NCERT Solutions for **Class 12 Physics Chapter 11 Dual Nature of Radiation Matter** is an essential resource material that is essential for your class 12 and entrance exam preparation. NCERT solution of Class 12 Physics Chapter 11 eases your work of notes preparation and revision. It consists of dual nature of radiation and matter important questions with answers important tables, exemplary questions and worksheets.

Dual nature of radiation and matter topic is very important when it comes to Class 12 Physics examination. Dual nature of radiation and matter NCERT SOLUTIONS pdf is provided here to help 12th standard students to prepare for the subject more efficiently. NCERT Solutions are prepared by expert teachers in such a way that it helps students understand the concepts of the chapter in an interactive manner.

## NCERT solutions for class 12 chapter 11 Dual Nature of Radiation Matter

Maxwell’s equations of electromagnetism and Hertz experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light. The NCERT solutions are one of the best tools to prepare for class 12 physics exam. Students must prepare this chapter well to score well in their examination. The Class 12 physics NCERT solutions are given below so that students can understand the concepts of this chapter in-depth.

#### Concepts involved in Chapter 11 Dual Nature of Radiation Matter

- Introduction
- Electron Emission
- Photoelectric Effect
- Hertz’s observations
- Hallwachs’ and Lenard’s observations

- Experimental Study Of Photoelectric Effect
- Effect of intensity of light on photocurrent
- Effect of potential on photoelectric current
- Effect of frequency of incident radiation on stopping potential

- Photoelectric Effect And Wave Theory Of Light Ex
- Einstein’s Photoelectric Equation: Energy Quantum Of Radiation
- Particle Nature Of Light: The Photon
- Wave Nature Of Matter

Some of the topics covered in this chapter are as follows. The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal. Energy (greater than the work function (φο ) required for electron emission from the metal surface can be supplied by suitably heating or applying a strong electric field or irradiating it by the light of suitable frequency. Some key points on the Dual Nature of Radiation Matter are given below.

Free electrons in a metal are free in the sense that they move inside the metal in a constant potential (This is only an approximation). They are not free to move out of the metal. They need additional energy to get out of the metal.

Observations on photoelectric effect imply that in the event of matter-light interaction, absorption of energy takes place in discrete units of hν. This is not quite the same as saying that light consists of particles, each of energy * hv*.

### Class 12 Physics NCERT Solutions Dual Nature of Radiation Matter Important Questions

**Q.11.1: Find the:**

**(a) Maximum frequency, and**

**(b) The minimum wavelength of X-rays produced by 30 kV electrons.**

** ****Ans:**

Electron potential, V = 30 kV **= 3 × 10 ^{4} V**

Hence, electron energy, **E = 3 × 10 ^{4} eV**

Where, e = Charge on one electron **= 1.6 × 10 ^{-19} C**

**(a)** Maximum frequency by the X-rays = ν

The energy of the electrons:

**E = hν**

Where,

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

Therefore,

= **= 7.24 x 10 ^{18} Hz**

Hence, 7.24 x 10^{18 }Hz is the maximum frequency of the X-rays.

**(b)** **The minimum wavelength produced:**

= **= 4.14 x 10 ^{-11} m = 0.0414 nm**

**Q.11.2: Cesium metal has a work function of 2.14 eV. Light with frequency 6 × 10 ^{14} Hz is incident on the metallic surface; hence there is a photoemission of electrons. Find out the:**

**(a) Max K.E of the electrons emitted,**

**(b) Stopping potential, **

**(c) Max speed of the photoelectrons emitted?**

** ****Ans:**

Work function of cesium, **= 2.14eV**

Frequency of light, **v = 6.0 x 10 ^{14} Hz**

**(a)** The max energy (kinetic) by the photoelectric effect:

K = hν –

Where,

h = Planck’s constant **= 6.626 x 10 ^{-34} Js**

Therefore,

K =

= 2.485 – 2.140 **= 0.345 eV**

Hence, 0.345 eV is the maximum kinetic energy of the emitted electrons.

**(b)** For stopping potential V_{o, }we can write the equation for kinetic energy as:

**K = eV _{o}**

Therefore, V_{o }=

=

**= 0.345 V**

Hence, 0.345 V is the stopping potential of the material.

