NCERT Solutions Class 12 Physics Chapter 11

NCERT Solutions Class 12 Physics Chapter 11- Dual Nature of Radiation and Matter

The NCERT Solutions Class 12 Physics Chapter 11 provided by Extramarks are a vital study material to make students well equipped for the important CBSE Class 12 Board examinations. The notes aim to help students gain an in-depth understanding of all concepts to avoid difficulty understanding higher and more complex topics. Every minute detail is explained with various techniques to make learning effective. Our solutions are based on the revised guidelines issued by the CBSE with reference to the NCERT’s latest syllabus. 

Dual Nature of Radiation and Matter is a crucial chapter from an exam point of view. Using Maxwell’s equations of electromagnetism and Hertz’s experiments on the generation and detection of EM waves, the chapter gives brief information about the wave nature of light. In addition, one will learn about the Electron Emission, Photoelectric Effect, and Einstein’s equation for Energy Quantum Of Radiation. With help from NCERT Solutions Class 12 Physics Chapter 11, students can solve many MCQs, the theoretical and numerical questions to score high marks.

Students may visit the Extramarks website to get all the latest news and information about the CBSE examinations. In addition to higher classes, Extramarks caters to the educational needs of primary and secondary school. Primary school students can find NCERT Solutions Class 1 to NCERT Solutions Class 5 on Extramarks. Secondary section NCERT Solutions Class 6 to NCERT Solutions Class 10 and NCERT Solutions Class 11 is also available for students for free. 

Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 11

In NCERT Solutions Class 12 Physics Chapter 11, students can expect many objective and subjective questions based on all topics of the chapter. From a basic introduction to the chapter to a detailed explanation of other topics such as observations on the photoelectric effect, absorption of energy, experimental study of the photoelectric effect, wave nature of matter, etc., are covered in this chapter. Studying this chapter will strengthen the foundation of the Dual Nature of the light and light-matter relationship for many competitive examinations such as NEET and JEE (Main and Advanced) exams. 

The key topics covered in chapter 11 are as follows: 

Exercise Topics
11.1

11.2

11.3

11.4

11.5

11.6

11.7

11.8

11.9

Introduction

Electron Emission

Photoelectric Effect

Experimental Study Of Photoelectric Effect.

Photoelectric Effect And Wave Theory Of Light 

Einstein’s Photoelectric Equation: Energy Quantum Of Radiation

Particle Nature Of Light: The Photon

Wave Nature Of Matter

Davisson and Germer Experiment

11.1 Introduction

In the introductory part, the dual wave nature of lights and the work function of a metal is discussed in this chapter. Students study phenomena such as diffraction, interference, and polarisation in this chapter. 

The NCERT Solutions Class 12 Physics Chapter 11 helps students get clarity and essential topics of the syllabus. In addition, students get a clear idea of what is present in the curriculum. Some of the main features of this chapter are as follows: 

  • Dual Nature of Radiation
  • Photoelectric Effect
  • Experimental study of Photoelectric effect
  • Einstein’s Photoelectric equation – Particle nature of light
  • Hertz and Lenard’s Observations
  • Matter waves – Wave nature of particles, and de Broglie relation

11.2 Electron Emission

This section helps students strengthen their understanding of the phenomenon of the emission of photoelectrons. In the NCERT Solutions Class 12 Physics Chapter 11, this sector covers all the basics and paves the way towards complex concepts. As a result, students get a good experience and the overall idea of the main topics. 

11.3 Photoelectric Effect

Within this section, students will be introduced to the phenomenon of the Photoelectric effect, which is defined as the conversion of light to electrical energy based on the law of conservation of energy. Students will also learn about the Factors affecting photoelectric current. Hertz’s observations and Hallwachs’ and Lenard’s observations based on the photoelectric emission are included in this part. 

11.4 Experimental Study Of Photoelectric Effect

This section is bifurcated into three parts:

  • Effect of intensity of light on photocurrent gives the relation between photocurrent and number of photoelectrons. 
  • The effect of potential on the photoelectric current gives information about saturation current.
  • Effect of frequency of incident radiation on stopping potential, which explains the variation of photoelectric current against the potential difference between the plates.

11.5 Photoelectric Effect And Wave Theory Of Light

In this section of NCERT Solutions Class 12 Physics Chapter 11, the wave picture of light is explained using the observations on photoelectric emission studied in the previous section. 

11.6 Einstein’s Photoelectric Equation: Energy Quantum Of Radiation

Students will learn about the relation between electromagnetic radiation and the photoelectric effect proposed by Albert Einstein. In this section of NCERT Solutions Class 12 Physics Chapter 11, students will also learn about Planck’s constant and quanta of radiation energy. 

There are many numerical questions asked on Einstein’s photoelectric equation, which is given by 

Kmax = ho. It will develop a better understanding of the relation of Stopping Potential (Vo) and Threshold Frequency (vo). 

