structure of Atom

If you are looking for **NCERT Solutions for Class 11 Chapter 2**, we’ve got you covered. Our subject-matter experts have prepared the NCERT Solutions for **Chapter 2 Chemistry Class 11** with accuracy to ensure that you ace your exams. You can go ahead and access **Class 11 Chemistry Chapter 2 Solutions **and refer to the questions and answers for free.

**Access NCERT Solutions for Class 11 Chemistry Chapter – 2 Structure of Atom **

**Chapter 2 – Structure of Atom **

**Chapter 2 Chemistry Class 11 Covers the Below Topics: **

- Discovery of Subatomic Particles
- Discovery of Electron
- Charge to Mass Ratio of Electron
- Charge on the Electron
- Discovery of Protons and Neutrons
- Atomic Models
- Thomson Model of Atom
- Rutherford’s Nuclear Model of Atom
- Atomic Number and Mass Number
- Isobars and Isotopes
- Drawbacks of the Rutherford Model
- Developments Leading to the Bohr’s Model of Atom
- Wave Nature of Electromagnetic Radiation
- Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory
- Photoelectric Effect
- Dual Behaviour of Electromagnetic Radiation
- Evidence for the quantized* Electronic Energy Levels: Atomic spectra
- Emission and Absorption Spectra
- Line Spectrum of Hydrogen
- The Spectral Lines for Atomic Hydrogen
- Bohr’s Model for Hydrogen Atom
- Explanation of Line Spectrum of Hydrogen
- Limitations of Bohr’s Model
- Towards Quantum Mechanical Model Of the Atom
- Dual Behaviour of Matter
- Heisenberg’s Uncertainty Principle
- Significance of the Uncertainty Principle
- Reasons for the Failure of the Bohr Model
- Quantum Mechanical Model Of Atom
- Hydrogen Atom and the Schrödinger Equation
- Orbitals and Quantum Numbers
- Shapes of Atomic Orbitals
- Energies of Orbital
- Filling of Orbitals in Atom
- Aufbau’s Principle
- Pauli Exclusion Principle
- Hund’s Rule of Maximum Multiplicity
- Electronic Configuration of Atoms
- Stability of Completely Filled and Half-Filled Subshells

**Introduction**

**Chapter 2 Chemistry Class 11** will give students an in-depth explanation of the structure of atoms. The chapter will focus on making the students learn about the existence, fundamentals, and characteristics of atoms, along with theorems related to atoms, features of the quantum mechanical model of atoms, Planck’s quantum theory, nature of electromagnetic radiation, the de Broglie relations and Heisenberg’s uncertainty principle, etc.

**Discovery of Subatomic Particles**

In this first segment of Chemistry Class 11 Chapter 2 – Structure of Atom, students will learn about the structure of atoms obtained after doing the experiments on electrical discharge through gases. **NCERT Solutions for Class 11 Chemistry Chapter 2 by Extramarks **has all the answers related to the Discovery of Subatomic particles.

**Discovery of Electron**

This segment will inform students about the discovery of electrons. Questions regarding the same may appear in the exams and you can refer to **NCERT Solutions**** **by Extramarks to prepare yourself.

**Charge to Mass Ratio of Electron**

The chapter covers details about the ratio of the charge of an electron to the mass of an electron. The entire concept is explained to the students with the help of experiments.

**Charge on the Electron**

This segment involves the interpretation of the value of the charge on an electron.

**Discovery of Protons and Neutrons**

This part of the chapter informs students about how protons and neutrons were discovered. To understand this better, students can refer to Extramarks **NCERT Solutions Class 11.**

**Atomic Models**

This section of the chapter discusses the structure of two different atomic models. The purpose behind proposing these is to explain the distribution of charged particles inside an atom.

**Thomson Model of Atom**

The Plum Pudding Model by scientist J.J Thomson is explained in this section of the chapter. To understand this model deeper, you can refer to **NCERT Solutions by Extramarks**.

**Rutherford’s Nuclear Model of Atom**

Scientist Rutherford proposed another model for the structure of a model of an atom. He suggested this after conducting experiments with his students. This section covers this model in detail.

**Atomic Number and Mass Number**

This section introduces students to the concept of atomic number and mass number of each element, and also shows how to calculate them. Atomic number and mass number are explained well by Extramarks experts. Extramarks solutions can be very useful for understanding these concepts from **Class 11 Chemistry Chapter 2.**

**Isobars and Isotopes**

This chapter discusses Isobars and Isotopes. **NCERT Solutions for Class 11 Chapter 2** by Extramarks are ideal to refer to for this part of the chapter. Isobars and Isotopes are different but the two are often confused with each other.

**Drawbacks of the Rutherford Model**

Rutherford did propose the structure of the model of an atom after running experiments with his students, however, there were many drawbacks to his model. Extramarks helps in understanding these drawbacks in depth.

**Developments Leading to the Bohr’s Model of Atom**

Did you know that Bohr’s Model was built upon Rutherford’s Model? The chapter covers all the aspects that led to the development of Bohr’s Model of Atom.

**Wave Nature of Electromagnetic Radiation**

This segment involves information about the discovery of the wave nature of electromagnetic radiation.

**Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory**

Planck’s Quantum Theory led him to win a Nobel Prize. The complex details of this theory can be difficult to understand but **Class 11 Chemistry Chapter 2 NCERT Solutions by Extramarks **can aid in reducing the burden to a great extent.

**Photoelectric Effect**

The photoelectric effect and the experiments that led to its discovery are discussed in this section.

**Dual Behaviour of Electromagnetic Radiation**

Electromagnetic Radiation has a dual nature. This problem of the dual nature of electromagnetic radiation and its concepts are discussed in the chapter.

**Evidence for the quantised* Electronic Energy Levels: Atomic spectra**

Atomic Spectra is explained in this segment of the chapter. In case of any doubts, it is advisable to refer to **NCERT Solutions for Class 11 Chapter 2** by Extramarks for guidance.

**Emission and Absorption Spectra**

The emission spectrum, which is a radiation effect, is explained in this section.

**Line Spectrum of Hydrogen**

The Line Spectrum of Hydrogen is also called the Emission Spectrum of Hydrogen. The reaction when light is passed through hydrogen is explained in detail in this section.

**The Spectral Lines for Atomic Hydrogen**

Different spectral lines named after scientists who discovered them are discussed in this part**. **

**Bohr’s Model for Hydrogen Atom**

The first explanation of the general feature of a hydrogen atom is discussed in this segment.

**Explanation of Line Spectrum of Hydrogen**

Bohr’s model is used here to explain the varied lines of the spectrum of hydrogen.

**Limitations of the Bohr’s Model**

Bohr’s Model has limitations and the same are discussed in this section.

**Towards Quantum Mechanical Model Of the Atom **

This part of the chapter talks about the different theories that tried to solve the limitations faced by Bohr’s Model.

**Dual Behaviour of Matter**

The dual behaviour of matter being discovered caused a huge leap in academics. The chapter discusses the concept in detail.

**Heisenberg’s Uncertainty Principle**

Heisenberg had put forward a theory that said it was uncertain to truly know the position or speed of an electron in an atom. The chapter discusses the uncertainty principle.

**Significance of the Uncertainty Principle**

This part explains the importance of Heisenberg’s Uncertainty Principle.

**Reasons for the Failure of the Bohr’s Model**

Bohr’s Model failed due to certain reasons and the same is expanded upon in this segment.

**Quantum Mechanical Model Of Atom**

A new structure of an atom is covered in this section of the chapter. This new structure of the model of an atom is called the Quantum Mechanical Model of Atom. **Class 11 Chemistry Chapter 2 NCERT Solutions by Extramarks **have explained all complex concepts easily.

**Hydrogen Atom and the Schrödinger Equation**

This segment has a clear explanation of hydrogen and the Schrödinger Equation.

**Orbitals and Quantum Numbers**

Quantum numbers help in distinguishing atomic orbitals. The chapter has an in-depth description of orbitals and quantum numbers.

**Shapes of Atomic Orbitals**

This section covers the factors that influence the shape of atomic orbitals. Students with doubts can refer to **Class 11 Chemistry Chapter 2 Solutions provided by Extramarks**.

**Energies of Orbital**

This part teaches students about how an electron’s energy in a hydrogen atom can be measured and discusses the factors it is dependent on.

**Filling of Orbitals in Atom**

This section of Chapter 2 of Chemistry Class 12 discusses how electrons fill atomic orbitals.

**Aufbau’s Principle**

Aufbau’s Principle states that electrons fill low-energy atomic orbitals first and then the high-energy atomic orbitals. Students will read more about the principle in the chapter.

**Pauli Exclusion Principle**

The Pauli Exclusion Principle says that no two electrons in an atom or a molecule have the same four electronic quantum numbers. This principle influences how the orbital space is filled by electrons.

**Hund’s Rule of Maximum Multiplicity**

This chapter talks about Hund’s Rule of Maximum Multiplicity. Hund’s Rule explains how the same subshell orbitals are filled by electrons.

**Electronic Configuration of Atoms**

The electronic configuration is the process of how electrons are distributed in the orbitals. This part of the chapter talks about the same.

**Stability of Completely Filled and Half-Filled Subshells**

The process of stability of electrons that have filled and half-filled subshells and causes of the same are a part of this chapter. The topic is a bit complex, so students must refer to **NCERT Solutions Class 11** by Extramarks to get a better understanding of the concept.