**(c)** Maximum speed of photoelectrons emitted **= ν**

Following is the kinetic energy relation:

K =

Where,

m = mass of electron **= 9.1 x 10 ^{-31} Kg**

=

=0.1104 x 10^{12}

Therefore, ν = 3.323 x 10^{5} m/s **= 332.3 km/s**

Hence**, **332.3 km/s is the maximum speed of the emitted photoelectrons.

** ****Question 11.3: The photoelectric cut-off voltage is 1.5 V. Find out the maximum kinetic energy of emitted photoelectrons?**

** ****Ans:**

Photoelectric cut-off voltage, **V _{o} = 1.5 V**

For emitted photoelectrons, the maximum kinetic energy is:

**K _{e} = eV_{o}**

Where,

e = charge on an electron **= 1.6 x 10 ^{-19} C**

Therefore, K_{e} = 1.6 x 10^{-19} x 1.5 **= 2.4 x 10 ^{-19} J**

Therefore, 2.4 x 10^{-19 }J is the maximum kinetic energy emitted by the photoelectrons.

**Question 11.4 Monochromatic light with wavelength 632.8 nm is produced using a He-Ne laser. The emitted power being 9.42 mW. Find the:**

**(a) Momentum and energy of each photon **

**(b) Number of photons/second; arrive at the target irradiated by this beam? (Assume the beam to have a uniform cross-section.)**

**(c) The speed that a hydrogen atom has to travel with so as to have the same momentum as that of the photon?**

** ****Ans:**

Monochromatic light having a wavelength, λ = 632.8 nm **= 632.8 × 10 ^{-9} m**

Given that the laser emits the power of, P = 9.42 mW **= 9.42 × 10 ^{-3} W**

Planck’s constant, **h = 6.626 × 10 ^{-34} Js**

Speed of light, **c = 3 × 10 ^{8} m/s**

Mass of a hydrogen atom, **m = 1.66 × 10 ^{-27} kg**

**(a)** The photons having the energy as:

E =

=

= 3.141 x 10^{-19} J

Therefore, each photon has a momentum of :

P =

=

**= 1.047 x 10 ^{-27} kg m/s**

** **

**(b)** Number of photons/second arriving at the target illuminated by the beam = n

Assuming the uniform cross-sectional area of the beam is less than the target area.

Hence, equation for power is written as:

P = nE

Therefore, n=

=

**= 3 x 10 ^{16} photons/s**

** **

**(c) **Given that, momentum of the hydrogen atom is equal to the momentum of the photon,

**P= 1.047 x 10 ^{-27} kg m/s**

Momentum is given as:

P=mv

Where,

ν = speed of hydrogen atom

Therefore, ν =

= **= 0.621 m/s**

** **** ****Question 11.5: The sunlight reaches the surface of the earth with energy flux of 1.388 × 10 ^{3} W/m^{2}. How many photons are incident on the Earth per second/square meter? Assume an average wavelength of 550 nm.**

** ****Ans:**

Sunlight reaching the surface of earth has an energy flux of

**= 1.388 × 10**

^{3}W/m^{2}Hence, power of sunlight per square metre, **P = 1.388 × 10 ^{3} W**

Speed of light, **c = 3 × 10 ^{8} m/s**

Planck’s constant, **h = 6.626 × 10 ^{-34} Js**

**= 550 x 10**is the average wavelength of the photons from the sunlight

^{-9}mNumber of photons per square metre incident on earth per second = n

Hence, the equation for power be written as:

**P = nE**

Therefore, n=

=

= **= 3.84 x 10 ^{21} photons/m^{2}/s**

Therefore, 3.84 x 10^{21 }are the photons that are incident on the earth per square meters.

**11.6: During an experiment on the photoelectric effect, the slope of the cut-off voltage and frequency of incident light is obtained to be 4.12 × 10 ^{-15} V s. Determine the value of Planck’s constant.**

** ****Ans:**

Given that the slope of cut-off voltage (V) versus frequency (v) being:

^{-15}Vs

V and frequency being related by the equation as:

**Hν = eV**

Where,

e = Charge on an electron = 1.6 x 10^{-19} C

h = Planck’s constant

Therefore, h = e x

= 1.6 x 10^{-19} x 4.12 x 10^{-15} **= 6.592 x 10 ^{-34} Js**

Therefore, 6.592 x 10^{-34} Js is the Planck’s constant that is determined from the above equation.

