CBSE Class 12 Physics Chapter-11 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-11 – Dual Nature of Radiation and Matter
CBSE Class 12 Physics Important Questions Chapter 11 – Dual Nature of Radiation and Matter
3 Marks Questions
1.The following table gives the values of work functions for a few sensitive metals.
| S. No. | Metal | Work function(eV) |
| 1. | Na | 1.92 |
| 2. | K | 2.15 |
| 3. | Mo | 4.17 |
If each of these metals is exposed to radiations of wavelength 3300nm, which of these will not exit photoelectrons and why?
Ans. That material will not emit photoelectrons whose work function is greater than the energy of the incident radiation.
E = 3.76 eV
Hence work function of 
is (4.17eV) which is greater than the energy of the incident radiation (= 3.76 eV) so will not emit photoelectrons.
2.Define threshold wavelength for photoelectric effect? Debroglie wavelength associated with an electron associated through a potential difference V is
? What will be the new wavelength when the accelerating potential is increase to 4V?
Ans. The maximum wavelength of radiation needed to cause photoelectric emission is known as threshold wavelength.
Or
3. An electron has kinetic energy equal to 100eV. Calculate (1) momentum (2) speed (3) Debroglie wavelength of the electron.
Ans. 
(1) (Momentum) 
(2) Speed 
(3) Debroglie wavelength 
4. (a) Define photoelectric work function? What is its unit?
(b) In a plot of photoelectric current versus anode potential, how does
(i) Saturation current varies with anode potential for incident radiations of different frequencies but same intensity?
(ii) The stopping potential varies for incident radiations of different intensities but same frequency.
(iii) Photoelectric current vary for different intensities but same frequency of radiations? Justify your answer in each case?
Ans. (a) The minimum amount of energy required to take out an electron from the surface of metal. It is measured in electron volt (eV).
(b) (i) Saturation current depends only on the intensity of incident radiation but is independent of the frequency of incident radiation.
(ii) Stopping potential does not depend on the intensity of incident radiations.
(iii) Photoelectric current is directly proportional to the intensity of incident radiations, provided the given frequency is greater than the threshold frequency.
5. Photoelectric work function of a metal is 1eV. Light of wavelength 
falls on it. What is the velocity of the effected photoelectron?
Ans. 
6.The wavelength of a photon and debroglie wavelength of an electron have the same value. Show that the energy of the photon is 
times the kinetic energy of electron where m, c, and h have their usual meanings?
Ans. Energy of a photon 
Kinetic energy 
of an electron
But de-broglie wavelength of an electron is given by
Dividing (1) by (2)
7.Draw a graph showing the variation of stopping potential with frequency of the incident radiations. What does the slope of the line with the frequency axis indicate. Hence define threshold frequency?
Ans. Slope of the graph 
Einstein photoelectric equation
Differentiating equation (1)
Thus slope is equal to the ratio of planck’s constant to the charge on electron.
Threshold frequency – The minimum values of frequency of the incident light below which photoelectric emission is not possible is called as threshold frequency.
8.Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Ans.Potential of the electrons, V= 30 kV 
V
Hence, energy of the electrons, E eV
Where,
e= Charge on an electron = 
(a)Maximum frequency produced by the X-rays = 
The energy of the electrons is given by the relation:
E = 
Where,
h= Planck’s constant 
Js
Hence, the maximum frequency of X-rays produced is
Hz.
(b)The minimum wavelength produced by the X-rays is given as:
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.
9. The work function of caesium metal is 2.14 eV. When light of frequency 
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.Work function of caesium metal, 
Frequency of light, 
(a)The maximum kinetic energy is given by the photoelectric effect as:
Where,
h= Planck’s constant = 
Hence, the maximum kinetic energy of the emitted electrons is
0.345 eV.
(b)For stopping potential
, we can write the equation for kinetic energy as:
Hence, the stopping potential of the material is 0.345 V.
(c)Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
Where,
m= Mass of an electron 
kg
Hence, the maximum speed of the emitted photoelectrons is
332.3 km/s.
10.The energy flux of sunlight reaching the surface of the earth is 
. 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.Energy flux of sunlight reaching the surface of earth,
Hence, power of sunlight per square metre, P
W
Speed of light, c
m/s
Planck’s constant, hJs
Average wavelength of photons present in sunlight, 
Number of photons per square metre incident on earth per second = n
Hence, the equation for power can be written as:
Therefore, every second, 
photons are incident per square metre on earth.
11. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 
V s. Calculate the value of Planck’s constant.
Ans.The slope of the cut-off voltage (V) versus frequency 
of an incident light is given as:
V is related to frequency by the equation:
Where,
e= Charge on an electron 
C
h= Planck’s constant
Therefore, the value of Planck’s constant is 
12.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.Power of the sodium lamp, P= 100 W
Wavelength of the emitted sodium light, 
= 589 nm
m
Planck’s constant, h
Js
Speed of light, c 
m/s
(a)The energy per photon associated with the sodium light is given as:
(b)Number of photons delivered to the sphere = n
The equation for power can be written as:
Therefore, every second, 
photons are delivered to the sphere.