**About Extramarks**

Extramarks is your go-to platform for past years’ question papers, sample papers, and other learning material. We provide study solutions for all the subjects for Class 1 – 12 so that students get a better understanding of the concepts.

**Class 11 Chemistry Chapter – 2 NCERT Solutions**

If you are looking for **NCERT Solutions for Class 11 Chapter 2**, we’ve got you covered. Our subject-matter experts have prepared the NCERT Solutions for **Chapter 2 Chemistry Class 11** with accuracy to ensure that you ace your exams. You can go ahead and access **Class 11 Chemistry Chapter 2 Solutions **and refer to the questions and answers for free.

**Access NCERT Solutions for Class 11 Chemistry Chapter – 2 Structure of Atom **

**Chapter 2 – Structure of Atom **

**Chapter 2 Chemistry Class 11 Covers the Below Topics: **

- Discovery of Subatomic Particles
- Discovery of Electron
- Charge to Mass Ratio of Electron
- Charge on the Electron
- Discovery of Protons and Neutrons
- Atomic Models
- Thomson Model of Atom
- Rutherford’s Nuclear Model of Atom
- Atomic Number and Mass Number
- Isobars and Isotopes
- Drawbacks of the Rutherford Model
- Developments Leading to the Bohr’s Model of Atom
- Wave Nature of Electromagnetic Radiation
- Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory
- Photoelectric Effect
- Dual Behaviour of Electromagnetic Radiation
- Evidence for the quantized* Electronic Energy Levels: Atomic spectra
- Emission and Absorption Spectra
- Line Spectrum of Hydrogen
- The Spectral Lines for Atomic Hydrogen
- Bohr’s Model for Hydrogen Atom
- Explanation of Line Spectrum of Hydrogen
- Limitations of Bohr’s Model
- Towards Quantum Mechanical Model Of the Atom
- Dual Behaviour of Matter
- Heisenberg’s Uncertainty Principle
- Significance of the Uncertainty Principle
- Reasons for the Failure of the Bohr Model
- Quantum Mechanical Model Of Atom
- Hydrogen Atom and the Schrödinger Equation
- Orbitals and Quantum Numbers
- Shapes of Atomic Orbitals
- Energies of Orbital
- Filling of Orbitals in Atom
- Aufbau’s Principle
- Pauli Exclusion Principle
- Hund’s Rule of Maximum Multiplicity
- Electronic Configuration of Atoms
- Stability of Completely Filled and Half-Filled Subshells

**Introduction**

**Chapter 2 Chemistry Class 11** will give students an in-depth explanation of the structure of atoms. The chapter will focus on making the students learn about the existence, fundamentals, and characteristics of atoms, along with theorems related to atoms, features of the quantum mechanical model of atoms, Planck’s quantum theory, nature of electromagnetic radiation, the de Broglie relations and Heisenberg’s uncertainty principle, etc.

**Discovery of Subatomic Particles**

In this first segment of Chemistry Class 11 Chapter 2 – Structure of Atom, students will learn about the structure of atoms obtained after doing the experiments on electrical discharge through gases. **NCERT Solutions for Class 11 Chemistry Chapter 2 by Extramarks **has all the answers related to the Discovery of Subatomic particles.

**Discovery of Electron**

This segment will inform students about the discovery of electrons. Questions regarding the same may appear in the exams and you can refer to **NCERT Solutions **by Extramarks to prepare yourself.

**Charge to Mass Ratio of Electron**

The chapter covers details about the ratio of the charge of an electron to the mass of an electron. The entire concept is explained to the students with the help of experiments.

**Charge on the Electron**

This segment involves the interpretation of the value of the charge on an electron.

**Discovery of Protons and Neutrons**

This part of the chapter informs students about how protons and neutrons were discovered. To understand this better, students can refer to Extramarks **NCERT Solutions Class 11.**

**Atomic Models**

This section of the chapter discusses the structure of two different atomic models. The purpose behind proposing these is to explain the distribution of charged particles inside an atom.

**Thomson Model of Atom**

The Plum Pudding Model by scientist J.J Thomson is explained in this section of the chapter. To understand this model deeper, you can refer to **NCERT Solutions by Extramarks**.

**Rutherford’s Nuclear Model of Atom**

Scientist Rutherford proposed another model for the structure of a model of an atom. He suggested this after conducting experiments with his students. This section covers this model in detail.

**Atomic Number and Mass Number**

This section introduces students to the concept of atomic number and mass number of each element, and also shows how to calculate them. Atomic number and mass number are explained well by Extramarks experts. Extramarks solutions can be very useful for understanding these concepts from **Class 11 Chemistry Chapter 2.**

**Isobars and Isotopes**

This chapter discusses Isobars and Isotopes. **NCERT Solutions for Class 11 Chapter 2** by Extramarks are ideal to refer to for this part of the chapter. Isobars and Isotopes are different but the two are often confused with each other.

**Drawbacks of the Rutherford Model**

Rutherford did propose the structure of the model of an atom after running experiments with his students, however, there were many drawbacks to his model. Extramarks helps in understanding these drawbacks in depth.

**Developments Leading to the Bohr’s Model of Atom**

Did you know that Bohr’s Model was built upon Rutherford’s Model? The chapter covers all the aspects that led to the development of Bohr’s Model of Atom.

**Wave Nature of Electromagnetic Radiation**

This segment involves information about the discovery of the wave nature of electromagnetic radiation.

**Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory**

Planck’s Quantum Theory led him to win a Nobel Prize. The complex details of this theory can be difficult to understand but **Class 11 Chemistry Chapter 2 NCERT Solutions by Extramarks **can aid in reducing the burden to a great extent.

**Photoelectric Effect**

The photoelectric effect and the experiments that led to its discovery are discussed in this section.

**Dual Behaviour of Electromagnetic Radiation**

Electromagnetic Radiation has a dual nature. This problem of the dual nature of electromagnetic radiation and its concepts are discussed in the chapter.

**Evidence for the quantised* Electronic Energy Levels: Atomic spectra**

Atomic Spectra is explained in this segment of the chapter. In case of any doubts, it is advisable to refer to **NCERT Solutions for Class 11 Chapter 2** by Extramarks for guidance.

**Emission and Absorption Spectra**

The emission spectrum, which is a radiation effect, is explained in this section.

**Line Spectrum of Hydrogen**

The Line Spectrum of Hydrogen is also called the Emission Spectrum of Hydrogen. The reaction when light is passed through hydrogen is explained in detail in this section.

**The Spectral Lines for Atomic Hydrogen**

Different spectral lines named after scientists who discovered them are discussed in this part**. **

**Bohr’s Model for Hydrogen Atom**

The first explanation of the general feature of a hydrogen atom is discussed in this segment.

**Explanation of Line Spectrum of Hydrogen**

Bohr’s model is used here to explain the varied lines of the spectrum of hydrogen.

**Limitations of the Bohr’s Model**

Bohr’s Model has limitations and the same are discussed in this section.

**Towards Quantum Mechanical Model Of the Atom **

This part of the chapter talks about the different theories that tried to solve the limitations faced by Bohr’s Model.

**Dual Behaviour of Matter**

The dual behaviour of matter being discovered caused a huge leap in academics. The chapter discusses the concept in detail.

**Heisenberg’s Uncertainty Principle**

Heisenberg had put forward a theory that said it was uncertain to truly know the position or speed of an electron in an atom. The chapter discusses the uncertainty principle.

**Significance of the Uncertainty Principle**

This part explains the importance of Heisenberg’s Uncertainty Principle.

**Reasons for the Failure of the Bohr’s Model**

Bohr’s Model failed due to certain reasons and the same is expanded upon in this segment.

**Quantum Mechanical Model Of Atom**

A new structure of an atom is covered in this section of the chapter. This new structure of the model of an atom is called the Quantum Mechanical Model of Atom. **Class 11 Chemistry Chapter 2 NCERT Solutions by Extramarks **have explained all complex concepts easily.

**Hydrogen Atom and the Schrödinger Equation**

This segment has a clear explanation of hydrogen and the Schrödinger Equation.

**Orbitals and Quantum Numbers**

Quantum numbers help in distinguishing atomic orbitals. The chapter has an in-depth description of orbitals and quantum numbers.

**Shapes of Atomic Orbitals**

This section covers the factors that influence the shape of atomic orbitals. Students with doubts can refer to **Class 11 Chemistry Chapter 2 Solutions provided by Extramarks**.

**Energies of Orbital**

This part teaches students about how an electron’s energy in a hydrogen atom can be measured and discusses the factors it is dependent on.

**Filling of Orbitals in Atom**

This section of Chapter 2 of Chemistry Class 12 discusses how electrons fill atomic orbitals.

**Aufbau’s Principle**

Aufbau’s Principle states that electrons fill low-energy atomic orbitals first and then the high-energy atomic orbitals. Students will read more about the principle in the chapter.

**Pauli Exclusion Principle**

The Pauli Exclusion Principle says that no two electrons in an atom or a molecule have the same four electronic quantum numbers. This principle influences how the orbital space is filled by electrons.

**Hund’s Rule of Maximum Multiplicity**

This chapter talks about Hund’s Rule of Maximum Multiplicity. Hund’s Rule explains how the same subshell orbitals are filled by electrons.

