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

NCERT Solutions for Class 12 Physics Chapter 11 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter is a vital reference material that is essential for your Class 12 and entrance exam preparations. The NCERT Solutions for Class 12 Physics Chapter 11 eases your work of preparing notes and revisions. The chapter consists of Dual Nature of Radiation and Matter important questions with answers, important tables, exemplary questions and worksheets.

The Dual Nature of Radiation and Matter topic is very important when it comes to the Class 12 Physics examination. The PDF of NCERT Solutions for Class 12 Physics for this chapter is provided here to help 12th standard students to prepare for the subject more efficiently. These NCERT Solutions for Class 12 are prepared by expert tutors in such a way that it helps students understand the concepts of the chapter easily and in an interactive manner.

Download NCERT Solutions Class 12 Physics Chapter 11 PDF:-Download Here

ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 01
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 02
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 03
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 04
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 05
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 06
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 07
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 08
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 09
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 10
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 11
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 12
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 13
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 14
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 15
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 16
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 17
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 18
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 19
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 20
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 21
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 22
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 23
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 24
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 25
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 26
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 27
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 28
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 29
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 30
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 31
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 32
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 33
ncert solutions may7 class 12 physics chapter 11 dual nature of radiation and matter 34

 

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 × 104 V

Hence, electron energy, E = 3 × 104 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=Ehv = \frac{E}{h}

= 1.6×1019×3×1046.626×1034\frac{1.6\times 10^{-19}\times 3 \times 10^{4}}{6.626\times 10^{-34}} = 7.24 x 1018 Hz

Hence,  7.24 x 1018 Hz is the maximum frequency of the X-rays.

(b) The minimum wavelength produced:

λ=cv\lambda =\frac{c}{v}

= 3×1087.24×1018\frac{3\times 10^{8}}{7.24\times 10^{18}} = 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 × 1014 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

(c) maximum speed of the emitted photoelectrons?

 Ans:

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

Frequency of light, v = 6.0 x 1014 Hz

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

K = hν – Φo\Phi _{o}

Where,

h = Planck’s constant = 6.626 x 10-34 Js

Therefore,

K = 6.626×1034×6×10141.6×1019\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, 0.345 eV is the maximum kinetic energy of the emitted electrons.

(b) For stopping potential Vo, we can write the equation for kinetic energy as:

K = eVo

Therefore, Vo = Ke\frac{K}{e}

= 0.345×1.6×10191.6×1019\frac{0.345\times 1.6\times 10^{-19}}{1.6\times 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 = 12mv2\frac{1}{2}mv^{2}

Where,

m = mass of electron = 9.1 x 10-31 Kg

v2=2Kmv^{2}=\frac{2K}{m}

= 2×0.345×1.6×10199.1×1031\frac{2\times 0.345\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}

=0.1104 x 1012

Therefore, ν = 3.323 x 105 m/s = 332.3 km/s

Hence332.3 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, Vo = 1.5 V

For emitted photoelectrons, the maximum kinetic energy is:

Ke = eVo

Where,

e = charge on an electron = 1.6 x 10-19 C

Therefore, Ke = 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 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 a uniform cross-section which 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 × 108 m/s

Mass of a hydrogen atom, m = 1.66 × 10-27 kg

(a) The photons having the energy as:

E = hcλ\frac{hc}{\lambda }

= 6.626×1034×3×108632.8×109\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 = hλ\frac{h}{\lambda }

= 6.626×1034632.8×109\frac{6.626\times 10^{-34}}{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= PE\frac{P}{E}

= 9.42×1033.141×1019\frac{9.42\times 10^{-3}}{3.141\times 10^{-19}}

= 3 x 1016 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 = pm\frac{p}{m}

= 1.047×10271.66×1027\frac{1.047\times 10^{-27}}{1.66\times 10^{-27}} = 0.621 m/s

  Question 11.5: The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. 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 the earth has an energy flux of

ϕ\phi = 1.388 × 103 W/m2

Hence, power of sunlight per square metre, P = 1.388 × 103 W

Speed of light, c = 3 × 108 m/s

Planck’s constant, h = 6.626 × 10-34 Js

λ\lambda = 550nm = 550 x 10-9is the average wavelength of the photons from the sunlight

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

Hence, the equation for power be written as:

P = nE

Therefore, n=PE\frac{P}{E}

= Pλhc\frac{P\lambda }{hc}

= 1.388×103×550×1096.626×1034×3×108\frac{1.388\times 10^{3}\times 550\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}} = 3.84 x 1021 photons/m2/s

Therefore, 3.84 x 1021 photons are incident on the earth per square meters.