11.7 Particle Nature Of Light: The Photon

This section of NCERT Solutions Class 12 Physics Chapter 11 summarizes the photon picture of electromagnetic radiation. It also explains the conservation of total energy and total momentum. Many solved examples are included in the NCERT textbook based on this concept. 

In the CBSE exams, various numerical questions are asked based on the formulas mentioned in this section. Extramarks NCERT Solutions Class 12 Physics Chapter 11 covers all the essential formulas for reference.

11.8 Wave Nature Of Matter

In this section, the wave nature of matter is explained using the Louis Victor de Broglie relation. He states the relation between wavelength (λ) and momentum (p) of a particle. There are various important formulas included in this part based on which students will have to solve numerical questions. This section of NCERT Solutions Class 12 Physics Chapter 11 includes a detailed explanation about the working and construction of a Photocell with a well-labelled diagram. Students will also learn about the Probability interpretation of matter waves with the help of sums. 

11.9 Davisson and Germer Experiment

Students will gain in-depth information about the wave nature of electrons through the Davisson and Germer Experiment. With the help of the circuit diagram in the NCERT Solutions Class 12 Physics Chapter 11, students can understand this topic well.

List of NCERT Solutions Class 12 Physics Chapter 11 Exercise & Answer Solutions

NCERT Solutions Class 12 Physics Chapter 11 is available on the Extramarks platform for free. It covers all the theoretical answers, stepwise derivations, and detailed explanations of numerical questions. The chapter 11 Physics class 12 NCERT solutions ensure a thorough understanding of all essential topics included in the chapter with graphical representation. To help students speed up their learning process, Extramarks provides NCERT Solutions for all secondary school classes.

Click on the below links to download practice questions and check the solutions for NCERT Solutions Class 12 Physics Chapter 11:

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NCERT Exemplar Class 12 Physics

The NCERT Exemplar Class 12 Physics is accessible on the Extramarks online platform for all students from CBSE affiliated schools. The notes are 100% reliable as each minute detail has a proper explanation lucidly. The NCERT Exemplar books of each subject are very important for school examination preparation and other competitive exams. If students want to attain high scores, it is advised to use NCERT Solutions. 

The ch 11 Physics class 12 carries high weightage. Therefore, it is important for students to grasp the knowledge of all concepts with the help of the worksheets and the best study materials provided by Extramarks. 

Key Features of Class 12 Physics Chapter 11 NCERT Solutions

The Extramarks NCERT solutions class 12 Physics chapter 11 key features include

  • At Extramarks, we provide concise and accurate NCERT solutions Class 12 Physics Chapter 11. These solutions are consistently updated and are based on the revised CBSE syllabus.
  • The notes are prepared by Subject matter experts having years of experience.
  • The NCERT Solutions Class 12 Physics Chapter 11 includes all minute details and is made in simple language for students to easily understand even the most complex concepts. 
  • Students can use these notes to analyze themselves based on their preparation level. 
  • Referring to the NCERT Solutions Class 12 Physics Chapter 11, students can apply their learnings to solve several numerical questions. 
  • The CBSE revision notes are arranged in a chapter-wise and page-wise format to ease accessibility. 

Q.1 Find the
(a) Maximum frequency, and
(b) Minimum wavelength of X-rays produced by 30 kV electrons.

Ans.

(a)  As maximum kinetic energy of photoelectrons is given by Kmax = eVo, where Vo is the stopping potential of anode.Thus using  Kmax =  eVο= hνmaxwe get νmax=  eVοh Whereh = 6.63 × 10-34 Js; e = 1.6 × 10-19 C Thusνmax = 1.6 × 10-19 C × 30,000 V6.63 × 10-34 Js  νmax = 7.24 × 1018 Hz (b) As  λmin = hceV= cνmax= 3 × 108 ms-17.24 × 1018 Hz λmin = 0.414 × 10-10 m= 0.041 nmHence, the minimum wavelength of X-rays producedis 0.041 nm .

Q.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the(a) Maximum kinetic energy of the emitted electrons,(b) Stopping potential, and(c) Maximum speed of the emitted photoelectrons?

Ans.

(a)Work function,ϕ = 2.14  eV= 2.14 × 1.6 × 10-19 J= 3.424 × 10-19 Jν = 6 × 1014 Hzh = 6.26 × 10-34 JsAs per Einstein’s photoelectric equation, maximumkinetic energy of photoelectrons isK.E = hνϕK.E = 6.626 × 10-34 Js × 6 × 1014 Hz 3.424 × 10-19 JK.E = 3.976 – 3.424 × 10-19 J= 0.552 × 10-19 JThus K.E in eV will be 0.552 x 10-19 J1.6 x 10-19 C = 0.345 eV= 0.35 eV(b)By the relationeVo = Maximum K.E eVo= 0.35 eV Vo = 0.35 V (c)As, K.Emax = 12mv2max= 0.35 eV= 0.552 × 10-19 JThus v2max = 0.552 × 10-19 J × 2m =0.552 × 10-19 J × 29.1 × 10-31 kgvmax =0.121318 × 1012 ms-1= 3.48307 × 105 ms-1= 348 kms-1

Q.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Ans.