**Electronic Configuration of Atoms**

The electronic configuration is the process of how electrons are distributed in the orbitals. This part of the chapter talks about the same.

**Stability of Completely Filled and Half-Filled Subshells**

The process of stability of electrons that have filled and half-filled subshells and causes of the same are a part of this chapter. The topic is a bit complex, so students must refer to **NCERT Solutions Class 11** by Extramarks to get a better understanding of the concept.

**About Extramarks**

Extramarks is your go-to platform for past years’ question papers, sample papers, and other learning material. We provide study solutions for all the subjects for Class 1 – 12 so that students get a better understanding of the concepts.

**Q.1 ****(i) Calculate the number of electrons which will together weigh one gram.**

**(ii) Calculate the mass and charge of one mole of electrons.**

**Ans.**

(i) Mass of one electron = 9.10939 × 10^{–31}kg

Number of electrons that weigh 9.10939 × 10^{–31}kg = 1

Number of electrons that will weigh 1 g or 1 × 10^{–3 }kg

=\frac{1}{9.10939\times {10}^{-31}kg}\times 1\times {10}^{-3}kg

= 0.1098 × 10^{28}

= 1.098 x 10^{27}

(ii) Mass of one electron = 9.10939 × 10^{–31}kg

Mass of one mole of electron =

(6.022 × 10^{23}) ×(9.10939 ×10^{–31}kg)

= 5.48 × 10^{–7}kg

Charge on one electron = 1.6022 × 10^{–19}coulomb

Charge on one mole of electron

= (1.6022 × 10^{–19}C) (6.022 × 10^{23})

= 9.65 × 10^{4}C

**Q.2 ****(i) Calculate the total number of electrons present in one mole of methane.**

**(ii) Find**

**(a) the total number and**

**(b) the total mass of neutrons in 7 mg of ^{14}C. (Assume that mass of a neutron =**

**1.675 × 10 ^{-27}kg).**

**(iii) Find**

**(a) the total number and**

**(b) the total mass of protons in 34 mg of NH _{3 }at STP. Will the answer change if the temperature and pressure are changed?**

**Ans.**

(i) Number of electrons present in 1 molecule of methane (CH_{4}) = {1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 ×10^{23} molecules of methane = 6.022 × 10^{23}× 10

= 6.022 × 10^{24}

(ii) (a) Number of atoms of ^{14}C in 1 mole= 6.023 × 10^{23}

Since, 1 atom of ^{14}C contains (14 – 6) i.e., 8 neutrons, the total number of neutrons in 14 g (or 1 mole) of ^{14}C is (6.023 × 10^{23}) × 8.

Therefore, 14 g of ^{14}C contains (6.023 × 10^{23}× 8) neutrons.

Therefore, number of neutrons in 7 mg of ^{14}C is

=\frac{6.023\times {10}^{23}\times 8\times 7\text{}mg}{14000\text{mg}}

= 2.4092 × 10^{21}

(b) Mass of one neutron = 1.67493 × 10^{–27}kg

Mass of total neutrons in 7 mg of ^{14}C =

= (2.4092 × 10^{21}) (1.67493× 10^{–27}kg)

= 4.0352 × 10^{–6}kg

(iii) (a) 1 mole of N of NH_{3} = {1(14) + 3(1)} g of NH_{3}

= 17 g of NH_{3} = 6.022× 10^{23 }molecules of NH_{3}

Total number of protons present in 1 molecule of NH_{3}

= {1(7) + 3(1)} =10

Number of protons in 6.023 × 10^{23} molecules of NH_{3}

= (6.023 × 10^{23}) (10)

= 6.023 × 10^{24}

=17 g of NH_{3 }contains (6.023 × 10^{24}) protons

Number of protons in 34 mg of NH_{3}

=\frac{6.022\times {10}^{24}\times 34\text{}mg}{17000\text{mg}}

= 1.2046 × 10^{22}

(b) Mass of one proton = 1.67493 × 10^{–27}kg

Total mass of protons in 34 mg of NH_{3}

= (1.67493 × 10^{–27}kg) (1.2046 × 10^{22})

= 2.0176 × 10^{–5}kg

The number of sub-atomic particles (protons, electrons, and neutrons) in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

**Q.3 ****How many neutrons and protons are there in the following nuclei?**

{}_{6}^{13}\mathrm{C,}{\text{}}_{8}^{16}\mathrm{O,}{\text{}}_{12}^{24}M\mathrm{g,}{\text{}}_{26}^{56}F\mathrm{e,}{\text{}}_{38}^{88}Sr

**Ans.**

Atomic mass of

{}_{6}^{13}C

= 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass – Atomic number)

= 13 – 6 = 7

Atomic mass of

{}_{8}^{16}O

=16

Atomic number = 8 and Number of protons = 8

Number of neutrons = (Atomic mass – Atomic number)

= 16 – 8 = 8

Atomic mass of

{}_{12}^{24}Mg

=24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass – Atomic number)

= 24 – 12 = 12

Atomic mass of

{}_{26}^{56}Fe

=56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass – Atomic number)

= 56 – 26 = 30

Atomic mass of

{}_{38}^{88}Sr

=88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass – Atomic number)

= 88 – 38 = 50

**Q.4 ****Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)**

**(i) Z = 17, A = 35**

**(ii) Z = 92, A = 233**

**(iii) Z = 4, A = 9**

**Ans.**

**Q.5 ****Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number**

\left(\overline{v}\right)

of the yellow light.

**Ans.**

λ= c/ν

or, ν = c/ λ ——— (1)

Here,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 10^{8 }m/s

λ= wavelength of yellow light = 580 nm = 580 × 10^{–9}m

Substituting the values in expression (1):

v=\frac{3\times {10}^{8}}{580\times {10}^{-9}}=5.17\times {10}^{14}{s}^{-1}

Thus, frequency of yellow light emitted from the sodium lamp is = 5.17 × 10^{14}s^{–1}.

Wave number of yellow light, ν = 1/λ.

\overline{v}=\frac{1}{{\mathrm{580\; x\; 10}}^{-9}}=1.72\times {10}^{6}{m}^{-1}

**Q.6 ****Find energy of each of the photons which**

**(i) correspond to light of frequency 3× 10 ^{15} Hz**

**(ii) have wavelength of 0.50 Å.**

**Ans.**

(i) Energy (E) of a photon is given by the expression,

E = hν

Where,

h = Planck’s constant = 6.626 × 10^{–34}Js

ν= frequency of light = 3 × 10^{15}Hz

Substituting the values in the given expression of E:

E= (6.626 x 10^{-34}) (3 x 10^{15})

=1.988 × 10^{–18}J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

E = h(c/λ)

h = Planck’s constant = 6.626 × 10^{–34}Js

c = velocity of light in vacuum = 3 × 10^{8}m/s

λ =0.50 Å = 0.50 x 10^{–10} m

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^{8}\right)}{0.50\times {10}^{-10}}=3.976\times {10}^{-15} E=3.98\times {10}^{-15}J

**Q.7 ****Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10 ^{–10 }s.**

**Ans.**

Frequency (ν) of light = 1/period

=\frac{1}{2.0\times {10}^{-10}s}=5.0\times {10}^{9}{s}^{-1}

Wavelength (λ) of light = c/ν

Where, c = velocity of light in vacuum = 3×10^{8} m/s

Substituting the value in the given expression of λ:

\lambda =\frac{3\times {10}^{8}}{5.0\times {10}^{9}}=6.0\times {10}^{-2}{m}^{\mathrm{}}

Wave number

\left(\overline{v}\right)

of light = 1/λ

=\frac{1}{6.0\times {10}^{-2}S}=1.66\times {10}^{1}{m}^{-1}=16.66{m}^{-1}

**Q.8 ****What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?**

**Ans.**

Energy (E) of a photon = hν

Energy (E_{n}) of ‘n’ photons = nhν

n={E}_{n}\lambda /hc

Where,

λ = wavelength of light = 4000 pm = 4000 ×10^{–12}m

c = velocity of light in vacuum = 3 × 10^{8} m/s

h = Planck’s constant = 6.626 × 10^{–34} Js

Substituting the values in the given expression of n, i.e.,

n={E}_{n}\lambda /hc

We get,

n=\frac{{E}_{n}\times \lambda}{h\times c}=\frac{1J\times 4\times {10}^{-9}m}{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}=2.012\times {10}^{16}photons

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10^{16}.

**Q.9 ****A photon of wavelength 4 × 10 ^{–7}m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photo electron (1 eV= 1.6020 × 10^{–19} J)**

**Ans.**

(i) Energy (E) of a photon= hν= hc/λ

Where, h = Planck’s constant = 6.626 × 10^{–34 }Js

c = velocity of light in vacuum = 3 × 10^{8} m/s

λ = wavelength of photon = 4 × 10^{–7 }m

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^{8}\right)}{4\times {10}^{-7}}=4.9695\times {10}^{-19}J

Hence, the energy of the photon is 4.97 × 10^{–19 }J.