11.6: In an experiment on the photoelectric effect, the slope of the cut-off
voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.

 Ans:

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

Vv\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 Vv\frac{V}{v}

= 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: A 100W 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 = 100W

Wavelength of the emitted sodium light, λ\lambda = 589nm

= 589 x 10-9 m

Planck’s constant, h = 6.626 x 10-34 Js

Speed of light, c = 3 x 108

(a)

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

E = hcλ\frac{hc}{\lambda }

E = 6.626×1034×3×108589×109\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{589\times 10^{-9}}

= 3.37 x 10-19 J = 3.37×10191.6×1019\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=PE\frac{P}{E}

= 1003.37×1019\frac{100}{3.37\times 10^{-19}} = 2.96 x 1020 photons/s

Therefore, 2.96 x 1020 photons are delivered every second to the sphere.

  Question 11.8: The threshold frequency for a certain metal is 3.3 x 1014 Hz. If the light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Ans:

Threshold frequency of the metal, vo= 3.3 × 1014 Hz.

Frequency of light incident on the metal, v= 8.2 × 1014 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 = Vo

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

eVo = h(ν – νo)

Vo = h(vvo)e\frac{h(v-v_{o})}{e}

= 6.626×1034×(8.2×10143.3×1014)1.6×1019\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 for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?

Ans:

Work function of the metal, Φo\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, λ\lambda = 330nm = 330  × 10-9 m

Speed of light, c = 3 × 108 m/s

The energy of the incident photon is given as:

E = hcλ\frac{hc}{\lambda }

= 6.626×1034×3×108330×109\frac{6.626 \times 10^{-34}\times 3\times 10^{8}}{330\times 10^{-9}}

= 6.0 x 10-19 J = 6.0×10191.6×1019\frac{6.0\times 10^{-19}}{1.6\times 10^{-19}} = 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 1014 Hz is incident in a metal surface. Electrons with a maximum speed of 6.0 x 105 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 10-9m

Maximum speed of the electrons, v = 6.0 x 105 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 vo, the relation for kinetic energy is written as:

12mv2\frac{1}{2}mv^{2} = h(ν – νo)

νo = ν – mv22h\frac{mv^{2}}{2h}

= 7.21 x 1014(9.1×1031)×(6×105)22×(6.626×1034)\frac{(9.1 \times 10^{-31})\times (6\times 10^{5})^{2}}{2 \times (6.626 \times 10^{-34})}

= 7.21 x 1014 – 2.472 x 1014= 4.738 x 1014 Hz

Therefore, 4.738 x 1014 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,

λ\lambda = 488nm = 488 x 10-9 m

Stopping potential of the photoelectrons, Vo = 0.38 V

1eV = 1.6 x 10-19 J

Therefore, Vo = 0.381.6×1019\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 108 m/s

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

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

eVo = hcλϕo\frac{hc}{\lambda } – \phi _{o}

 

Φo\Phi _{o}  = hcλ\frac{hc}{\lambda } – eVo

 

= 6.6×1034×3×1081.6×1019×488×1091.6×1019×0.381.6×1019\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, 2.16eV is the work function of 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 10-19C

(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:

12mv2\frac{1}{2}mv^{2} = eV

v2=2eVmv^{2}=\frac{2eV}{m}

Therefore, v = 2×1.6×1019×569.1×1031\sqrt{\frac{2\times 1.6\times 10^{-19}\times 56}{9.1\times 10^{-31}}}

= 19.69×1012\sqrt{19.69\times 10^{12}}

= 4.44 x 106 m/s

The momentum of each accelerated electron is given as:

p = mv

= 9.1 x 10-31 x 4.44 x 106= 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:

λ\lambda = 12.27V\frac{12.27}{\sqrt{V}} Ao

= 12.2756\frac{12.27}{\sqrt{56}} x 10-19 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, EK = 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:

Ek12mv2\frac{1}{2}mv^{2}

Where, v = speed of the electron

Therefore, v2=2eEkmv^{2}=\sqrt{\frac{2eE_{k}}{m}}

= 2×1.6×1019×1209.1×1031\sqrt{\frac{2\times 1.6\times 10^{-19}\times 120}{9.1\times 10^{-31}}}