**Question 11.7: Energy is radiated uniformly in all directions by a sodium lamp of power 100W. The lamp is contained in the centre of a large sphere.This sphere absorbs the entire sodium light incident on it. The wavelength of sodium light is given as 589 nm.**

** (a) Determine the energy per photon associated with the sodium light?**

** (b) At what rate is the photons incident onto the sphere?**

**Ans:**

Power of the sodium lamp **P = 100W**

Wavelength of the emitted sodium light,

**= 589 x 10 ^{-9} m**

Planck’s constant, **h = 6.626 x 10 ^{-34} Js**

**Speed of light, c = 3 x 10 ^{8}**

**(a) **

The energy per photon associated with the sodium light is given as:

E =

E =

= 3.37 x 10^{-19} J = **= 2.11 eV**

**(b) **

Number of photons delivered to the sphere = n

The equation for power can be written as:

**P = nE**

Therefore, n=

= ^{20} photons/s

Therefore, 2.96 x 10^{20 }photons are delivered every second to the sphere.

** **** ****Question 11.8: The threshold frequency for a certain metal is 3.3 x 10 ^{14} Hz. If the light of frequency 8.2 x 10^{14} is incident on the metal, predict the cut-off voltage for the photoelectric emission.**

**Ans:**

Threshold frequency of the metal, **v _{o}= 3.3 × 10^{14} Hz.**

Frequency of light incident on the metal, **v= 8.2 × 10 ^{14} Hz**

Charge on an electron, **e = 1.6 × 10 ^{-19} C**

Planck’s constant, **h = 6.626 × 10 ^{-34} Js**

Cut-off voltage for the photoelectric emission from the metal **= V _{o}**

The equation for the cut –off energy is given as:

**eV _{o} = h(ν – ν_{o})**

V_{o} =

= **= 2.0291 V**

** **** ****Question 11.9: The work function of a certain metal is 4.2eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?**

**Ans:**

Work function of the metal, **=4.2eV**

Charge on an electron, **e = 1.6 x 10 ^{-19} C**

Planck’s constant, **h = 6.626 × 10 ^{-34} Js**

Wavelength of the incident radiation, **= 330 × 10 ^{-9} m**

Speed of light, **c = 3 × 10 ^{8} m/s**

The energy of the incident photon is given as:

E =

=

= 6.0 x 10^{-19} J = **= 3.76 eV**

The energy of the incident radiation is less than the work function of the metal. Hence, there is no photoelectric emission taking place.

** ****Question 11.10: Light of frequency 7.21 x 10 ^{14} Hz is incident in a metal surface. Electrons with a maximum speed of 6.0 x 10^{5} m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?**

**Ans:**

Frequency of the incident photon, **ν = 488nm = 488 x 10 ^{-9}m**

Maximum speed of the electrons, **v = 6.0 x 10 ^{5} m/s**

Planck’s constant, **h= 6.626 x 10 ^{-34} Js**

Mass of an electron, **m = 9.1 x 10 ^{-31} Kg**

For threshold frequency v_{o}, the relation for kinetic energy is written as:

_{o})

ν_{o }= ν –

= 7.21 x 10^{14} –

= 7.21 x 10^{14} – 2.472 x 10^{14} **= 4.738 x 10 ^{14} Hz**

Therefore, 4.738 x 10^{14} Hz is the threshold frequency for the photoemission of the electrons.

** ****Question 11.11: 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 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.**

**Ans:**

Wavelength of light produced by the argon laser,

**= 488 x 10**

^{-9}mStopping potential of the photoelectrons, **V _{o} = 0.38 V**

**1eV = 1.6 x 10 ^{-19} J**

Therefore, V_{o} =

Planck’s constant, **h = 6.6 x 10 ^{-34} Js**

Charge on an electron, **e = 1.6 x 10 ^{-19} C**

Speed of light, **c = 3 x 10 ^{8} m/s**

Using Einstein’s photoelectric effect, following is the relation for the work function:

eV_{o} = _{o}

=

= 2.54 – 0.38 **= 2.16 eV**

Therefore, 2.16eV is the work function of the the material with which the emitter is made.