(ii) The kinetic energy of emission E_{k }is given by

= hν – hν_{0} = (E–W) eV

Where W is the work function. Let us place value of E and W in electron volt in the above equation, we get

{E}_{k}=\left(\frac{4.9695\times {10}^{-19}}{1.6020\times {10}^{-19}}\right)eV-2.13eV

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the following expression:

\begin{array}{l}\frac{1}{2}m{v}^{2}=h\nu -h{\nu}_{0}\\ or\text{v}=\sqrt{\left\{2\left(h\nu -h{\nu}_{0}\right)/m\right\}}\end{array}

Where, (hν – hν_{0}) is kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Multiply kinetic energy with 1.6020 × 10^{–19} to convert into joules. Substitute these values in the given expression of v:

\begin{array}{l}v=\sqrt{\frac{2\times \left(0.9720\times 1.6020\times {10}^{-19}\right)J}{9.10939\times {10}^{-31}kg}}\\ =\sqrt{0.3418\times {10}^{12}{m}^{2}{S}^{-2}}\end{array}

v = 5.84 × 10^{5} ms^{–1}

Hence, the velocity of the photo electron is 5.84 × 10^{5 }ms^{-1}.

**Q.10 ****Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol ^{–1}**

**Ans.**

Energy of sodium (E) = N_{A}hcλ

=\frac{\left({6.023510}^{23}mo{l}^{-1}\right)\left({6.626510}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{242\times {10}^{-9}m}

= 4.947 × 10^{5} J mol^{–1}

= 494 kJ mol^{-1}

**Q.11 ****A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 mm. Calculate the rate of emission of quanta per second.**

**Ans.**

Power of bulb, P = 25 Watt = 25 Js^{–1}

Energy of one photon, E = hν = hc/λ

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{\left(0.57\times {10}^{-6}m\right)}

E = 34.87 × 10^{–20 }J

Rate of emission of quanta per second, R

=\frac{25}{34.87\times {10}^{-20}}=7.169\times {10}^{19}{s}^{-1}

**Q.12 ****Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν _{0}) and work function (W_{0}) of the metal.**

**Ans.**

Threshold wavelength of radian (λ_{0}) =6800 Å

= 6800 × 10^{–10 }m

Threshold frequency (ν_{0}) of the metal= c/λ_{0}

=\frac{3\times {10}^{8}m{s}^{-1}}{6.8\times {10}^{-7}m}

= 4.41 × 10^{14 }s^{-1}

Thus, the threshold frequency (ν_{0}) of the metal is 4.41 × 10^{14 }s^{–1}

Hence, work function (W_{0}) of the metal = hν_{0}

=\left(6.626\times {10}^{-34}Js\right)\left(4.41\times {10}^{14}{s}^{-1}\right)=2.922\times {10}^{-19}J

**Q.13 ****What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?**

**Ans.**

Transition of hydrogen atom from an energy level with n_{i }= 4 to n_{f }= 2 will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

\Delta E=2.18\times {10}^{-18}J\left(\frac{1}{{n}_{i}^{2}}-\frac{1}{{n}_{f}^{2}}\right)=h\nu

Substituting the values in the given expression of E:

\Delta E=2.18\times {10}^{-18}J\left(\frac{1}{{4}^{2}}-\frac{1}{{2}^{2}}\right) \Delta E=2.18\times {10}^{-18}J\left(\frac{\left(1-4\right)}{16}\right) \Delta E=2.18\times {10}^{-18}J\left(\frac{-3}{16}\right)

E = – (4.0875 × 10^{–19}J)

The negative sign indicates the energy of emission.

Wavelength of light emitted, λ = hc/E

Substitute these values in the given expression of λ:

\lambda =\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{\left(4.0875\times {10}^{-19}\right)} \lambda =486.3\times {10}^{-9}m=486\text{}nm

**Q.14 ****How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).**

**Ans.**

The expression of energy is given by,

{E}_{n}=\frac{-\left(2.18\times {10}^{-18}\right){Z}^{2}}{{n}^{2}}

Z = atomic number of the atom

n = principal quantum number

For ionisation from n_{1}= 5 to n_{2 }= ∞

\Delta E={E}_{\propto}-{E}_{5}

=\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left({1}^{2}\right)}{{(\propto )}^{2}}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left({1}^{2}\right)}{{5}^{2}}\right\}\right)

=\left(2.18\times {10}^{-18}J\right)\left(\frac{1}{\left({5}^{2}\right)}\right)\text{}Since,\text{}\frac{1}{\propto}=0

= 0.0872 x 10^{-18} J

ΔE = 8.72 x 10^{-20} J

Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10^{–20}J.

Energy required for n_{1 }= 1 to n = ∞,

ΔE’ = E_{ ∞} – E_{1}

=\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left({1}^{2}\right)}{{(\propto )}^{2}}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left({1}^{2}\right)}{{1}^{2}}\right\}\right) =\left(2.18\times {10}^{-18}J\right)\left(1-0\right)

= 2.18 x 10^{-18} J

On comparing the ionisation energies,

ΔE/ΔE’ =2.18 x 10^{-18} J / 8.72 x 10^{-20} J = 25

Thus, the energy required to remove an electron from n=1 orbit in a hydrogen atom is 25 times the energy required to remove an electron from n = 5 orbit.

**Q.15 ****What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?**

**Ans.**

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible

Thus, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The numbers of spectral lines produced when an electron in the n^{th} level drops down to the ground state is given by n (n – 1)/2.

When n = 6, the Number of spectral lines = 6 (6 – 1)/2 = 15

**Q.16**

**(i) The energy associated with the first orbit in the hydrogen atom is –2.18 ×10 ^{–18 }J atom^{–1}. What is the energy associated with the fifth orbit?**

**(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.**

**Ans.**

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

{E}_{5}=\frac{-\left(2.18\times {10}^{-18}\right)}{{5}^{2}}=\frac{-2.18\times {10}^{-18}}{25}

E_{5 }= – 8.72 × 10^{–20 }J

(ii) Radius of Bohr’s n^{th} orbit for hydrogen atom is given by,

r_{n }= (0.0529 nm) n^{2}

For n = 5, let us calculate radius for hydrogen atom for fifth orbit

r_{5 }= (0.0529 nm) (5)^{2}

r_{5 }= 1.3225 nm

**Q.17 ****Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.**

**Ans.**

For the Balmer series, n_{i }= 2. Thus, the expression of wave number

\left(\overline{v}\right)

is given by

\overline{v}=\left(\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{n}_{f}^{2}}\right)\left(1.097\times {10}^{7}\right){m}^{-1} \overline{v}=1/\lambda

Wave number is inversely proportional to wavelength of transition. Hence, the longest wavelength transition, wave number has to be the smallest.

For wave number to be minimum, n_{f} should be minimum. For the Balmer series, a transition from n_{i }= 2 to n_{f }= 3 is allowed. Hence, by taking n_{f }= 3, we get:

\begin{array}{l}\overline{v}=\left(1.097\times {10}^{7}\right)\left(\frac{1}{{\left(2\right)}^{2}}-\frac{1}{{\left(3\right)}^{2}}\right)\\ \overline{v}=\left(1.097\times {10}^{7}\right)\left(\frac{1}{4}-\frac{1}{9}\right)\\ \overline{v}=\left(1.097\times {10}^{7}\right)\left(\frac{9-4}{36}\right)\end{array}

ν = 1.5236 × 10^{6 }m^{–1}

**Q.18 ****What is the energy in Joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10 ^{–11 }ergs.**

**Ans.**

Let us convert ground state electron energy into joules.

Ground state electron energy = – 2.18 × 10^{–11 }ergs

= – 2.18 × 10^{–11 }× 10^{–7 }J

= – 2.18 × 10^{–18 }J

Energy (E) of the n^{th} Bohr orbit of an atom is given by,

{E}_{n}=\frac{-\left(2.18\times {10}^{-18}\right){Z}^{2}}{{n}^{2}}

Where, Z = atomic number of the atom

Energy required for shifting the electron from n = 1 to n = 5 is given as:

ΔE = E_{5} -E_{1}

=\frac{-\left(2.18\times {10}^{-18}\right){\left(1\right)}^{2}}{{5}^{2}}-\left(2.18\times {10}^{-18}\right) =\left(2.18\times {10}^{-18}\right)\left(1-\frac{1}{25}\right) =\left(2.18\times {10}^{-18}\right)\left(\frac{24}{25}\right)

= 2.0928 x 10^{-18} J

Wave length of emitted light = hc/E

=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^{8}\right)}{\left(2.0928\times {10}^{-18}\right)}

= 9.498 x 10^{-8} m

**Q.19 ****The electron energy in hydrogen atom is given by E _{n }= (–2.18 × 10^{–18})/n^{2 }J. Calculate the energy required to remove an electron completely from the n= 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?**

**Ans.**

Given,

{E}_{n}=-\frac{2.18\times {10}^{-18}}{{n}^{2}}J

Energy required for ionization from n = 2 is given by,

= 0.545 x 10^{-18 }J

\Delta E={E}_{\propto}-{E}_{2} =\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)}{{(\propto )}^{2}}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)}{{2}^{2}}\right\}\right) =\left(\frac{2.18\times {10}^{-18}}{4}-0\right)

ΔE = 5.45 x 10^{-19} J

λ = hc/ΔE

Here, λ is the longest wavelength causing the transition.

\lambda =\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^{8}\right)}{\left(5.45\times {10}^{-19}\right)}=3.647\times {10}^{-7}m

= 3647 x 10^{-10} m

= 3647 Å

**Q.20 ****Calculate the wavelength of an electron moving with a velocity of 2.05 ×10 ^{7 }ms^{–1}.**

**Ans.**

According to de Broglie’s equation,

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of particle

v= velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

\begin{array}{l}\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(2.05\times {10}^{7}m{s}^{-1}\right)}\\ \lambda =3.548\times {10}^{-11}m\end{array}

Hence, the wavelength of the electron moving with a velocity of 2.05 × 10^{7 }ms^{–1 }is 3.548 × 10^{–11 }m.

