The solutions cover electron emission, photoelectric effect, Einstein’s photoelectric equation, photon properties, de Broglie’s hypothesis, Davisson-Germer experiment, and Heisenberg’s uncertainty principle.
The solutions explain the photoelectric effect through the emission of electrons from a metal surface when suitable-frequency light falls on it. Einstein’s photoelectric equation is used to relate the energy of incident photons to the maximum kinetic energy of photoelectrons.
Work function is the minimum energy needed to remove an electron from a metal surface. Threshold frequency is the minimum frequency required for photoelectric emission, while stopping potential is the potential needed to stop the most energetic photoelectrons.
NCERT Solutions explain that the wave theory could not account for important observations of the photoelectric effect, such as the dependence of photoelectric emission on frequency and the immediate emission of photoelectrons.
The de Broglie hypothesis is used to determine the wavelength of a moving particle. NCERT Solutions apply the relation between de Broglie wavelength and the momentum of the particle.
The Davisson-Germer experiment provided experimental evidence for the wave nature of electrons through electron diffraction.
Yes, NCERT Solutions Class 12 Physics Chapter 11 help strengthen important concepts such as the photoelectric effect, Einstein’s photoelectric equation, photon properties, de Broglie wavelength, and the wave nature of matter. These concepts provide a useful foundation for JEE and NEET Physics preparation.
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NCERT Solutions Class 12 Physics Chapter 11 - Dual Nature of Radiation and Matter
In Chapter 11 titled as "The Dual Nature of Radiation and Matter" of Class 12 Physics, Light and Matter are presented as having a duality, contrary to what was believed in classical mechanics, where every object was said either to be a particle or a wave. The phenomenon known as Photoelectric Effect was observed originally by Hertz and Hallwachs; later, it was explained by Einstein using his photon theory, which defined the nature of light in terms of quanta or photons. de Broglie's wave hypothesis helps us define matter's wave behaviour.
Chapter 11 has been designed to provide students with all the information they need to get ready for CBSE board examinations, both JEE and NEET. It contains both conceptual and mathematical elements and uses NCERT Solutions as a tool to help connect classical and quantum mechanics. Students will be encouraged to use the mathematical relationships proven using experimental results and devised in NCERT Solutions to explore the concepts of Physics contained within this unit of study.
1.0Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: Key Concepts
Class 12 Physics Chapter 11, Dual Nature of Radiation and Matter, explains the particle nature of light and the wave nature of matter. The chapter covers important experiments and concepts that show how light and matter can exhibit both wave-like and particle-like properties. The main concepts covered in this chapter include:
Electron Emission: Explains how electrons are released from a metal surface through thermionic emission, field emission, and photoelectric emission.
Photoelectric Effect: Covers the experimental setup and explains how the intensity, frequency, and potential difference affect the emission of photoelectrons from a metal surface.
Einstein’s Photoelectric Equation: Explains the photoelectric effect using the photon concept and shows the relationship between the energy of incident light and the maximum kinetic energy of emitted electrons.
Particle Nature of Light: Explains the concept of photons and their properties, including photon energy, momentum, and zero rest mass.
Wave Nature of Matter: Introduces de Broglie’s hypothesis and explains how moving particles can have wave-like properties. The de Broglie wavelength is related to the momentum of a moving particle.
Davisson-Germer Experiment: Explains how the wave nature of electrons was confirmed through electron diffraction.
Heisenberg’s Uncertainty Principle: Introduces the idea that the position and momentum of a microscopic particle cannot both be known with unlimited accuracy at the same time.
2.0NCERT Solutions Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter : Detailed Solutions
SOLVED EXAMPLES
Monochromatic light of frequency 6.0×1014Hz is produced by a laser. The power emitted is 2.0×10−3W.
(a) What is the energy of a photon in the light beam?
(b) How many photons per second, on an average, are emitted by the source?
Sol.
(a) Each photon has an energy
E=hν=(6.63×10−34Js)(6.0×1014Hz)=3.98×10−19J
(b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P=NE. Then
N=EP=3.98×10−19J2.0×10−3W=5.0×1015 photons per second
The work function of caesium is 2.14 eV. Find
(a) the threshold frequency for caesium, and
(b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.
Sol.
(a) For the cut-off or threshold frequency, the energy hν0 of the incident radiation must be equal to work function ϕ0, so that
Thus, for frequencies less that this threshold frequency, no photoelectrons are ejected.
(b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy eV0 by the retarding potential V0.Einstein's Photoelectric equation is
eV0λ=hv−ϕ0=λhc−ϕ0 or λ=(eV0+ϕ0)hc=(0.60eV+2.14eV)(6.63×10−34Js)×(3×108m/s)=(2.74eV)19.89×10−26Jm=2.74×1.6×10−19J19.89×10−26Jm=454nm
What is the de-Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s ?