Given, Vο = 1.5 VThus maximum kinetic energyK.Emax = eV0= 1.6 × 10-19 C × 1.5 VK.Emax = 2.4 × 10-19 J 

Q.4 Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Ans.

aGiven,λ = 632.8  nm= 632.8 × 10-9 m Power, P = 9.42  mW= 9.42 × 10-3 WAs energy of each photon is E = hcλE = 6.626 × 10-34 Js × 3 × 108 ms-1632.8 × 10-9 m = 3.14 × 10-19 JMomentum of photon p = hλ = 6.626 × 10-34 Js 632.8 × 10-9 mp = 1.05 ×10-27 kgms-1bIf N is number of photons emitted per second, then power P transmitted in the beam equals Ntimes the energy per photon so that N = PE= 9.42 × 10-3 W3.14 × 10-19 J= 3 × 1016 photons per second cAs mass of hydrogen atom is equivalent to mass of proton which is 1.67 x 10-27p = mvv = pmThus, v = 1.05×10-27 kgms-11.67×10-27 kg = 0.63 m s-1

Q.5 The energy flux of sunlight reaching the surface of the earth is 1.388 x 103 Wm-2. How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Ans.

Given, λ = 550 nm = 550 × 10 -9 nm h = 6 .626 × 10 -34 Js c = 3 × 10 8 ms -1 Energyflux,Φ = 1 .388 × 10 3 Wm -2 Hence,powerofsunlightpersquaremetre,P = 1 .388 × 10 3 Wm -2 Thus number of photons incident per square metre of earth per second will be N = P E = hc 1 .388 × 10 3 Wm -2 × 550 × 10 -9 m 6 .626 ×10 -34 Js × 3 × 10 8 ms -1 = 3 .85 × 10 21 photons/m 2 s

Q.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10-15 Vs. Calculate the value of Planck’s constant.

Ans.

The slope of the cut-off voltage versus frequency of incidentlight isVον = 4.12 × 10-15 VsAs per photoelectric equationeVο = hνϕthus, the slope of the graph in this case is Vον = hewhich implies that he = 4.12 × 10-15 VsThus,h = 4.12 × 10-15 Vs × e = 4.12 × 10-15 Vs × 1.6 × 10-19 CHence, h = 6.592 × 10-34 Js  

Q.7 A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light?
(b) At what rate are the photons delivered to the sphere?

Ans.

(a) Power of sodium lamp, P = 100 W λ = 589 nm= 589 × 10-9 mEnergy of photonE = hcλ = 6.626 × 10-34 Js × 3 × 108 ms-1589 × 10-9 mThus, E = 3.38 × 10-19 J= 2.11 eV (b)Rate at which the photons are delivered is equal to numberof photons delivered per second .Thus, N = PE  = 100 kgms-13.38 × 10-19 J N = 2.96 × 1020 photons s-1

Q.8 The threshold frequency for a certain metal is 3.3 x 1014 Hz. If light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Ans.

Given that threshold frequencyνo = 3.3 × 1014 HzIncident frequencyν = 8.2 × 1014 Hz h = 6.626 × 10-34 Js e = 1.6 × 10-19 CUsing photoelectric equation, eVo = hν – hνoVo = he ννo Vo=6.626 × 10-34Js 1.6 × 10-19 C × 8.2 × 1014 Hz – 3.3 × 1014HzThus, Vo= 2.03 V

Q.9 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Ans.

Work function, ϕο = 4.2 eV = 4.2 × 1.6 × 10-19 J= 6.72 × 10-19 JAs per definition of work functionϕο = hνο Thus,νο = Wh= 6.72 × 10-19 J 6.626 × 10-34  JsHence,νο = 1.01 × 1015 HzAs λο = cνο= 3 × 108ms11.01 × 1015 HzThus λο = 2.97 × 10-7 m= 297 × 10-9 m= 297 nmSince threshold wavelength λο < incident wavelength λ,which implies that,Threshold frequency  νο > ν, therefore the photoelectricemission for this radiation cannot take place.

Q.10 Light of frequency 7.21 x 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 105 ms-1 are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Ans.

As per photoelectric equation 12 mv2 max = hν – hν0hν0 = hν - 12mv2maxThusν0 = hν - 12mv2maxh = 7.21 × 1014 J – 12 × 9.1 × 10-31 kg × 6 × 105ms12 6.626 × 10-34 Js = 7.21 × 1014 Hz – 2.47 × 1014 Hz= 4.74 × 1014 Hz 

Q.11 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.

Ans.

