**‘NCERT Solutions for Class 12 Physics Chapter 11‘ PDF Quick download link** is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 12 Physics Chapter 1 Exercise Solution’ using the download button.

## NCERT Class 12 Physics Textbook Chapter 11 With Answer PDF Free Download

### Chapter 11: Dual Nature of Radiation and Matter

Maxwell’s equations of electromagnetism and Hertz’s experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light.

Towards the same period at the end of the 19th century, experimental investigations on the conduction of electricity (electric discharge) through gases at low pressure in a discharge tube led to many historic discoveries.

The discovery of X-rays by Roentgen in 1895, and of electrons by J. J. Thomson in 1897, were important milestones in the understanding of atomic structure.

It was found that at sufficiently low pressure of about 0.001 mm of the mercury column, a discharge took place between the two electrodes on applying the electric field to the gas in the discharge tube.

A fluorescent glow appeared on the glass opposite to cathode.

The color of the glow of the glass depended on the type of glass, it being yellowish-green for soda glass.

The cause of this fluorescence was attributed to the radiation which appeared to be coming from the cathode.

These cathode rays were discovered, in 1870, by William Crookes who later, in 1879, suggested that these rays consisted of streams of fast-moving negatively charged particles.

The British physicist J. J. Thomson (1856-1940) confirmed this hypothesis. By applying mutually perpendicular electric and magnetic fields across the discharge tube, J. J.

Thomson was the first to determine experimentally the speed and the specific charge [charge to mass ratio (e/m )] of the cathode ray particles.

They were found to travel with speeds ranging from about 0.1 to 0.2 times the speed of light (3 ×108 m/s).

The presently accepted value of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be independent of the nature of the material/metal used as the cathode (emitter), or the gas introduced in the discharge tube. This observation suggested the universality of the cathode ray particles.

Around the same time, in 1887, it was found that certain metals, when irradiated by ultraviolet light, emitted negatively charged particles having small speeds. Also, certain metals when heated to a high temperature were found to emit negatively charged particles.

The value of e/m of these particles was found to be the same as that for cathode ray particles.

These observations thus established that all these particles, although produced under different conditions, were identical in nature. J. J. Thomson, in 1897, named these particles like electrons and suggested that they were fundamental, universal constituents of matter.

For his epoch-making discovery of the electron, through his theoretical and experimental investigations on the conduction of electricity by gasses, he was awarded the Nobel Prize in Physics in 1906.

In 1913, the American physicist R. A. Millikan (1868-1953) performed the pioneering oil-drop experiment for the precise measurement of the charge on an electron.

He found that the charge on an oil droplet was always an integral multiple of an elementary charge, 1.602 × 10 –19 C.

Millikan’s experiment established that electric charge is quantized. From the values of charge (e) and specific charge (e/m ), the mass (m) of the electron could be determined.

Author | NCERT |

Language | English |

No. of Pages | 28 |

PDF Size | 808 KB |

Category | Physics |

Source/Credits | ncert.nic.in |

### NCERT Solutions Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

**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}*v*=*h**E*

= \frac{1.6\times 10^{-19}\times 3 \times 10^{4}}{6.626\times 10^{-34}}6.626×10−341.6×10−19×3×104 **= 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:**\lambda =\frac{c}{v}*λ*=*v**c*

= \frac{3\times 10^{8}}{7.24\times 10^{18}}7.24×10183×108 **= 4.14 x 10 ^{-11} m = 0.0414 nm**

**Q.11.2: The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10 ^{14} Hz is incident on the metal surface, photoemission of electrons occurs. What is the**

**(a) the maximum kinetic energy of the emitted electrons**

**(b) Stopping potential**

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

** ****Ans:**

Work function of caesium, \Phi _{o}Φ*o* **= 2.14eV**

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

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

K = hν – \Phi _{o}Φ*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}}1.6×10−196.626×10−34×6×1014 – 2.14

= 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 }= \frac{K}{e}*e**K*

= \frac{0.345\times 1.6\times 10^{-19}}{1.6\times 10^{-19}}1.6×10−190.345×1.6×10−19

**= 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 = \frac{1}{2}mv^{2}21*m**v*2

Where,

m = mass of electron **= 9.1 x 10 ^{-31} Kg**v^{2}=\frac{2K}{m}

*v*2=

*m*2

*K*

= \frac{2\times 0.345\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}9.1×10−312×0.345×1.6×10−19

=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 in a certain experiment is 1.5 V.What is the maximum kinetic energy of photoelectrons emitted?**

** ****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 of wavelength 632.8 nm is produced by ahelium-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 average, arrive at a target irradiated by this beam? (Assume the beam to have a uniform cross-section that is less than the target area)**

**(c) How fast does a hydrogen atom have to travel in order 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 }*λ**h**c*

= \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{632.8\times 10^{-9}}632.8×10−96.626×10−34×3×108

= 3.141 x 10^{-19} J

Therefore, each photon has a momentum of :

P = \frac{h}{\lambda }*λ**h*

= \frac{6.626\times 10^{-34}}{632.8\times 10^{-9}}632.8×10−96.626×10−34

**= 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}*E**P*

= \frac{9.42\times 10^{-3}}{3.141\times 10^{-19}}3.141×10−199.42×10−3

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

**(c) **Given that, the 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}*m**p*

= \frac{1.047\times 10^{-27}}{1.66\times 10^{-27}}1.66×10−271.047×10−27 **= 0.621 m/s**

NCERT Class 12 Physics Textbook Chapter 11 With Answer PDF Free Download