**Q.21 ****The mass of an electron is 9.1 × 10 ^{–31 }kg. If its K.E. is 3.0 × 10^{–25 }J, calculate its wavelength.**

**Ans.**

From de Broglie’s equation,

λ = h/mv

Given,

Kinetic energy (K.E.) of the electron = 3.0 × 10^{–25 }J.

K.E = ½ mv^{2}

\begin{array}{l}Velocity(v)=\sqrt{2K.E./m}\\ =\sqrt{\frac{2\left(3.0\times {10}^{25}J\right)}{9.10939\times {10}^{-31}kg}}\end{array}

v = 812 ms^{-1}

Substituting the value in the expression of λ:

\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(812m{s}^{-1}\right)}

λ= 8.9627 x 10^{-7} m

Hence, the wavelength of the electron is 8.9627 × 10^{–7 }m.

**Q.22 ****Which of the following are isoelectronic species, i.e., those having the same number of electrons? Na ^{+}, K^{+}, Mg^{2+}, Ca^{2+}, S^{2–}, Ar**

**Ans.**

Number of electrons in Na^{+ }= 11–1 = 10

K^{+ }= 19–1 = 18

Mg^{2+}= 12–2 = 10

Ca^{2+ }= 20–2 = 18

S^{2– }= 16+2 = 18

Ar = 18

Hence Na^{+ }and Mg^{2+} are isoelectronic species and

K^{+}, Ca^{2+}, S^{2–}, Ar are isoelectronic species.

**Q.23 ****(i)Write the electronic configurations of the following ions: (a) H ^{–} (b) Na^{+}(c) O^{2–}(d) F^{–}**

**(ii) What are the atomic numbers of elements whose outermost electrons are represented by**

**(a) 3s ^{1} (b) 2p^{3 }and (c) 3p^{5}?**

**(iii)Which atoms are indicated by the following configurations?**

**(a) [He] 2s ^{1}**

**(b) [Ne] 3s ^{2}3p^{3}**

**(c) [Ar] 4s ^{2}3d^{1}.**

**Ans.**

(i) (a) H^{–} ion:

The electronic configuration of H atom is 1s^{1}.

A negative charge on hydrogen indicates the gain of electron by it.

Electronic configuration of H^{–} = 1s^{2}

(b) Na^{+} ion:

The electronic configuration of Na atom is 1s^{2}2s^{2}2p^{6}3s^{1}.

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of Na^{+} =1s^{2} 2s^{2} 2p^{6} 3s^{0}

=1s^{2} 2s^{2} 2p^{6}

(c)O^{2–} ion:

The electronic configuration of O atom is 1s^{2} 2s^{2} 2p^{4}

A dinegative charge on the species indicates that two electrons are gained by it.

Electronic configuration of O^{2–}ion = 1s^{2} 2s^{2} 2p^{6}

(d)F^{–}ion:

The electronic configuration of F atom is 1s^{2 }2s^{2 }2p^{5}.

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of F^{– }ion = 1s^{2 }2s^{2 }2p^{6}

(ii) (a) 3s^{1}:

Completing the electron configuration of the element as = 1s^{2} 2s^{2} 2p^{6} 3s^{1 }

Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11

Atomic number of the element = 11

(b)2p^{3}:

Completing the electron configuration of the element as

= 1s^{2}2s^{2}2p^{3}

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c) 3p^{5}:

Completing the electron configuration of the element as

= 1s^{2}2s^{2}2p^{5 }

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a)[He] 2s^{1}:

The electronic configuration of the element is

[He] 2s^{1}= 1s^{2}2s^{1}.

Atomic number of the element = 3

The element is lithium (Li).

b)[Ne] 3s^{2}3p^{3} :

The electronic configuration of the element is

[Ne]3s^{2}3p^{3 }= 1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}

Atomic number of the element = 15

Hence, the element with the electronic configuration

[Ne] 3s^{2}3p^{3} is phosphorus (P).

(c)[Ar] 4s^{2}3d^{1}:

The electronic configuration of the element is [Ar]4s^{2}3d^{1}= 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{1}.

Atomic number of the element = 21

Hence, the element with the electronic configuration

[Ar] 4s^{2}3d^{1} is scandium (Sc).

**Q.24 ****What is the lowest value of n that allows g- orbitals to exist?**

**Ans.**

For g-orbital, l (Azimuthal quantum number) = 4,

As for any value of principal quantum number(n), the Azimuthal quantum number (l) can have a value from zero to (n–1).

∴ For l= 4, minimum value of n= 5

**Q.25 ****An electron is in one of the 3d orbitals. Give the possible values of n, l and m _{l} for this electron.**

**Ans.**

For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (m_{l}) = –2,–1, 0, 1, 2

**Q.26 ****An atom of an element contains 29 electrons and 35 neutrons. Deduce**

**(i) the number of protons and (ii) the electronic configuration of the element**.

**Ans.**

(i) For a neutral atom, number of protons is equal to the number of electrons. Thus the number of protons in the atom of given element = 29.

(ii)The electronic configuration of the atom is

= 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}.

**Q.27 ****Give the number of electrons in the species H _{2}^{+}, H_{2}, and O_{2}^{+}.**

**Ans.**

Number of electrons present in hydrogen molecule (H_{2})

=1+1 = 2

Number of electrons in H_{2}^{+} = 2-1= 1

Number of electrons in H_{2}= 1 + 1 = 2

Number of electrons present in oxygen molecule (O_{2}) = 8 + 8 = 16

Number of electrons in O_{2}^{+} = 16 – 1 = 15

**Q.28 ****(i) An atomic orbital has n = 3. What are the possible values of l and m _{l}?**

**(ii) List the quantum numbers (m _{l} and l) of electrons for 3d orbital.**

**(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f**

**Ans.**

(i) n = 3 (Given)

For a given value of n, l can have values from 0 to

(n– 1).

For n= 3, l= 0, 1, 2

For a given value of l, m_{l} can have (2l+ 1) values

For l= 0, m= 0

l= 1, m= – 1, 0, 1

l= 2, m= – 2, – 1, 0, 1, 2

Thus, for n= 3 l= 0, 1, 2 m_{0}= 0

m_{1}= – 1, 0, 1 m_{2}= – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l= 2. For a given value of l, m_{l} can have (2l+ 1) values i.e., 5 values.

∴ For l = 2, m_{l} = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l= 1.

For a given value of n, l can have values from zero to (n– 1).

∴ For l =1, the minimum value of n is 2.

Similarly, for f-orbital, l = 3.

For l= 3, the minimum value of n is 4.

Hence, 1p and 3f do not exist.

**Q.29 ****Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n= 1, l = 0; (b)n= 3; l=1 (c)n = 4; l= 2; (d)n= 4; l=3**

**Ans.**

(a) For n= 1, l= 0 (Given), the orbital is 1s.

(b)For n= 3 and l= 1, the orbital is 3p.

(c) For n= 4 and l= 2, the orbital is 4d.

(d) For n= 4 and l= 3, the orbital is 4f.

**Q.30 ****Explain, giving reasons, which of the following sets of quantum numbers are not possible.**

\begin{array}{l}a)\text{}n=0\text{}l=0\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\\ b)\text{}n=1\text{}l=0\text{}{m}_{l}=0\text{}{m}_{s}=-1/2\\ c)\text{}n=1\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\\ d)\text{}n=2\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=-1/2\\ e)\text{}n=3\text{}l=3\text{}{m}_{l}=-3\text{}{m}_{s}=+1/2\\ f)\text{}n=3\text{}l=1\text{}{m}_{l}=0\text{}{m}_{s}=+1/2\end{array}

**Ans.**

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c)The given set of quantum numbers is not possible.

For a given value of n, ‘l’ can have values from zero to (n– 1). For n= 1, l= 0 and not 1.

(d)The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible.

For n= 3, l= 0 to (3 – 1),i.e., l= 0 to 2 or 0, 1, 2

(f)The given set of quantum numbers is possible.

**Q.31 ****How many electrons in an atom may have the following quantum numbers?**

(a) n= 4, m_{s}=–1/2 and (b) n=3, l=0

**Ans.**

(a) Total number of electrons in an atom for a value of n= 2n^{2}

For n=4, total number of electrons= 2(4)^{2} =32.

The given element has a fully filled orbital as

1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10}

Hence, all the electrons are paired.

Number of electrons (having n= 4 and m_{s}=–1/2) = 16

(b) n = 3, l= 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n= 3 and l= 0 is 2.