Given,λ = 488 × 10-9 mstopping potential of photoelectron = 0.38 eVν = cλ = 3 × 108 ms-1488 × 10-9 m Thus, ν = 6.15 × 1014 HzUsing photoelectric equationeV0 = hν – WWhere W is work function of the material, we get W = hν – eV0 W = 6.626 × 10-34 Js × 6.15 × 1014 Hz1.6 × 10-19 C × 0.38 eV W = 3.467 × 10-19 JThus W in eV will be 3.467 × 10-19 J1.6 × 10-19 C= 2.167 eV

Q.12 Calculate the
(a) momentum, and
(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Ans.

(a) Given, V = 56 V, thus for an electron of mass “m”, charge “e” on being accelerated by potential of 56 V, the momentum “p” will be p = 2 meV p = 2 × 9 .1 × 10 -31 kg × 1 .6 × 10 -19 C × 56 V p = 1.630 .72 × 10 -50 kgms -1 p = 4 .038 × 10 -24 kgms -1 (b) As per de – Broglie’s relation wavelength of an electron will be λ = 12.27 V Å Thus, λ = 12.27 56 V Å λ = 1.64 Å = 0.164nm

Q.13 What is the
(a) momentum
(b) speed and
(c) de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Ans.

(a) Given K.E = 120 eV = 120 ×1.6× 10 -19 =1 .92 × 10 -17 J thus for an electron of mass “m”, charge “e”, and the momentum “p” will p = 2mE = 2 × 9 .1 × 10 -31 kg × 1 .92 × 10 -17 J = 34 .94 × 10 -48 kg ms -1 = 5 .91 × 10 -24 kgms -1 (b) The speed of an electron of mass “m” and momentum “p” will be v = p m Thus,v = 5 .91 × 10 -24 kgms -1 9 .1 × 10 -31 kg = 6 .5 × 10 6 ms -1 (c) de-Broglie wavelength “λ” of an electron of momentum “p” will be, λ = h p λ = 6 .626 × 10 -34 Js 5 .91 × 10 -24 kgms -1 = 1 .121 × 10 -10 m = 1.121 Å = 0.1121nm

Q.14 The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron and
(b) a neutron, would have the same de-Broglie wavelength.

Ans.

(a) For an electron, λ = h 2 m e E squaring both sides λ 2 = h 2 2 m e E E = h 2 2 m e λ 2 E = 6 .626 × 10 -34 Js 2 2 × 9 .1 × 10 -31 kg × 589 × 10 -9 m 2 = 6 .95 × 10 -25 J (b) For a neutron, E = h 2 2m n λ 2 = 6 .626 × 10 -34 Js 2 2 × 1 .67 × 10 -27 kg × 589 × 10 -9 m 2 E = 3 .79 × 10 -28 J

Q.15 What is the de-Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 kms-1,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 ms-1, and
(c) a dust particle of mass 1.0 x 10-9 kg drifting with a speed of 2.2 ms-1?

Ans.

(a) Given mass of the bullet, m = 0.040 kg Velocity, v = 1 .0 kms -1 = 1000 ms -1 Thus, wavelength λ = h mv = 6 .626 × 10 -34 Js 0 .040 kg × 1000 ms -1 = 1 .66 × 10 -35 m (b) For a ball of mass, m = 0.060 kg Velocity, v = 1 .0 ms -1 λ = h mv = 6 .626 × 10 -34 Js 0.060 kg × 1.0 ms -1 λ = 1 .10 × 10 -32 m (c) For a dust particle of mass, m = 1 .0 x 10 -9 kg Velocity, v = 2 .2 ms -1 λ = h mv = 6 .626 × 10 -34 Js 1 .0 × 10 -19 kg ×2 .2 ms -1 λ = 3 .01 × 10 -25 m

Q.16 An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta
(b) the energy of the photon and
(c) the kinetic energy of electron.

Ans.

(a) Momentum of an electron and photon,with same wavelength, will be same as p = h λ = 6 .626 × 10 -34 Js 1 × 10 -9 m Thus, p = 6 .626 × 10 -2 5 kgms -1 (b) Energy of photon, E = hc λ = 6 .626 × 10 -34 Js × 3 × 10 8 ms -1 1 × 10 -9 m Thus, E= 1 .99 × 10 -16 J Energy of photon in eV = 1 .99 × 10 -16 1 .6 × 10 -16 keV E = 1.24 keV (c) Kinetic energy of electron = p 2 2m = 6 .626 × 10 -25 kgms -1 2 2 × 9 .1 × 10 -31 kg K.E = 2 .412 × 10 -19 J = 2 .412 × 10 -19 1 .6 × 10 -19 eV = 1.51 eV

Q.17 (a) For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.40 x 10-10 m?
(b) Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of

3 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakeaacaGGGcWaaSaaaeaacaGGGcGaae4maaqaaiaabkdaaaaaaa@3CD0@

kT at 300 K.

Ans.