**Question 11.12: Calculate the**

**(a) momentum, and**

**(b) the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.**

**Ans:**

Potential difference, **V = 56V**

Planck’s constant, **h = 6.6 x 10 ^{-34} Js**

Mass of an electron, **m = 9.1x 10 ^{-31} Kg**

Charge on an electron, **e = 1.6 x 10 ^{-19}C**

**(a)** At equilibrium, the kinetic energy of each electron is equal to the accelerating potential i.e., we can write the relation of velocity (v) of each electron as:

Therefore, v =

=

= 4.44 x 10^{6} m/s

The momentum of each accelerated electron is given as:

p = mv

= 9.1 x 10^{-31} x 4.44 x 10^{6} **= 4.04 x 10 ^{-24} Kg m/s**

Therefore, 4.04 x 10^{-24} Kg m/s is the momentum of each electron.

**(b)** de Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

^{o}

= ^{-19} m **= 0.1639 nm**

Therefore, 0.1639 nm is the de Broglie wavelength of each electron.

** **** ****Question 11.13: What is the:**

**(a) Momentum,**

**(b) Speed, and**

**(c) De Broglie wavelength of an electron with a kinetic energy of 120 eV.**

** ****Ans: **

Kinetic energy of the electron, **E _{K} = 120 eV**

Planck’s constant, **h = 6.6 × 10 ^{-34} Js**

Mass of an electron, **m = 9.1 × 10 ^{-31} Kg**

Charge on an electron, **e = 1.6 × 10 ^{-19} C**

**(a)** For an electron, we can write the relation for kinetic energy as:

E_{k} =

Where, v = speed of the electron

Therefore,

=

=

= 6.496 **× **10^{6} m/s

Momentum of the electron, p = mv = 9.1 **× **10^{-31} **× **6.496 **× **10^{6}

= 5.91 **× **10^{-24} kg m/s

Therefore, 5.91 × 10^{-24} Kg m/s is the momentum of the electron.

**(b)** **speed of the electron, v = 6.496 × 10 ^{6} m/s**

**(c)** de Broglie wavelength of an electron having a momentum p, is given as:

= ^{-10} m **= 0.112 nm**

Therefore, 0.112 nm the de Broglie wavelength of the electron.

** **** ****Question 11.14: 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:**

Wavelength of light of a sodium line, ^{-9} m

Mass of an electron, m_{e} = 9.1 x 10^{-31} Kg

Mass of a neutron, m_{n }= 1.66 x 10^{-27} Kg

Planck’s constant, h = 6.6 x 10^{-34} Js

**(a)** For the kinetic energy K, of an electron accelerating with the velocity v, we have the relation:

K = **(1)**

We have the relation for de Broglie wavelength as:

Therefore, **(2)**

**Substituting equation (2) in equation (1), we get the relation:**

K =

= **(3)**

=

= 6.9 x 10^{-25} J = **= 4.31 x 10 ^{-6} eV **

**Hence, the kinetic energy of the electron is 6.9 x 10 ^{-25} J**

**(b)** Using equation (3), we can write the relation for the kinetic energy of the neutron as:

=

=

= 3.78x 10 ^{-28} J

=

= 2.36 x 10^{-9} eV **= 2.36 neV**

The neutron has the kinetic energy of 3.78 x 10^{-28} J or 2.36 neV.

** ****Question 11.15: 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 ^{-9} kg drifting with a speed of 2.2 m/s?**

**Ans:**

**(a)** Mass of the bullet, **m = 0.040 Kg**

Speed of the bullet, v = 1.0 km/s **= 1000 m/s**

Planck’s constant, **h = 6.6 x 10 ^{-34} Js**

de Broglie wavelength of the bullet is given by the relation :

=

**= 1.65 x 10 ^{-35} m**

**(b)** Mass of the ball, **m = 0.060 Kg**

Speed of the ball, **v = 1.0 m/s**

de Broglie wavelength of the ball is given by the relation:

=

=

**= 1.1 x 10 ^{-32} m**

**(c)** Mass of the dust particle, **m = 1 x 10 ^{-9} Kg**

speed of the dust particle,** v = 2.2 m/s**

de Broglie wavelength of the dust particle is given by the relation:

