NCERT solutions for class 12 physics chapter 11 – dual nature of radiation and matter is very important when it comes to Class 12 Physics examination. The class 12 NCERT solutions for physics chapter 11 is provided here to help 12th standard students to prepare for the subject more efficiently. The 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. Check the class 12 physics chapter 11 pdf provided below.

The 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 is one of the best tool to prepare for physics exam in class 12 Class. Students must prepare this chapter well to score well in their examination. The physics NCERT solutions are given below so that students can understand the concepts of this chapter in depth.

**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, \(v = \frac{E}{h}\)

= \(\frac{1.6\times 10^{-19}\times 3 \times 10^{4}}{6.626\times 10^{-34}}\) **= 7.24 x 10 ^{18} Hz**

**Hence, the max frequency of X-rays = 7.24 x 10 ^{18 }Hz**

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

\(\lambda =\frac{c}{v}\)

= \(\frac{3\times 10^{8}}{7.24\times 10^{18}}\) **= 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, \(\Phi _{o}\) **= 2.14eV**

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

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

K = hν – \(\Phi _{o}\)

Where,

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

Therefore,

K = \(\frac{6.626\times 10^{34}\times 6\times 10^{14}}{1.6\times 10^{-19}}\) – 2.14

= 2.485 – 2.140 **= 0.345 eV**

**Hence, the kinetic energy (max) of the electrons emitted is 0.345 eV**

** **

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

**K = eV _{o}**

Therefore, V_{o }= \(\frac{K}{e}\)

= \(\frac{0.345\times 1.6\times 10^{-19}}{1.6\times 10^{-19}}\)

**= 0.345 V**

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

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

Hence, the relation with kinetic energy:

K = \(\frac{1}{2}mv^{2}\)

Where,

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

\(v^{2}=\frac{2K}{m}\)

= \(\frac{2\times 0.345\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}\)

=0.1104 x 10^{12}

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

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

** **

** ****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**

The maximum kinetic energy of the emitted photoelectrons is given as:

**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, the max K.E emitted by the photoelectrons is 2.4 x 10 ^{-19 }J.**

** **

** ****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 = \(\frac{hc}{\lambda }\)

= \(\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{632.8\times 10^{-9}}\)

= 3.141 x 10^{-19} J

Therefore, each photon has a momentum of :

P = \(\frac{h}{\lambda }\)

= \(\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{632.8\times 10^{-9}}\)

**= 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= \(\frac{P}{E}\)

= \(\frac{9.42\times 10^{-3}}{3.141\times 10^{-19}}\)

**= 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 = \(\frac{p}{m}\)

= \(\frac{1.047\times 10^{-27}}{1.66\times 10^{-27}}\) **= 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

\(\phi\) **= 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**

Average wavelength of photons present in sunlight,

\(\lambda\) = 550nm **= 550 x 10 ^{-9}m**

Number of photons per square metre incident on earth per second = n

Hence, the equation for power be written as:

**P = nE**

Therefore, n=\(\frac{P}{E}\)

= \(\frac{P\lambda }{hc}\)

= \(\frac{1.388\times 10^{3}\times 550\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}}\) **= 3.84 x 10 ^{21} photons/m^{2}/s**

**Therefore, every second, 3.84 x 10 ^{21} photons are incident on 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:

\(\frac{V}{v}\) = 4.12 x 10^{-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 \(\frac{V}{v}\)

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

**Therefore, Planck’s constant is determined to be 6.592 x 10 ^{-34} Js by 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, \(\lambda\) = 589nm

**= 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 = \(\frac{hc}{\lambda }\)

E = \(\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{589\times 10^{-9}}\)

= 3.37 x 10^{-19} J = \(\frac{3.37\times 10^{-19}}{1.6\times 10^{-19}}\) **= 2.11 eV**

**(b) **

Number of photons delivered to the sphere = n

The equation for power can be written as:

**P = nE**

Therefore, n=\(\frac{P}{E}\)

= \(\frac{100}{3.37\times 10^{-19}}\) = 2.96 x 10^{20} photons/s

**Therefore, every second, 2.96 x 10 ^{20} photons are delivered 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} = \(\frac{h(v-v_{o})}{e}\)