= 42.198×1012\sqrt{42.198\times 10^{12}}

= 6.496 × 106 m/s

Momentum of the electron, p = mv = 9.1 × 10-31× 6.496 × 106

= 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 × 106 m/s

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

λ=hp\lambda =\frac{h}{p}

= 6.6×10345.91×1024\frac{6.6\times 10^{-34}}{5.91\times 10^{-24}}  = 1.116 x 10-10 m = 0.112 nm

Therefore, 0.112 nm is 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, λ\lambda = 589nm = 589 x 10-9 m

Mass of an electron, me = 9.1 x 10-31 Kg

Mass of a neutron, mn = 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 = 12mv2\frac{1}{2}mv^{2} . . . . . . . . . . . . . . . (1)

We have the relation for de Broglie wavelength as:

λ=hmev\lambda = \frac{h}{m_{e}v}

Therefore, v2=h2λ2me2v^{2} = \frac{h^{2}}{\lambda ^{2}m_{e}^{2}} . . . . . . . . . . . (2)

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

K = 12\frac{1}{2} meh2λ2me2\frac{m_{e}h^{2}}{\lambda ^{2}m_{e}^{2}}

= h22λ2me\frac{h^{2}}{2\lambda^{2}m_{e} } . . . . . . . . . . . . (3)

= (6.6×1034)22×(589×109)2×9.1×1031\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 = 6.9×10251.6×1019\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:

= h22λ2mn\frac{h^{2}}{2\lambda ^{2}m_{n}}

= (6.6×1034)22×(589×109)2×1.66×1027\frac{(6.6\times 10^{-34})^{2}}{2\times (589\times 10^{-9})^{2}\times 1.66\times 10^{-27}}

= 3.78x 10 -28 J

= 3.78×10281.6×1019\frac{3.78\times 10^{-28}}{1.6\times 10^{-19}}

= 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 :

λ=hmv\lambda = \frac{h}{mv}

= 6.6×10340.040×1000\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:

= λ=hmv\lambda = \frac{h}{mv}

= 6.6×10340.060×1\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:

= λ=hmv\lambda = \frac{h}{mv}

= 6.6×10342.2×1×109\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 λe\lambda _{e} and a photon λp\lambda _{p} ,

λe\lambda _{e} = λp\lambda _{p} = λ\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:

λ=hp\lambda = \frac{h}{p}

 

p=hλp=\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 = 6.63×10341×109\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 = hcλ\frac{hc}{\lambda }

Where,

Speed of light, c = 3 x 108 m/s

Therefore, E = 6.63×1034×3×1081×109×1.6×1019\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 = 12p2m\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 = 12×(6.63×1025)29.1×1031\frac{1}{2}\times \frac{(6.63\times 10^{-25})^{2}}{9.1\times 10^{-31}}

= 2.415 x 10-19 J

= 2.415×10191.6×1019\frac{2.415\times 10^{-19}}{1.6\times 10^{-19}} = 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, λ\lambda = 1.40 x 10-10 m

Mass of a neutron, mn = 1.66 x 10-27 Kg

Planck’s constant, h = 6.63 x 10-34 Js

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

K = 12mnv2\frac{1}{2}m_{n}v^{2} ……… (1)

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

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

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

K = 12mnh2λ2mn2\frac{1}{2}\frac{m_{n}h^{2}}{\lambda ^{2}m_{n}^{2}}

= h22λ2mn\frac{h^{2}}{2\lambda ^{2}m_{n}}

= (6.63×1034)22×(1.40×1010)2×1.66×1027\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

= 6.75×10211.6×1019\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 m2 s-2 K-1

Average kinetic energy of the neutron:

K’ =  32\frac{3}{2} kT

= 32\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:

λ=h2Kmn\lambda ‘=\frac{h}{\sqrt{2K’m_{n}}}

Where,

mn = 1.66 x 10-27 Kg

h = 6.63 x 10-34 Js

K’ = 6.21 x 10-21 J

Therefore, λ=6.63×10342×6.21×1021×1.66×1027\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, 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 = hvc\frac{hv}{c} = hλ\frac{h}{\lambda }

 

λ=hp\lambda = \frac{h}{p} ………. (i)