(a) λ = 1 .40 × 10 -10 m h = 6 .63 × 10 -34 Js m = 1 .67 × 10 -27 kg Thus, λ = h 2mE E = h 2 2mλ 2 E = 6 .63×10 -34 Js 2 2 × 1 .67 × 10 -27 kg × 1 .4×10 -10 m 2 = 6 .7 × 10 -21 J Energy in eV will be E = 6 .7 × 10 -21 J 1 .6 × 10 -19 eV = 4 .19 × 10 -2 eV (b) As average kinetic energy E = 3 2 kT wavelength,λ = h 2mE = h 3mkT where,k = 1 .38 × 10 -23 J K -1 and temperature T = 300 K Thus, λ = 6 .63 × 10 -34 Js 3 × 1 .67 × 10 -27 kg × 1 .38 × 10 -23 J K -1 × 300 K λ = 1 .456 × 10 -10 m

Q.18 Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon).

Ans.

The de-Broglie wavelength of photon is λ1 = hp As for a photon  p = hνcThus, λ1 = hchνc= cν = λ= wavelength of electromagnetic radiation .

Q.19 What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules a this temperature. (Atomic mass of nitrogen = 14.0076 u).

Ans.

As for nitrogen, rms velocity, v = 3kT m Thus, de-Broglie wavelength of nitrogen molecule will be λ = h mv = h 3mkT For Nitrogen molecule, Mass, m = 2 × 14.0076 u = 28.01521u = 28.01521 × 1 .66 × 10 -27 kg Hence,λ = 6 .63 × 10 -34 Js 3 × 28.0152 × 1 .66 × 10 -27 kg × 1 .38 × 10 -23 JK 1 × 300 K λ = 0 .275 × 10 -10 m = 0.0275 nm

Q.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., Its e/m is given to be 1.76 × 1011 Ckg-1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Ans.

As given,V = 500 V;em = 1.76 × 1011 C kg-1Thus, using eV = 12mv2 or v = 2eVmv = 2 × 1.76 × 1011 C kg-1 × 500 V = 1.33 × 107 ms-1b Using the same formula for V = 10 MV = 10 × 106 Vv =2 × 1.76 × 1011 C kg-1 × 10×106 V =3.52 × 1018 ms-1 So , v=1.88 × 109 ms-1The formula employed in part a is not valid. The expression for energy  12mv2 can only be used in the non-relativistic limit.

For very high speed are concerned, the relative domains comesinto consideration. For this we use relativistic formula as given:Here mass,  m = m01-v2c2  is to be consideredSo that, r = m0veB1 - v2 c2

Q.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 106 ms-1 is subject to a magnetic field of 1.30 × 10-4 T normal to beam velocity. What is the radius of circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

Ans.

a As given v = 5.20 × 10 6 ms -1 ; B = 1.30 × 10 -4 T, e m = 1.76 × 10 11 C kg -1 As the condition for a moving electron to trace a circle under the effect of magnetic field is, qvB = mv 2 r or r = mv qB = mv eB q = e Hence, r = 5.20 × 10 6 m s 1 1.76 × 10 11 Ckg 1 × 1.30 × 10 -4 T = 2.27 × 10 -1 m = 22.7 cm b Energy, E = 20 MeV = 20 × 10 6 eV × 1 .6 x 10 -19 J The energy of the electron is given by: E = 1 2 mv 2 v = 2 × 20 × 10 6 eV × 1.6 × 10 -19 C 9 × 10 -31 kg = 2.67 × 10 9 ms -1 Whichisgreaterthanthevelocityoflight.

Q.22 An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (10-2 mm of Hg). A magnetic field of 2.83 x 10-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method) Determine e/m from the data.

Ans.

As given, r = 12.0 cm = 0.12 mAs kinetic energy gained by an electron under the effect of anode potential is  12mv2 = eVWhich implies that mv2 = 2 eV    .………(1)Also, mv2r =  evBThus, 2eVr = evB v = 2 Vr B v = 2 × 100 V 0.12 m × 2.83 × 10-4 T v = 5.89 × 106 ms-1From eqn …..1em = v22V =    5.89 × 106ms12      2 × 100 V = 1.73 × 1011 Ckg-1

Q.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at

0.45 Å MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaaIWaGaaiOlaiaabsdacaqG1aGaaeiiaiaadwmaaaa@3DD1@

. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.

Ans.

(a)As given λmin = 0.45 Å = 0.45 × 10-10 m Thus maximum energy of a photon will beEmax = hcλ= 6.626 × 10-34 Js × 3 × 108 ms-1 0.45 × 10-10 m = 4.42 × 10-15 J = 4.42 × 10-15 1.6 × 10-16   ke V= 27.61 keV (b)In X-ray tube for getting X- ray photons of 27.6 keV, the kinetic energy of atleast 27.6 keV is required. So, acceleratingvoltage of 30 keV is required to applied across the X-ray tube toget X-ray photons.

Q.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ -rays of equal energy. What is the wavelength associated with each γ-ray?
(1 BeV = 109 eV)

Ans.