=

=

**= 3.0 x 10 ^{-25} m**

**Question 11.16: 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 the electron.**

**Ans:**

Wavelength of an electron

**= 1 x 10**

^{-9}m**Planck’s constant, h = 6.63 x 10 ^{-34} Js**

**(a)** The momentum of an elementary particle is given by de Broglie relation:

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

Therefore, p =

= 6.63 x 10^{-25} Kg m/s

**(b)** The energy of a photon is given by the relation:

E =

Where,

Speed of light, c = 3 x 10^{8} m/s

Therefore, E =

= 1243.1 eV = 1.243 keV

**Therefore, the energy of the photon is 1.243 keV.**

**(c)** The kinetic energy (K) of an electron having momentum p, is given by the relation:

K =

Where, m = Mass of the electron = 9.1 x 10^{-31 }Kg

p = 6.63 x 10^{-25} Kg m/s

Therefore, K =

= 2.415 x 10^{-19} J

= **= 1.51 eV**

1.51eV is the kinetic energy of the electron.

**Question 11.17:**

**(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) kT at 300 K.**

**Ans:**

**(a)** de Broglie wavelength of the neutron, ^{-10} m

Mass of a neutron, m_{n} = 1.66 x 10^{-27} Kg

Planck’s constant, h = 6.6 x 10^{-34} Js

Kinetic energy (K) and velocity (v) are related as:

K = **……… (1)**

de Broglie wavelength (

**………. (2)**

Using equation (2) and equation (1), we get:

K =

=

=

**= 6.75 x 10 ^{-21} J**

=

**= 4.219 x10 ^{-2} eV**

Hence, the kinetic energy of the neutron is 6.75 x 10^{-21} J or 4.219 x 10^{-2} eV.

**(b)** Temperature of the neutron, **T = 300K**

**Boltzmann constant, k = 1.38 x 10 ^{-23} Kg m^{2} s^{-2} K^{-1}**

Average kinetic energy of the neutron:

K’ =

= ^{-23} x 300

**= 6.21 x 10 ^{-21} J**

The relation for the de Broglie wavelength is given as:

Where,

**m _{n} = 1.66 x 10^{-27} Kg**

**h = 6.6 x 10 ^{-34} Js**

K’ = 6.75 x 10^{-21} J

Therefore,

= 1.46 **× **10^{-10} m **= 0.146 nm**

Therefore, 0.146nm is the de Broglie wavelength of the neutron.

**Question 11.18: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).**

**Ans:**

The momentum of a photon having energy (hv) is given as:

p = **………. (i)**

Where,

**c = speed of light**

**h = Planck’s constant**

De Broglie wavelength of the photon is given as:

But, **p = mv**

Therefore, **……………(ii)**

Where, m = mass of the photon

v = velocity of the photon

From equation (i) and (ii) it can be concluded that the wavelength of the electromagnetic radiation and the de Broglie wavelength of the photon are equal.

**Question 11.19: 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 at this temperature. (Atomic mass of nitrogen = 14.0076 u)**

** ****Ans:**

Temperature of the nitrogen molecule, **T = 300 K**

Atomic mass of nitrogen **= 14.0076 u**

Hence, mass of the nitrogen molecule, m = 2 × 14.0076 **= 28.0152 u**

But, **1** **u = 1.66 × 10 ^{-27} kg**

Therefore, **m = 28.0152 ×1.66 × 10 ^{-27} kg**

**Planck’s constant, h = 6.63 × 10 ^{-34} Js**

**Boltzmann constant, k = 1.38 × 10 ^{-23} J/K**

We have the expression that relates mean kinetic energy _{rms} ) as:

V_{rms} =

For nitrogen molecule, the de Broglie wavelength is given as:

=

= 0.028 x 10^{-9} m **= 0.028 nm**

Therefore, the nitrogen molecule is 0.028 nm is the de Broglie wavelength.

Dual Nature of Radiation Matter along with other concepts are very crucial for your CBSE class 12 board examination. It is very important for the students to understand the topics deeply to avoid difficulty in understanding higher concepts.

BYJU’S provide you with complete NCERT class 12 Physics solutions to make you well equipped for the all-important CBSE Class 12 Board examination. BYJU’S also provides downloadable pdf of Chapter wise NCERT solutions for class 12 for subjects along with exemplar problems.