**Q.32 ****Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.**

**Ans.**

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of electron is given by:

mvr=nh/2\pi \text{}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots}(1)

Where,

n = 1, 2, 3…

According to de Broglie’s equation:

\begin{array}{l}\lambda =h/mv\\ Or,\text{mv}=h/\lambda \text{}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots}(2)\end{array}

Substituting the value of ‘mv’ from equation (2) in equation (1):

\begin{array}{l}hr/\lambda =nh/2\pi \\ Or,\text{2}\pi \text{r}=n\lambda \text{}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots}(3)\end{array}

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

**Q.33 ****What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n= 4 to n= 2 of He ^{+} spectrum?**

**Ans.**

For He^{+} ion, expression of the wave number

\left(\overline{v}\right)

associated with the Balmer transition, n=4 to n=2

Where, n_{1}= 2, and n_{2}=4

Z = atomic number of helium

\overline{v}=\frac{1}{\lambda}=R{Z}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right) =4R\frac{\left(4-1\right)}{16} \begin{array}{l}\overline{v}=1/\lambda =3R/4\text{}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots}(2)\\ \lambda =4/3R\end{array}

Comparing above two equations the desired transition for hydrogen will have the same wavelength as that of He^{+}.

\begin{array}{l}=R{\left(1\right)}^{2}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)=\frac{3R}{4}\\ =\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)=\frac{3}{4}\text{}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots}(1)\end{array}

By hit and trail method, the equality given by equation (1) is true only when

n_{1}= 1 and n_{2}= 2.

The transition for n_{2}= 2 to n_{1}= 1 in hydrogen spectrum would have the same wavelength as Balmer transition n= 4 to n = 2 of He^{+} spectrum.

**Q.34 ****Calculate the energy required for the process**

H{e}^{+}(g)\to H{e}^{2+}(g)+{e}^{-}

**The ionization energy for the H atom in the ground state is 2.18 × 10 ^{–18 }J atom^{–1}.**

**Ans.**

Energy associated with hydrogen-like species is given by,

{E}_{n}=-2.18\times {10}^{-18}\left({Z}^{2}/{n}^{2}\right)J

For ground state of hydrogen atom,

ΔE = E_{ ∞}–E_{1 }

=0-\left[-2.18\times {10}^{-18}\left\{\frac{{\left(2\right)}^{2}}{{\left(1\right)}^{2}}\right\}\right]J

ΔE= 2.18x 10^{-18 }J

For the given process,

H{e}^{+}(g)\to H{e}^{2+}(g)+{e}^{-}

An electron is removed from n= 1 to n= ∞.

ΔE = E_{ ∞}-E_{1}

=0-\left[-2.18\times {10}^{-18}\left\{\frac{{\left(2\right)}^{2}}{{\left(1\right)}^{2}}\right\}\right]J

ΔE= 8.72 x 10^{-18} J

The energy required for the process= 8.72 x 10^{-18} J.

**Q.35 ****If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.**

**Ans.**

1 m= 100 cm, or 1cm= 10^{-2} m

Length of the scale = 20 cm= 20 × 10^{–2}m

Diameter of a carbon atom = 0.15 nm= 0.15 × 10^{–9}m

One carbon atom occupies 0.15 × 10^{–9}m

Number of carbon atoms that can be placed in a straight line

=\frac{20\times {10}^{-2}m}{0.15\times {10}^{-9}m}=133.33\times {10}^{7}

= 1.33 x 10^{9}

**Q.36 ****2 × 10 ^{8}atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.**

**Ans.**

Number of carbon atoms present = 2 × 10^{8}

Length of the given arrangement = 2.4 cm

Diameter of carbon atom

=\frac{2.4\times {10}^{-2}m}{2\times {10}^{8}m}=1.2\times {10}^{-10}

Radius of carbon atom = (Diameter/2)

=\frac{1.2\times {10}^{-10}m}{2}=6.0\times {10}^{-11}m

**Q.37 ****The diameter of zinc atom is 2.6Å Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.**

**Ans.**

(a) Radius of zinc = (Diameter/2)

=2.6/2 Å

\begin{array}{l}=1.3\times {10}^{-10}m\\ =130\times {10}^{-12}m=130\text{}pm\end{array}

(b) Length of the arrangement = 1.6 cm

= 1.6 x 10^{-2} m Diameter of zinc atom = 2.6 × 10^{–10}m

Number of zinc atoms present in the arrangement

\begin{array}{l}=\frac{1.6\times {10}^{-2}m}{2.6\times {10}^{-10}m}\\ =0.6153\times {10}^{8}m\\ =6.153\times {10}^{7}m\end{array}

**Q.38 ****A certain particle carries 2.5 × 10 ^{–16}C of static electric charge. Calculate the number of electrons present in it.**

**Ans.**

Charge on one electron = 1.6022 × 10^{–19} C

= 1.6022 ×10^{–19 }C charge is carried by 1 electron

Number of electrons carrying a charge of 2.5 × 10^{–16} C

=\frac{2.5\times {10}^{-16}C}{1.6022\times {10}^{-19}C}=1.560\times {10}^{3}C=1560C

**Q.39 ****In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10 ^{–18}C, calculate the number of electrons present on it.**

**Ans.**

Charge on the oil drop = – 1.282 ×10^{–18} C

Charge on one electron = – 1.6022 × 10^{–19} C

∴Number of electrons present on the oil drop

\begin{array}{l}=\frac{-1.282\times {10}^{-18}C}{-1.6022\times {10}^{-19}C}\\ =0.8001\times {10}^{1}=8.0\end{array}

**Q.40 ****In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?**

**Ans.**

A thin foil of lighter atoms cannot give the same results as given with the foil of heavier atoms.

The lighter atoms will not cause enough deflection of α- particles because they would be able to carry very little positive charge.

**Q.41 ****Symbols**

{}_{35}^{79}Br\text{and}{\text{}}^{79}Br

**can be written, whereas symbols**

{}_{79}^{35}B\mathrm{r\; and}\text{}{\text{}}^{35}Br

**are not acceptable. Answer briefly.**

**Ans.**

The convenient way of representing an element with its atomic mass (A) and atomic number (Z) is

{}_{Z}^{A}X

Atomic mass of Br is 79, but atomic number Z is 35, thus the symbol

{}_{35}^{79}Br

is acceptable but

{}_{79}^{35}Br

is not acceptable.

Again, the atomic number of an element is constant, but the atomic mass of an element depends

upon the relative abundance of its isotopes, so

{}^{79}Br

can be written but

{}^{35}Br

can not be written.

**Q.42 ****An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.**

**Ans.**

Let the number of protons in the element be x. Number of neutrons in the element = x + 31.7% of x

= x + 0.317 x = 1.317 x

According to the question,

Mass number of the element = 81

(Number of protons + number of neutrons) = 81

X=\frac{81}{2.317}=34.95

X= 35

The number of protons in the element is 35.

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

The atomic symbol of the element is

{}_{35}^{81}Br

**Q.43 ****An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.**

**Ans.**

Let the number of electrons in the ion carrying a negative charge be x.

Then, number of neutrons present = x+ 11.1% of x

= x + 0.111x = 1.111x

Number of electrons in the neutral atom = (x– 1)

(When an ion carries a negative charge, it carries an extra electron).

Number of protons in the neutral atom = (x – 1)

Mass number of the ion = 37

So, (x– 1) + 1.111x= 37

Or, 2.111x= 38 or, x= 18

The symbol of the ion is

{}_{17}^{37}C{l}^{-}

**Q.44 ****An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.**

**Ans.**

Let the number of electrons present A^{3+ }be x.

Number of neutrons in it = x + 30.4% of x = 1.304x

Since the ion is tri-positive,

Number of electrons in neutral atom = x+ 3 = Number of protons in neutral atom

Mass number of the ion = 56

\begin{array}{l}\left(x+3\right)+\left(1.304x\right)=56\\ or,\text{}2.304x=53\\ X=\frac{53}{2.304}\text{}\text{}or,\text{}X=23\end{array}

Number of protons = x + 3 = 23 + 3 = 26

The symbol of the

{}_{26}^{56}F{e}^{3+}

**Q.45 ****Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.**

**Ans.**

The increasing order of frequency is as follows:

Radiation from FM radio < radiation from microwave oven < amber light < X- rays < Cosmic rays

Frequency is inversely proportional to the wavelength.

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < amber light < radiation from microwave ovens < radiation of FM radio.

**Q.46 ****Nitrogen laser produces a radiation at a wave length of 337.1 nm. If the number of photons emitted is 5.6 × 10 ^{24}, calculate the power of this laser.**

**Ans.**

Power of laser = Energy with which it emits photons

=E=\frac{Nhc}{\lambda}

Where,

N= number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy (E):

E=\frac{\left(5.4\times {10}^{24}\right)\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{337.1\times {10}^{-9}m}

Energy (E):

E = 0.32 × 10^{7}J

= 3.2 × 10^{6}J

The power of the laser is 3.2 × 10^{6}J.

**Q.47 ****Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.**

**Ans.**

Wavelength of radiation emitted = 616 nm

= 616 × 10^{–9}m (Given)

\nu =c/\lambda

Where,

c= velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression (ν)

\begin{array}{l}\nu =\frac{3.0\times {10}^{8}m/s}{616\times {10}^{-9}m}\\ =4.87\times {10}^{8}\times {10}^{9}\times {10}^{-3}{s}^{-1}\end{array}

= 4.87 × 10^{14}s^{–1}

Frequency of emission (ν) = 4.87 × 10^{14}s^{-1}

(b)\text{Velocity of radiation},\text{}\left(\text{c}\right)\text{}=\text{3}.0\text{}\times \text{1}{0}^{\text{8}}{\text{ms}}^{\u2013\text{1}}

Distance travelled by this radiation in 30 s

= (3.0 × 10^{8}ms^{–1}) (30 s)

= 9.0 × 10^{9}m

(6.626 × 10^{–34}Js) (4.87 × 10^{14}s^{–1})

(c) Energy of quantum (E) = 32.27 × 10^{–20}J

=

\begin{array}{l}\text{}\\ (d)\text{Energy of one photon}\left(\text{quantum}\right)\text{}=\text{32}.\text{27}\times \text{1}{0}^{\u2013\text{2}0}\text{J}\end{array}

Therefore, 32.27 × 10^{–20}J of energy is present in 1 quantum.