As given energy carried by the pair of γ-rays = 10.2 BeV Hence, energy of each γ-ray is E = hcλλ = hcE = 6.626 × 10-34 Js × 3 × 108 ms-18.16 × 10-10 J λ = 2.43×10-16 m  E = 10.2 BeV2 = 5.1 BeV = 5.1×109 eV = 5.1 × 109 eV × 1.6×10-19 J = 8.16 ×10-10 J  

Q.25 Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10-10 Wm-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.

Ans.

(a)Number of photons emitted per second, n = Power of transmitter Energy of each photonAs, Power of transmitter is P = 10 k W = 104 W and λ = 500 mUsing ,E = hcλ = (6.626 × 10-34 Js × 3 × 108 ms-1)500 m = 3.98 × 10-28 J Thus,  n = PE= 104 W  3.98 × 10-28 J = 2.52 × 1031 s-1As the energy of a radio photon is exceedingly small and the number of photons emitted per second in a radio beamis enormously large.(b)This is quite a small number, but still large enough to be counted by uscomparison of cases a and b tells us that our eye cannot count the number of photons individually.

Q.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~ 105 Wm-2) red light of wavelength 6328 Å produced by a He-Ne laser?

Ans.

Here , λu = 2271 Å = 2271 × 10-10 mV0 = 1.3 VUsing the photoelectric equation we get,eV0 = hν – W or,  W = hν – eV0 W = hcλ= 6.63 × 10-34 Js × 3 × 10-8 ms-12271 × 10-10 m – 1.6 × 10-19 C × 1.3 V= 6.67 × 10-19 J= 6.67 × 10-191.6 × 10-19  eV =  4.2 eVLet ν0 be the threshold frequeny of the metal .W = hν0ν0 = Wh= 6.67 × 10-19 J6.63 × 10-34 Js= 1.0 × 1015 Hz Also, for red light, λr = 6328 Å = 6328 × 10-10 mThus, frequency of red light νr = cλr = 3 × 108 ms-16328 × 10-10 m = 4.74 × 1014 Hz .Since , νr < ν0 , therefore , the photocell will not respond​to this red light , produced by laser .

Q.27 Monochromatic radiation of wavelength 640.2 nm (1 nm = 10-9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Ans.

As given wavelength from neon lamp is λ = 640.2 nm = 640 .2 × 10 -9 m Using photoelectric equation eV 0 = hν – W = hc λ W W = hc λ – eV 0 = 6 .626 × 10 -34 Js × 3 × 10 8 ms -1 640 .2 × 10 -9 m _ 1 .6×10 -19 C × 0.54V = 3 .10 × 10 -19 J – 8 .64 × 10 -20 J = 3.10 × 0.864 × 10 -19 J = 2 .24 × 10 -19 J Thus, Work function W = 2 .24 × 10 -19 1 .6 × 10 -19 eV = 1.40 eV Also, for wavelength line of an iron source, λ = 4.272 nm = 427 .2 × 10 -9 m, Using, W = hc λ – eV 0 eV 0 = hc λ – W V 0 = hc W e = 6 .626 × 10 -34 Js × 3 × 10 8 ms -1 1 .6 × 10 -19 C × 427 .2 × 10 -9 m 2 .24 × 10 -19 J 1 .6 × 10 -19 C = 2.91V – 1.40 V = 1.51 V Hence,newstoppingpotentialis1.51V.

Q.28

A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:λ1 = 3650 Å , λ2 = 4047 Å , λ3 = 4358 Å,λ4 = 5461 Å , λ5 = 6907 Å .The stopping voltages, respectively, were measured to be:V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 VaDetermine the value of Planck’s constant h.b Estimate the threshold frequency and work function for the material.

Ans.

(a) From the Einstein photoelectric equation, eV o = hν – W Which implies hc λ – eV o = W As, ν 1 = c λ 1 = 3 × 10 8 ms -1 3650 × 10 -10 m = 8 .22 × 10 14 Hz Similarly, ν 2 = c λ 2 = 3 × 10 8 ms -1 4047 × 10 -10 m = 7 .41 × 10 14 Hz ν 3 = c λ 3 = 3 × 10 8 ms -1 4358 × 10 -10 m = 6 .88 × 10 14 Hz ν 4 = c λ 4 = 3 × 10 8 ms -1 5461 × 10 -10 m = 5 .49 × 10 14 Hz ν 5 = c λ 5 = 3 ×10 8 ms -1 6907 × 10 -10 m = 4 .34 × 10 14 Hz A graph can be plotted between V 0 and ν which comes out to be a straight line. Then, h e = ΔV Δν = 1.28 V – 0.16 V 8.22 – 5.50 × 10 14 Hz Thus, h = 1 .6 × 10 -19 C 1.12V 2 .726 × 10 14 Hz = 6 .57 × 10 -34 Js (b) Now using , hc λ 2 – eV 02 = W W = hc λ 2 eV 02 Thus,W = 6 .6 × 10 -34 Js × 3× 10 8 ms -1 4047 × 10 -10 m – 1 .6 × 10 -19 C × 0.95 V = 4 .89 × 10 -19 J – 1 .52 × 10 -19 J = 3 .37 × 10 -19 J = 3 .37 × 10 -19 1 .6 × 10 -19 eV = 2.11 eV Also, ν 0 = W h = 3 .37 × 10 -19 J 6 .6 × 10 -34 Js = 5 .11 × 10 14 Hz

Q.29 The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; MO: 4.17 eV;
Ni : 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Ans.