Where,

λ\lambda = wavelength of the electromagnetic radiation

c = speed of light

h = Planck’s constant

De Broglie wavelength of the photon is given as:

λ=hmv\lambda = \frac{h}{mv}

But, p = mv

Therefore, λ=hp\lambda = \frac{h}{p}  ……………(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 (32kT)(\frac{3}{2}kT) of the nitrogen molecule with the root mean square speed (Vrms ) as:

12mvrms2\frac{1}{2}mv_{rms}^{2} = (32kT)(\frac{3}{2}kT)

Vrms = 3kTm\sqrt{\frac{3kT}{m}}

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

λ=hmvrms\lambda =\frac{h}{mv_{rms}} = h3mkT\frac{h}{\sqrt{3mkT}}

= 6.63×10343×28.0152×1.66×1027×1.38×1023×300\frac{6.63\times 10^{-34}}{\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.

Question 11.20: (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg–1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Answer:

(a) Potential difference of the evacuated tube = 500 V

Specific charge of the electron, e/m = 1.76 × 1011 C kg–1

Kinetic energy = (1/2) mv2 = eV

Speed of the emitted electron, v = (2Ve/m)1/2

= (2 x 500 x 1.76 x 1011)1/2  

= 1.32 x 107 m/s

(b) Collector potential, V = 10 MV = 10 x 106 V.

Speed of electron = v = (2Ve/m)1/2

= (2 x 107 x 1.76 x 1011)1/2

= 1.88 x 109 m/s

This answer is not correct. Since the value is greater than the speed of light (c). The expression (1/2) mv2 for energy should be used in the non -relativistic limit.i.e., v << c.

In the relativistic limits, the total energy is given as

E = mc2

Here,

m is the relativistic mass

m = m0 (1- v2/c2)1/2

m0 = mass of the particle at rest

Kinetic energy is given as

K = mc2 – m0c2

Question 11.21: (a) A monoenergetic electron beam with an electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011C kg–1.
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

Answer:

Speed of the electron, v= 5.20 × 106 m s–1

Magnetic field experienced by the electron, B = 1.30 × 10–4 T

Specific charge, e/m = 1.76 x 1011 Ckg-1

Here,

e = charge on the electron =  1.6 x 10-19 C

m = mass of the electron = 9.1 x 10-31 kg-1

Force exerted on the electron is given as

F=ev×BF = e\left | \vec{v}\times \vec{B} \right |

= evBsinθ

θ is the angle between the magnetic field and the velocity of the beam. The magnetic field is normal to the direction of the beam.

θ = 900

F = evB

The normal magnetic field provides the centripetal force.

Therefore, evB = mv2/r

r = mv/eB = v/(e/m)B

= (5.20 x 106)/ (1.76 x 1011) x (1.30 x 10-4) = 0.227 m = 22.7 cm

Therefore, the radius of the circular path is 22.7 cm

(b) Energy of the electron beam, E = 20 Mev = 20 x 106 x 1.6 x 10-19 J

The energy of the electron beam is, E = (1/2) mv2

⇒ v=2Emv =\sqrt{\frac{2E}{m}}

 

v=2×20×106×1.6×10199.1×1031=2.652×109m/sv =\sqrt{\frac{2\times 20\times 10^{6}\times1.6\times 10^{-19} }{9.1\times 10^{-31}}} = 2.652 \times 10^{9}m/s

The result is greater than the speed of light. Therefore, it is wrong. The expression (1/2) mv2 for energy should be used in the non -relativistic limit.i.e., v << c.

In the relativistic limits, the total mass is given as

m = m0 (1- v2/c2)1/2

m0 = mass of the particle at rest

Therefore, the radius of the circular path is

r = mv/eB

r=m0veBc2v2c2r = \frac{m_{0}v}{eB\sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}

Question 11.22: An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

Answer:

Potential of the collector, V= 100 V

Magnetic field experienced by the electron, B =2.83 × 10–4 T

Radius of the circular orbit, r = 12 cm = 12.0 x 10-2 m

Kinetic energy, (1/2)mv2 = eV

v2 = 2eV/m ——–(1)

The magnetic field that curves the path of the electron provides the centripetal force

evB = mv2/r

eB = mv/r

v = eBr/m ——–(2)

Substituting (2) in (1)