Number of quanta in 2 J of energy =

\nu =\frac{2J}{32.27\times {10}^{-20}J}

= 6.19 x 10^{18} = 6.2 x 10^{18}

**Q.48 ****In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10 ^{–18}J from the radiations of 600 nm, calculate the number of photons received by the detector.**

**Ans.**

From the expression of energy of one photon (E)

E=hc/\lambda

Where,

λ = wavelength of radiation

h= Planck’s constant

c= velocity of radiation

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{600\times {10}^{-9}m}

E = 3.313 x 10^{-19} J

Energy of one photon = 3.313 × 10^{–19}J

Number of photons received with 3.15 × 10^{–18}J energy

=\frac{315\times {10}^{-18}J}{3.313\times {10}^{-19}J}

=9.5

**Q.49 ****Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 10 ^{15}, calculate the energy of the source.**

**Ans.**

Frequency of radiation (ν) = 1/ (2.0x 10^{-9} s)

= 5.0 x 10^{8} s^{-1}

Energy (E) of source = N h ν

Where,

N = number of photons emitted

h = Planck’s constant

ν = frequency of radiation

Substituting the values in the given expression of (E):

E= (2.5 × 10^{15}) (6.626 × 10^{–34}Js) (5.0 × 10^{8}s^{–1})

= 8.282 × 10^{–10}J

Thus the energy of the source = 8.282x 10^{-10} J

**Q.50 ****The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.**

**Ans.**

λ_{1}= 589nm= 589 x 10^{-9} m

λ_{2} =589.6nm= 589.6 x 10^{-9} m

\begin{array}{l}{v}_{1}=c/{\lambda}_{1}=\frac{3.0\times {10}^{8}m{s}^{-1}}{589\times {10}^{-9}m}=5.093\times {10}^{14}{s}^{-1}\\ {v}_{2}=c/{\lambda}_{2}=\frac{3.0\times {10}^{8}m{s}^{-1}}{589\times {10}^{-9}m}=5.088\times {10}^{14}{s}^{-1}\end{array}

ΔE= E_{2} – E_{1}

= h (ν_{2} – ν_{1})

= (6.626 x 10^{-34} Js) (5.093 – 5.088) x 10^{14 }s^{-1}

= 3.31 x 10^{-22} J

**Q.51 ****The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. (c) If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.**

**Ans.**

Work function (W_{0}) for caesium atom is 1.9 eV.

(a)From the expression

W_{0} = hc/λ_{0 }

we get,

{\lambda}_{0}=\text{hc}/{\text{W}}_{0}

λ_{0}= threshold wavelength

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of (λ_{0}):

We get,

\begin{array}{l}{\lambda}_{0}=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{1.9\times 1.602\times {10}^{-19}J}\\ {\lambda}_{0}=6.53\times {10}^{-7}m\end{array}

The threshold wavelength is 653 nm.

(b) From the expression W_{0} = hν_{0}, we get

ν_{0} = W_{0}/h

Where, ν_{0} = Threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν_{0}:

{v}_{0}=\frac{1.9\times 1.602\times {10}^{-19}J}{6.626\times {10}^{-34}Js}

(1 eV = 1.602 × 10^{–19}J)

ν_{0} = 4.593 × 10^{14 }s^{–1}

The threshold frequency of radiation (ν_{0}) is

4.593 × 10^{14 }s^{–1}.

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy of ejected electron = h (ν–ν_{0})

\begin{array}{l}=hc\left(\frac{1}{\lambda}-\frac{1}{{\lambda}_{0}}\right)\\ =\left(6.626\times {10}^{-34}Js\right)\left(3.0\times {10}^{8}m{s}^{-1}\right)\left(\frac{{\lambda}_{0}-\lambda}{\lambda {\lambda}_{0}}\right)\\ =\left(1.9878\times {10}^{-26}Jm\right)\left[\frac{\left(653-500\right)\times {10}^{-9}m}{\left(653\right)\left(500\right)\times {10}^{-18}{m}^{2}}\right]\\ =\frac{\left(1.9878\times {10}^{-26}Jm\right)\left(153\times {10}^{9}\right)}{\left(653\right)\left(500\right)}J\end{array}

= 9.3149 × 10^{–20} J

Kinetic energy of the ejected photoelectron

= 9.3149 × 10^{–20}J

Since, K.E. = ½ mv^{2}

= 9.3149 x 10^{-20} J

\begin{array}{l}v=\sqrt{\frac{2\left(9.3149\times {10}^{-20}J\right)}{9.10939\times {10}^{-31}kg}}\\ =\sqrt{2.0451\times {10}^{11}{m}^{2}{s}^{-2}}\end{array}

v = 4.52 × 10^{5 }ms^{–1}

The velocity of the ejected photoelectron is 4.52×10^{5}ms^{–1}

**Q.52 ****Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.**

λ (nm) | 500 | 450 | 400 |

v x 10^{-5} (cm s^{-1}) |
2.55 | 4.35 | 5.35 |

**Ans.**

(a) Assuming the threshold wavelength

\begin{array}{l}{\lambda}_{0}nm=\left({\lambda}_{0}\times {10}^{-9}m\right),\\ h\left(\nu -{\nu}_{0}\right)=\frac{1}{2}m{v}^{2}\end{array}

Three different equalities can be formed by the given values as:

\begin{array}{l}=hc\left(\frac{1}{\lambda}-\frac{1}{{\lambda}_{0}}\right)=\frac{1}{2}m{v}^{2}\\ hc\left(\frac{1}{500\times {10}^{-9}}-\frac{1}{{\lambda}_{0}\times {10}^{-9}}\right)=1/2m{\left(2.55\times {10}^{+3\mathrm{}}m{s}^{-1}\right)}^{2}\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{500}-\frac{1}{{\lambda}_{0}}\right)=1/2m{\left(2.55\times {10}^{+3}m{s}^{-1}\right)}^{2}\text{}\mathrm{\dots \dots \dots \dots \dots \dots .}(1)\\ Similarly,\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{450}-\frac{1}{{\lambda}_{0}}\right)=1/2m{\left(4.35\times {10}^{+3}m{s}^{-1}\right)}^{2}\text{}\mathrm{\dots \dots \dots \dots \dots \dots .}(2)\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{400}-\frac{1}{{\lambda}_{0}}\right)=1/2m{\left(5.35\times {10}^{+3}m{s}^{-1}\right)}^{2}\text{}\mathrm{\dots \dots \dots \dots \dots \dots .}(3)\\ Dividng\text{equation (3) by equation (1):}\\ \frac{\left[\frac{{\lambda}_{0}-400}{400{\lambda}_{0}}\right]}{\left[\frac{{\lambda}_{0}-500}{500{\lambda}_{0}}\right]}=\frac{{\left(5.35\times {10}^{3}m{s}^{-1}\right)}^{2}}{{\left(2.55\times {10}^{3}m{s}^{-1}\right)}^{2}}\\ \frac{\left(5{\lambda}_{0}-2000\right)}{4{\lambda}_{0}-2000}={\left(\frac{5.35}{2.55}\right)}^{2}=\frac{28.6225}{6.5025}\\ \frac{\left(5{\lambda}_{0}-2000\right)}{4{\lambda}_{0}-2000}=4.40177\\ \end{array}

Solving this equation we get the value of λ_{0} = 539.68 » 540 nm

(b) Substituting this value in equation (3),we get,

\begin{array}{l}\frac{h\times 3\times {10}^{8}}{{10}^{-9}}\left(\frac{1}{400}-\frac{1}{540}\right)=\frac{1}{2}\left(9.11\times {10}^{-31}\right){\left(5.35\times {10}^{3}\right)}^{2}\\ h=6.70\times {10}^{-38}Js\end{array}

Note: The value of h is not the correct value of Planck’s constant. It is solved according to the values given in NCERT. The correct value of Planck’s constant is

6.626\times {10}^{-34}Js

**Q.53 ****The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.**

**Ans.**

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W_{0}) of radiation and its kinetic energy (K.E) i.e.,

E = W_{0} + K.E.

Or, W_{0 }= E – K.E.

Energy of incident photon (E)= hc/λ

Where,

c = velocity of radiation

h = Planck’s constant

λ= wavelength of radiation

Substituting the values in the given expression of E:

\begin{array}{l}E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{256.7\times {10}^{-9}m}\\ =7.744\times {10}^{-19}J\\ =\frac{7.744\times {10}^{-19}}{1.602\times {10}^{-19}}eV\end{array}

E= 4.83 eV.