Here, wavelength of incident radiation is λ = 3300 Å = 3300 ×10 -10 m Distance of the laser from the photocell,r = 1m New distance, r’= 50cm = 0.5m Using the relation, E = hν = hc λ = 6 .626 × 10 -34 Js × 3 × 10 8 ms -1 3300 × 10 -10 m = 6 .02 × 10 -19 J = 6 .02 × 10 -19 1 .6 × 10 -19 eV = 3.76 eV Since, the energy E of the incident photon of light is greater than the work functions of Na and K, thus photoelectric emission will occur in these metals. As the work function of MO and Ni is greater than the incident energy, thus emission will not occur in these metals. If laser is brought closer, the intensityof radiation increases but this will not affect the result regarding Mo and Ni. However photoelectric current from Na and K will increase in proportion to intensity.

Q.30 Light of intensity 10-5 Wm-2 falls on a sodium photo-cell of surface area 2 cm2 . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Ans.

Intensity of incident light, I = 10-5 Wm-2Surface area of sodium photocell, A = 2 cm2= 2 × 10-4 m2Work function of the metal, W = 2 eV= 2 × 1.6 × 10-19 JNumber of layers of sodium that absorbs theincident energy, n = 5As we know that the effective atomic area ofa sodium atom, Ae = 10-20 m2Number of conduction electrons in five layers= 5 × area of one layereffective atomic area= 5 × 2 × 10-4m210-20m2 = 1017The answer obtained implies that the time of emission of electron is very large and is not in agreement with the observed time of emission which is approximately 10-9 s. Thus, wave-picture of radiation is in disagreement with photo-electric emission.

Q.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate, voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1Å , which is of the order of inter-atomic spacing in the lattice) (me = 9.1 × 10-31 kg).

Ans.

λ = 1 A0= 10-10 mFor X ray,  EX = hcλ = 6.626 × 10-34 Js × 3 × 108 ms-110-10 m= 1.99 × 10-15 J = 1.99 × 10-151.6 × 10-19 eV = 1.24 × 104 eV = 12.4 KeVFor the electron, as per de Broglie’s relation ,λ = hP = h2 me EeSquaring both the sides,Ee = h22meλ2= 6.626 × 10-34 Js2 2 × 9.11 × 10-31 kg × 10-10 m2 Ee = 2.41 × 10-17 J = 2.41 × 10-171.6 × 10-19 eV = 151 eVClearly, the energy of photon is much greaterthan the energy of electron .

Q.32 (a) Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. ( mn = 1.675 × 10-27 kg)
(b ) Obtain the de-Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Ans.

(a) Kinetic energy, E = 150 eV = 150 × 1 .6 × 10 -19 J = 2 .4 × 10 -17 J Using , λ = h 2 m n E λ = 6 .626 × 10 -34 Js 2 × 1 .675 × 10 -27 kg × 2 .4 × 10 -17 J = 2 .34 × 10 -12 m Theinter-atomicspacing ~1 Åi.e., 10 -10 misabout ahunderedtimesgreater than the wave length.Hence,aneutronbeamof energy150eVisnotsutiablefordiffractionexperiments. (b)Here T = 27 o C = 300 K λ = h 3 m n KT = 6 .626 × 10 -34 Js 3 × 1 .675 × 10 -27 kg × 1 .38 × 10 -23 J mole -1 K -1 × 300 K = 1 .45 × 10 -10 m = 1.45Å

Q.33 An electron microscope uses electron accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc,) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Ans.

Accelerating voltage, V = 50 kV = 50 × 10 3 V Wavelength of electron λ e = h 2 meV = 6 .626 × 10 -34 Js 2 × 9 .1 × 10 -31 kg × 1 .6 × 10 -19 C × 50 × 10 3 V = 5 .49 × 10 -12 m = 5 .49 × 10 -12 × 10 10 Å = 0.0549 Å Also, for yellow light, λ y = 5900 Å = 5900 × 10 -10 m Now Resolving power, µ α 1 λ Resolving power of electron microscope Resolving power of optical microscope = λ y λ e = 5900 × 10 -10 m 0 .0549 × 10 -10 m = 1 .07 × 10 5 Thus,theresolvingpowerofanelectronmicroscope isnearly 10 5 timesthatofanopticalmicroscope.

Q.34 The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15 m or less. This structure was first probe in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).

