Dual Nature of Radiation and Matter is a conceptual and high-weightage chapter in Class 12 Physics that explains the particle and wave nature of light and matter. This chapter covers important topics such as photoelectric effect, Einstein’s photoelectric equation, threshold frequency, de Broglie wavelength, and wave nature of matter, which are frequently asked in CBSE board exams and competitive exams like JEE and NEET.
NCERT Solutions for Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear derivations, graphs, and solved numericals, helping students understand modern physics concepts clearly and score well in board examinations
NCERT Solutions for Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter
Q.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Q.
Q.
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.
Q.
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.
Q.
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).
Q.
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?
Q.
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).
Q.
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?
Q.
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?
Q.
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.
Q.
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?
Q.
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?
Q.
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.
Q.
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)
Q.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at
. 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.
Q.
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.
Q.
(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?
Q.
Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon).
Q.
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.
Q.
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:
but while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ) has no physical significance. Why?
NCERT Solutions for Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter
Q. 1) Figure shows a plot of stopping potential V0 versus the frequency ν of the incident radiation for two photosensitive materials A and B.
(a) Which material has a higher work function?
(b) What is the value of work function for each material?
(c) The slope of the plot for both materials is the same. What does this represent?
Ans:
(a) Material B has a higher work function because its threshold frequency is higher.
(b) Work function of a material is given by:
Work function = hν0
For material A:
φA = hν0A
For material B:
φB = hν0B
(c) The slope of the stopping potential versus frequency graph represents h/e, where h is Planck’s constant and e is the charge of an electron.
Q. 2) The work function of cesium is 1.9 eV. Calculate:
(a) The threshold wavelength.
(b) The threshold frequency.
Ans:
Work function, φ = 1.9 eV
= 1.9 × 1.6 × 10−19 J
= 3.04 × 10−19 J
(a) Threshold frequency:
φ = hν0
ν0 = φ / h
= (3.04 × 10−19) / (6.63 × 10−34)
= 4.59 × 1014 Hz
(b) Threshold wavelength:
λ0 = c / ν0
= (3 × 108) / (4.59 × 1014)
= 6.53 × 10−7 m
Q. 3) Monochromatic radiation of frequency 2.2 × 1015 Hz is incident on a metal surface having work function 4.42 eV. Calculate:
(a) The maximum kinetic energy of the emitted photoelectrons.
(b) The stopping potential.
Ans:
Energy of incident photon, E = hν
= 6.63 × 10−34 × 2.2 × 1015
= 1.46 × 10−18 J
Work function, φ = 4.42 eV
= 4.42 × 1.6 × 10−19
= 7.07 × 10−19 J
(a) Maximum kinetic energy:
Kmax = E − φ
= (1.46 × 10−18 − 7.07 × 10−19)
= 7.53 × 10−19 J
(b) Stopping potential:
V0 = Kmax / e
= (7.53 × 10−19) / (1.6 × 10−19)
= 4.7 V
FAQs: Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter
Q1. Is Dual Nature of Radiation and Matter important for exams?
Yes, it is a high-weightage chapter for CBSE, JEE, and NEET.
Q2. Which topics are most important in this chapter?
Photoelectric effect, Einstein’s equation, and de Broglie wavelength.
Q3. Are numericals asked from this chapter?
Yes, photoelectric effect and de Broglie wavelength numericals are common.
Q4. Are derivations important here?
Yes, derivations related to Einstein’s photoelectric equation are frequently asked.
Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals and graphs.