Sol.
(a) For the electron :
Mass m=9.11×10−31kg,
speed v=5.4×106m/s.
Then, momentum
p=mv=9.11×10−31×5.4×106p=4.92×10−24kgm/s
de Broglie wavelength,
λλ=ph=4.92×10−24kgm/s6.63×10−34Js=0.135nm
(b) For the ball :
Mass m′=0.150kg, speed v′=30.0m/s
Then momentum
p′=m′v′=0.150×30.0p′=4.50kgm/s
de-Broglie wavelength λ′=p′h
λ′=4.50kgm/s6.63×10−34Js=1.47×10−34m
The de-Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10−19 times the size of the proton, quite beyond experimental measurement.
EXERCISE QUESTIONS WITH SOLUTIONS
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Sol. (a) hvmax =eV
The work function of caesium metal is 2.14 eV. When light of frequency 6×1014Hz 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 ?
Sol. (a) Kmax =hν−ϕ
=(1.6×10−196.62×10−34×6×1014−2.14)eV
or Kmax =0.34eV=0.54×10−19J
(b) Kmax =eV0=0.34eV
or V0=0.34 volt
(c) 21mevmax 2=Kmax
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Sol.Kmax =eV0=1.5eV
=1.5×1.6×10−19J=2.4×10−19J
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?
Sol. Here,
λP=632.8nm=632.8×10−9m=9.42mW=9.42×10−3W
(a) Energy of each photon,
E=λhc=632.8×10−96.63×10−34×3×108=3.14×10−19kJ
Momentum of each photon,
p=λh=632.8×10−96.63×10−34=1.05×10−27kgms−1.
(b) Number of photons arriving per second at the target,
N=EP=3.14×10−199.42×10−3=3×10+16 photons per second.
(c) Momentum of a hydrogen atom = Momentum of a photon or mv=p
∴ Velocity,
v=mp=1.67×10−27kg1.05×10−27kgms−1=0.63ms−1
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10−15Vs. Calculate the value of Planck's constant.
Sol. Here, ΔvΔV=4.12×10−15V.s.
e=1.6×10−19C
Planck constant
h=ΔvΔV×e=(4.12×10−15)(1.6×10−19)
So, h=6.592×10−34 J.s.
The threshold frequency for a certain metal is 3.3×1014Hz. If light of frequency 8.2×1014Hz is incident on the metal, predict the cut off voltage for the photoelectric emission.
Sol. ∵Kmax =hν−hν0
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Sol. here W0=4.2eV,λ=330nm=330×10−9m Energy of incident photon,
As the energy of incident photon is less than the work function of the metal, there will be no photoelectric emission.
Light of frequency 7.21×1014Hz is incident on a metal surface. Electrons with a maximum speed of 6.0×105m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Sol. Here,
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 emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V.
Find the work function of the material from which the emitter is made.
Sol.
∵∴ or Kmax=λhc−ϕ0eV0=4881240eV−ϕ0⇒ϕ0=2.54eV−0.38eVϕ0=2.16eV
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0×10−9kg drifting with a speed of 2.2 m/s ?
Sol. (a) de Broglie wavelength
λ=mvh=0.04×(1×103)6.62×10−34=1.7×10−35m
(b) de Broglie wavelength
λ=mvh=0.060×16.62×10−34=1.1×10−32m
(c) de Broglie wavelength
λ=mvh=1×10−9×2.26.62×10−34=3×10−25m
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Get NCERT Solutions for Class 12 Physics for every chapter, covering exercise answers, useful formulas, and step-by-step solutions to help students understand topics and solve questions.
4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 11
Easy Graph Interpretation: The solutions explain important graphs from the photoelectric effect, including the effect of light intensity on photocurrent and the relationship between frequency and stopping potential. This helps students understand what each graph shows and answer graph-based questions correctly.
Clear Understanding of Work Function: The difference between work function and threshold frequency is explained clearly. This helps students use the correct values and formulas while solving numerical problems.
Step-by-Step Derivations: Important derivations, such as the de Broglie wavelength of an electron accelerated through a potential difference VV, are explained step by step with simple algebraic calculations.
Photoelectric Effect and Photon Theory: The solutions explain why the wave theory of light could not explain the photoelectric effect and how Einstein’s photon theory explains the observations. This is useful for conceptual and descriptive questions in CBSE examinations.
Important Terms in Answers: Key terms such as work function, threshold frequency, stopping potential, photoelectron, photon, de Broglie wavelength, and matter waves are used correctly in the answers. This helps students write clear answers using the important terms expected in Class 12 Physics examinations.
Table of Contents
1.0Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: Key Concepts
2.0NCERT Solutions Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter : Detailed Solutions