= \(\frac{6.626\times 10^{-34}\times(8.2\times 10^{14}-3.3\times 10^{14})}{1.6\times 10^{-19}}\) **= 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, \(\Phi _{o}\) **=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, \(\lambda\) = 330nm **= 330 × 10 ^{-9} m**

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

The energy of the incident photon is given as:

E = \(\frac{hc}{\lambda }\)

= \(\frac{6.626 \times 10^{-34}\times 3\times 10^{8}}{330\times 10^{-9}}\)

= 6.0 x 10^{-19} J = \(\frac{6.0\times 10^{-19}}{1.6\times 10^{-19}}\) **= 3.76 eV**

**It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take 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:

\(\frac{1}{2}mv^{2}\) = h(ν – ν_{o})

ν_{o }= ν – \(\frac{mv^{2}}{2h}\)

= 7.21 x 10^{14} – \(\frac{(9.1 \times 10^{-31})\times (6\times 10^{5})^{2}}{2 \times (6.626 \times 10^{-34})}\)

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

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

** **

** ****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,

\(\lambda\) = 488nm **= 488 x 10 ^{-9} m**

Stopping potential of the photoelectrons, **V _{o} = 0.38 V**

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

Therefore, V_{o} = \(\frac{0.38}{1.6\times 10^{-19}}\) eV

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**

From Einstein’s photoelectric effect, we have the relation involving the work function ,

\(\Phi _{o}\) of the material of the emitter as:

eV_{o} = \(\frac{hc}{\lambda } – \phi _{o}\)

\(\Phi _{o}\) = \(\frac{hc}{\lambda }\) eV_{o}

= \(\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{1.6\times 10^{-19}\times 488\times 10^{-9}} – \frac{1.6\times 10^{-19}\times 0.38}{1.6\times 10^{-19}}\)

= 2.54 – 0.38 **= 2.16 eV**

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

**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:

\(\frac{1}{2}mv^{2}\) = eV

\(v^{2}=\frac{2eV}{m}\)

Therefore, v = \(\sqrt{\frac{2\times 1.6\times 10^{-19}\times 56}{9.1\times 10^{-31}}}\)

= \(\sqrt{19.69\times 10^{12}}\)

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

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

\(\lambda\) = \(\frac{12.27}{\sqrt{V}}\) A^{o}

= \(\frac{12.27}{\sqrt{56}}\) x 10^{-19} m **= 0.1639 nm**

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

** **

** ****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} = \(\frac{1}{2}mv^{2}\)

Where, v = speed of the electron

Therefore, \(v^{2}=\sqrt{\frac{2eE_{k}}{m}}\)

= \(\sqrt{\frac{2\times 1.6\times 10^{-19}\times 120}{9.1\times 10^{-31}}}\)

= \(\sqrt{42.198\times 10^{12}}\)

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

**(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:

\(\lambda =\frac{h}{p}\)

= \(\frac{6.6\times 10^{-34}}{5.91\times 10^{-24}}\) = 1.116 x 10^{-10} m **= 0.112 nm**

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

** **

** ****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, \(\lambda\) = 589nm = 589 x 10^{-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 = \(\frac{1}{2}mv^{2}\) . . . . . . . . . . . . . . . **(1)**

We have the relation for de Broglie wavelength as:

\(\lambda = \frac{h}{m_{e}v}\)

Therefore, \(v^{2} = \frac{h^{2}}{\lambda ^{2}m_{e}^{2}}\) . . . . . . . . . . . **(2)**

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

K = \(\frac{1}{2}\) \(\frac{m_{e}h^{2}}{\lambda ^{2}m_{e}^{2}}\)

= \(\frac{h^{2}}{2\lambda^{2}m_{e} }\) . . . . . . . . . . . . **(3)**

= \(\frac{(6.6\times 10^{-34})^{2}}{2\times (589\times 10^{-9})^{2}\times 9.1\times 10^{-31}}\)

= 6.9 x 10^{-25} J = \(\frac{6.9\times 10^{-25}}{1.6\times 10^{-19}}\) **= 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:

= \(\frac{h^{2}}{2\lambda ^{2}m_{n}}\)

= \(\frac{(6.6\times 10^{-34})}{2\times (589\times 10^{-9})^{2}\times 1.66\times 10^{-27}}\)