Ans.

Rest mass energymoc2 = 0.511 MeV= 0.511 × 1.6 × 10-13 J = 8.18 × 10-14 JUsing de-Broglie’s relation , λ = hp p = hλ = 6.626 × 10-34 Js10-15  m = 6.626 × 10-19 kgms-1As total energy of the particle isE = p2c2 + moc22= 6.626 × 10-19 kgms-12× 3 × 108 ms-12 + 8.18 × 10-14 J2 = 1.99 × 10-10 J = 1.99 × 10-101.6 × 10-19 eV = 1.24 × 109 eV , 109 eV = 1 BeV= 1.24 BeVThus, the energy of the proton ejected out of thelinear accelerator is of the order of a few BeV.

Q.35 Find the typical de-Broglie wavelength associated with a He atom in helium gas at room temperature ( 27ºC ) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

Ans.

Given, T = 270C= 270C + 273 = 300 K P = 1 atm= 1.01 × 105 N m-2Mass of helium atom, m = 4Avogadro’s number = 4 6.02 × 1023 g = 6.64 × 10-24 g Orm = 6.64 × 10-27 kg ‘ As, λ = h 3 mkT =  6.626 × 10-34 Js3 × 6.64 × 10-27 kg × 1.38 × 10-23 J mole1 K1 × 300 K= 7.3 × 10-11 m = 7.3 × 10-11 × 1010 Å = 0.73 ÅAlso, the mean separation r0= VN1/3 = kTP1/3 = 1.38 × 10-23 J mole1 K1 × 300 K1.01 × 105 Pa1/3 = 4.1 × 10-261/3m = 3.45 × 10-9 m = 3.45 × 10-9 × 1010 Å = 3.45 Åλro = 0.73 × 10-10m3.4 × 10-9 m= 0.021  Clearly,  ro >> λ 

Q.36 Compute the typical de-Broglie wavelength of an electron in a metal at 27ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10-10 m.

Ans.

Given, T =  27 oC= 27 oC + 273 = 300 KFor an electron in metal momentum, p = 3 m k TNow,λ = h3 mkT = 6.626 × 10-34 Js3 × 9.1 × 10-31 kg × 1.38 × 10-23 J mole-1 K-1 × 300 K = 6.23 × 10-9 m λro = 6.23 × 10-9 m2 × 10-10 m= 31.2 where, ro is the mean separation between two electrons in a metal. Clearly λ >> ro

Q.37 Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e;(-1/3) e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m ? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:

E= hv, p=hλ

but while the value of λ is physically significant, the value of ν  (and therefore, the value of the phase speed

νλ

) has no physical significance. Why?

Ans.

(a)In case of the Millikan oil drop experiment, the charge on the electron is measured. The electron revolves outside the nucleus and each has a charge “e”. Thus, we do not observe the fractional charges.i.e. + 23e or – 13e (b)The energy acquired by the electronwhen accelerated through V volt is12mv2 = eV v = 2e VmSimilarly, the force on the electron in an electric field is Fe = ma = eE a = e E mAnd force on electron in a magnetic field fm= evB = mv2r or v = erB m Clearly, the dynamic of electrons is determined by emin all the above relations instead of “e” alone .(c) At low pressure, the density of gas decreases and mean free path of gas molecules increases. Hence, ions do not collide frequently and are able to reach the respective electrodes to constitute a current. At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not have a chance to reach the respective electrodes to constitute a current because of their recombination.(d)Work function in fact is the energy required to knock out the electron from highest filled level of conduction band, there are different energy levels which collectively form a continuous band of levels. Therefore, different amounts of energy are required to bring the electrons out of the different levels. Electrons once omitted have different kinetic energies according to the energy supplied to the emitter.(e)Since, frequency for a given matter wave remains constant for different layers of the matter but wavelength changes so is more significant than. The phase speed is not significant as group speed is given byd(1λ) = dE dp= dp22mdp= pm is physically significant .

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FAQs (Frequently Asked Questions)

1. What concepts are included in the NCERT Solutions Class 12 Physics Chapter 11?

Students will get a detailed explanation of various concepts like the Compton effect and photoelectric effect. Furthermore, there is an explanation of Einstein’s photoelectric equation, Davisson and Germer’s experiment, De Broglie equation, Hertz’s observation, Hallwachs’ and Lenard’s experiment.

2. What is the marking system of Class 12 Physics Chapter 11?

Students preparing for the CBSE 12th Exam are advised to know the chapter-wise weightage and marking system of all subjects. The entire Physics Class 12 syllabus is bifurcated into 9 Units. Unit 7-  Dual Nature of Radiation and Matter and Unit 8- Atoms and Nuclei together carry 12 marks. Maximum 2 to 3 objective or subjective questions of 6-8 marks based on the 11th chapter will be asked in the exams. The class 12 Physics chapter 11 NCERT solution enlists all the questions which are most likely to be asked in the examination.