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,

K.E = 0.35 V

Work function, W_{0}= E – K.E = 4.83 eV – 0.35 eV

= 4.48 eV

**Q.54 ****If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of 1.5 × 10 ^{7}ms^{–1}, calculate the energy with which it is bound to the nucleus.**

**Ans.**

Energy of incident photon (E) is given by,

\begin{array}{l}E=hc/\lambda \\ E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{150\times {10}^{-12}m}\end{array}

= 1.3252 x 10^{-15} J = 13.252 x 10^{-16} J

Energy of the electron ejected (K.E) = ½ mv^{2}

= ½ (9.10939 x 10^{-31} kg)(1.5 x 10^{7} ms^{-1})^{2}

= 10.2480 × 10^{–17} J

= 1.025 × 10^{–16 }J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E – K.E.

= 13.252 × 10^{–16 }J – 1.025 × 10^{–16 }J = 12.227 × 10^{–16 }J

=\frac{12.227\times {10}^{-16}}{1.602\times {10}^{-19}}eV

= 7.6 x 10^{3} eV

**Q.55 ****Emission transitions in the Paschen series end at orbit n= 3 and start from orbit n and can be represented as ν= 3.29 × 10 ^{15}(Hz) [1/3^{2}– 1/n^{2}] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.**

**Ans.**

**Q.56 ****Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.**

**Ans.**

The radius of the n^{th} orbit of hydrogen-like particles is given by,

\begin{array}{l}r=\frac{0.529{n}^{2}}{Z}\AA \\ r=\frac{52.9{n}^{2}}{Z}pm\end{array}

For radius (r_{1}) = 1.3225 nm = 1.3225 × 10^{–9}m

= 1322.5 × 10^{–12}m = 1322.5 pm

\begin{array}{l}{n}_{1}^{2}=\frac{{r}_{1}Z}{52.9}\\ Or,\text{}{n}_{1}^{2}=\frac{1322.5Z}{52.9}\\ Similarly\\ {n}_{2}^{2}=\frac{211.6Z}{52.9}\\ \frac{{n}_{1}^{2}}{{n}_{2}^{2}}=\frac{1322.5}{211.6}\\ \frac{{n}_{1}^{2}}{{n}_{2}^{2}}=6.25\\ \frac{{n}_{1}}{{n}_{2}}=2.5\\ \frac{{n}_{1}}{{n}_{2}}=\frac{25}{10}=\frac{5}{2}\end{array}

Or, n_{1} = 5 and n_{2} = 2

Thus, the transition is from the 5^{th} orbit to the 2^{nd} orbit. It belongs to the Balmer series.

Wave number

\left(\overline{v}\right)

or the transition is given by,

\begin{array}{l}=1.097\times {10}^{7}{m}^{-1}\left(\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right)\\ =1.097\times {10}^{7}{m}^{-1}\left(\frac{21}{100}\right)\end{array}

= 2.3037 x 10^{6} m^{-1}

Wavelength (λ) associated with the emission transition is given by,

λ =1

/

=\frac{1}{2.303\times {10}^{6}{m}^{-1}}

= 0.434 ×10^{–6} m

λ = 434 nm

It lies in the visible region.

**Q.57 ****Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is 1.6 × 10 ^{6}ms^{–1}, calculate de Broglie wavelength associated with this electron.**

**Ans.**

From de Broglie’s equation,

\begin{array}{l}\lambda =h/mv\\ \lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(1.6\times {10}^{6}m{s}^{-1}\right)}\end{array}

λ= 4.55 x 10^{-10} m, or λ = 455 pm

de Broglie’s wavelength associated with the electron is 455 pm.

**Q.58 ****Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.**

**Ans.**

From de Broglie’s equation,

\begin{array}{l}\lambda =h/mv\\ v=h/m\lambda \end{array}

Where,

v= velocity of particle (neutron)

h= Planck’s constant

m= mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of velocity (v),

v=\frac{6.626\times {10}^{-34}Js}{\left(1.67493\times {10}^{-27}kg\right)\left(800\times {10}^{-12}m\right)}

= 4.94 × 10^{2}ms^{–1}

V = 494 ms^{–1}

Velocity associated with the neutron = 494 ms^{–1}

**Q.59 ****The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10 ^{5}ms^{–1}. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.**

**Ans.**

**Q.60 ****If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is**

h/4π_{m}×0.05 nm, is there any problem in defining this value?

**Ans.**

From Heisenberg’s uncertainty principle,

\Delta x\times \Delta p=h/4\pi \text{}\to \Delta P=\frac{1}{\Delta x}\times \frac{h}{4\pi}

Where,

Δx= uncertainty in position of the electron

Δp= uncertainty in momentum of the electron

Substituting the values in the expression in the last equation:

\begin{array}{l}\Delta p=\frac{1}{0.002nm}\times \frac{6.626\times {10}^{-34}Js}{4\times 3.14}\\ \Delta p=\frac{1}{0.002nm}\times \frac{6.626\times {10}^{-34}Js}{4\times 3.14}\end{array}

= 2.637 × 10^{–23}Jsm^{–1}

Δp = 2.637 × 10^{–23 }kgms^{–1}

Or, uncertainty in the momentum of the electron

= 2.637× 10^{–23 }kgms^{–1}.

Given that actual momentum

=\frac{h}{4{\pi}_{m}\times 0.05nm}

Substituting the values, we get

=\frac{6.626\times {10}^{-34}Js}{4\times 3.14\times 5.0\times {10}^{-11}m}

= 1.055 × 10^{–24 }kgms^{–1}

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

**Q.61 ****The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:**

- n=4, l=2, m
_{l}=-2, m_{s}=-1/2 - n=3, l=2, m
_{l}=1, m_{s}=+1/2 - n=4, l=1, m
_{l}=0, m_{s}=+1/2 - n=3, l=2, m
_{l}=-2, m_{s}=-1/2 - n=3, l=1, m
_{l}=-1, m_{s}=+1/2 - n=4, l=1, m
_{l}= 0, m_{s}=+1/2

**Ans.**

- For n = 4 and l= 2, the orbital occupied is 4d.
- For n = 3 and l= 2, the orbital occupied is 3d.
- For n = 4 and l= 1, the orbital occupied is 4p.
- For n=3 and l= 2, the orbital occupied is 3d.
- For n=3 and l= 1, the orbital occupied is 3p.
- For n=4 and l= 1, the orbital occupied is 4p.

Therefore, the increasing order of energies is

5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d)

**Q.62 ****The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?**

**Ans.**

In a multi-electron atom the effective nuclear charge experienced by an electron is dependent upon the distance between the nucleus and the orbital in which the electron is present. If the distance increases the effective nuclear charge decreases.

Among the given orbitals, the 4p orbital is the farthest from the nucleus, so it experiences the lowest attraction force by the nucleus. Again the electrons in the 4p orbital are shielded by electrons present in the 2p and 3p-orbitals along with the s-orbital. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge.

**Q.63 ****Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f,(iii) 3d and 3p**

**Ans.**

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. It is inversely proportional to the distance between the orbital and the nucleus.

- The electrons present in the 2s orbital experience more effective nuclear charge than the 3s orbital because 2s orbital is closer to nucleus than 3s orbital.
- 4d orbital experiences greater nuclear charge than 4f orbital, since 4d orbital is closer to nucleus.
- 3p orbital is closer to the nucleus than 3d orbital, hence it will experience greater effective nuclear charge.

**Q.64 ****The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?**

**Ans.**

Nuclear charge is directly proportional to the effective nuclear charge. The higher the atomic number higher will be the nuclear charge.

Silicon has 14 protons while aluminium has 13 protons in the nucleus. Hence, silicon has a larger nuclear charge of +14 than aluminium which has a nuclear charge of +13. The electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

**Q.65 ****Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.**

**Ans.**

(a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}

The orbital picture of P can be represented as:

Phosphorus has three unpaired electrons.

(b) Silicon (Si):

Atomic number = 14

The electronic configuration of Si is: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}

The orbital picture of Si can be represented as:

Silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number= 24

The electronic configuration of Cr is:

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{5}

The orbital picture of Cr can be represented as:

Chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number= 26

The electronic configuration is:

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{6}

The orbital picture of Fe can be represented as:

Iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number= 36

The electronic configuration is:

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}

The orbital picture of Kr can be represented as:

Krypton has no unpaired electrons, all orbitals are fully occupied.

**Q.66 ****a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m _{s} value of –1/2 for n= 4?**

**Ans.**

(a) Given n=4,

For a given value of ‘n’, ‘l’ can have values from 0 to (n-1).

Thus, l= 0,1,2,3

Thus, four sub-shells are associated with n=4, which are s, p, d and f.

(b) The maximum number of electrons present in n^{th }shell =2n^{2}. For n=4, the maximum capacity of electrons

= 2×4^{2} = 32.

Total number of orbitals = n^{2 }= 4^{2}=16

If each orbital is taken fully, then it will have 1 electron with m_{s }value of –1/2.

Number of electrons with m_{s} value of (–1/2) is 16.

**Q.67 ****If the velocity of the electron in Bohr’s first orbit is 2.19 × 10 ^{6}ms^{–1}, calculate the de Broglie wavelength associated with it.**

**Ans.**

According to de Broglie’s equation,

\lambda =h/mv

Where,

λ = wavelength associated with the electron

h= Planck’s constant

m = mass of electron

v = velocity of electron

Substituting the values in the expression of λ

\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(2.19\times {10}^{6}m{s}^{-1}\right)}

= 3.32 x 10^{-10} m= 3.32 x 10^{-10} m x (100/100)

= 332 x 10^{-12} m

λ = 332 pm

Wavelength associated with the electron = 332 pm

## FAQs (Frequently Asked Questions)

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