= 3.78x 10 ^{-28} J

= \(\frac{3.78\times 10^{-28}}{1.6\times 10^{-19}}\)

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

**Hence, the kinetic energy of the neutron is 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 :

\(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{0.040\times 1000}\)

**= 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:

= \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{0.060\times 1}\)

**= 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:

= \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{2.2\times 1\times 10^{-9}}\)

**= 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 \(\lambda _{e}\) and a photon \(\lambda _{p}\) ,

\(\lambda _{e}\) = \(\lambda _{p}\) = \(\lambda\) = 1 nm **= 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:

\(\lambda = \frac{h}{p}\)

\(p=\frac{h}{\lambda }\)

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 = \(\frac{6.63\times 10^{-34}}{1\times 10^{-9}}\)

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

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

E = \(\frac{hc}{\lambda }\)

Where,

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

Therefore, E = \(\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{1\times 10^{-9}\times 1.6\times 10^{-19}}\)

= 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 = \(\frac{1}{2}\frac{p^{2}}{m}\)

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

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

Therefore, K = \(\frac{1}{2}\times \frac{(6.63\times 10^{-25})^{2}}{9.1\times 10^{-31}}\)

= 2.415 x 10^{-19} J

= \(\frac{2.415\times 10^{-19}}{1.6\times 10^{-19}}\) **= 1.51 eV**

**Hence, the kinetic energy of the electron is 1.51 eV.**

**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, \(\lambda\) = 1.40 x 10^{-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 = \(\frac{1}{2}m_{n}v^{2}\) **……… (1)**

de Broglie wavelength (\(\lambda\)) and velocity (v) are related as:

\(\lambda = \frac{h}{m_{n}v}\) **………. (2)**

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

K = \(\frac{1}{2}\frac{m_{n}h^{2}}{\lambda ^{2}m_{n}^{2}}\)

= \(\frac{h^{2}}{2\lambda ^{2}m_{n}}\)

= \(\frac{(6.63 \times 10^{-34})^{2}}{2\times (1.40\times 10^{-10})^{2} \times 1.66\times 10^{-27} }\)

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

= \(\frac{6.75\times 10^{-21}}{1.6\times 10^{-19}}\)

**= 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’ = \(\frac{3}{2}\) kT

= \(\frac{3}{2}\) x 1.38 x 10^{-23} x 300

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

The relation for the de Broglie wavelength is given as:

\(\lambda ‘=\frac{h}{\sqrt{2K’m_{n}}}\)

Where,

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

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

K’ = 6.75 x 10^{-21} J

Therefore, \(\lambda ‘=\frac{6.63\times 10^{-34}}{\sqrt{2\times 6.21\times 10^{-21}\times 1.66\times 10^{-27}}}\)

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

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

** **

** ****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 = \(\frac{hv}{c}\) = \(\frac{h}{\lambda }\)

\(\lambda = \frac{h}{p}\) **………. (i)**

Where,

\(\lambda\) = wavelength of the electromagnetic radiation

**c = speed of light**

**h = Planck’s constant**

De Broglie wavelength of the photon is given as:

\(\lambda = \frac{h}{mv}\)

But, **p = mv**

Therefore, \(\lambda = \frac{h}{p}\) **……………(ii)**

Where, m = mass of the photon

v = velocity of the photon

**Hence, it can be concluded from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.**

**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 \((\frac{3}{2}kT)\) of the nitrogen molecule with the root mean square speed (V_{rms} ) as:

\(\frac{1}{2}mv_{rms}^{2}\) = \((\frac{3}{2}kT)\)

V_{rms} = \(\sqrt{\frac{3kT}{m}}\)

Hence, the de Broglie wavelength of the nitrogen molecule is given as:

\(\lambda =\frac{h}{mv_{rms}}\) = \(\frac{h}{\sqrt{3mkT}}\)

= \(\frac{6.63\times 10^{-14}}{\sqrt{3\times 28.0152\times 1.66\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}\)

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

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

CBSE is one of the popular educational board in India. The Central Board of Secondary Education follow the NCERT syllabus to conducts its class 10th and class 12th board examinations. As such, the NCERT solutions that we are offering will help students get a clarity on all the topics in this chapter.

Some of the topics discussed 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 strong electric field or irradiating it by light of suitable frequency. Some key points on Dual Nature of Radiation Matter is 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.