2eVm=e2B2r2m2\frac{2eV}{m}=\frac{e^{2}B^{2}r^{2}}{m^{2}}

 

em=2VB2r2\frac{e}{m}=\frac{2V}{B^{2}r^{2}}

 

em=2×100(2.83×104)2(12×102)2=1.73×1011Ckg1\frac{e}{m}=\frac{2\times 100}{(2.83\times 10^{-4})^{2}(12\times 10^{-2})^{2}} = 1.73 \times 10^{11}Ckg^{-1}

Therefore, the specific charge ration e/m is 1.73 x 1011 Ckg-1

Question 11. 23: (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

Answer:

(a) Wavelength produced by the X-ray tube, λ= 0.45 Å = 0.45 x 10-10 m

Speed of light, c = 3 x 108 m/s

Planck’s constant, h = 6.626 x 10-34 Js

The maximum energy of a photon is given as

Emax = hc/λmin

= (6.626 x 10-34)(3 x 108 m/s)/(0.45 x 10-10 m x 1.6 x 10-19)

= 19.878 x 10-26/0.72 x 10-29

= 27.60 x 103 eV = 27.6 keV

(b) To incident electron should have an energy of 27.6 keV to get a X-ray of 27.6 keV. Therefore, the accelerating voltage of the order of 30 keV is required for producing X-rays.

Question 11.24:In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV)

Answer:

Total energy of the electron-positron pair, E= 10.2 BeV = 10.2 x 109 eV = 10.2 x 109 x 1.6 x 10-19 J

Hence the energy of each γ-ray , E’ = E/2 = (10.2 x 109 x 1.6 x 10-19 )/2 = 8.16 x 10-10 J

Energy and wavelength relation is given as,

E’ = hc/λ

Therefore, λ = hc/E’

Here, h = 6.626 x 10-34 Js

c = 3 x 108 m/s

λ=6.626×1034×3×1088.16×1010=2.436×1016m\lambda = \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{8.16\times 10^{-10}}= 2.436 \times 10^{-16}m

The wavelength associated with each γ-ray is 2.436 x 10-16 m

Question 11.25: Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry
much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10–10 W m–2). Take the area of the pupil
to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz

Answer:

(a) Power of the medium wave transmitter, P = 10 kW = 104 W

Energy emitted by the transmitter per second, E = 104

Wavelength of the radio waves, λ = 500 m

The energy of the wave is given as, E’ = hc/λ

E’ = (6.6 x 10-34 x 3 x 108)/500

= 3.96 x 10-28 J

Let n be the number of photons emitted by the transmitter

nE’ = E

n = E/E’

= 104/(3.96 x 10-28 )

= 0.2525 x 1032

The energy E’ of the radio photon is very less, but the number of photons emitted is large. The total energy of the radio ways can be considered as continuous and the existence of the minimum quantum energy can be ignored.

(b) Intensity of the light perceived by the human eye, I =  10–10 W m–2

Area of the pupil, A = 0.4 cm2 = 0.4 x 10-4 m2

Frequency of the white light, ν = 6 x 1014 Hz

h = Planck’s constant = 6.6 x 10-34 Js

Energy of the emitted photon, E = hν

= 6.6 x 10-34  x6 x 1014

= 3.96 x 10-19 J

Let n be the total number of photons falling per unit area per unit time. The total energy per unit for n photons is

E = n x 3.96 x 10-19 J/s/m2

Total energy per unit for n photons is equal to the intensity of the light.

E = I

I = n x 3.96 x 10-19 J/s/m2

n = I/3.96 x 10-19

= 10-10/3.96 x 10-19

= 2.52 x 108 m2/s

The total number of photons entering the pupil is given as,

nA = 2.52 x 108 x 0.4 x 10-4

= 1.008 x 104 s-1

The number is large. So the human eye can never count the number of individual photons.

Question 11.26: Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser?

Answer:

Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m

Stopping potential of the metal, V0 = 1.3 V

Planck’s constant, h = 6.6 × 10−34 J

Charge on an electron, e = 1.6 × 10−19 C

From photoelectric effect we have the photon-electron relation as

Work function of the metal, Φ0 = hν – eV0

= (hc/λ) – eV0

6.6×1034×3×1082271×10101.6×1019×1.3\frac{6.6\times 10^{-34}\times 3\times 10^{^{8}}}{2271\times 10^{-10}}-1.6 \times 10^{-19}\times 1.3

= 8.72 x 10-19 – 2.08 x 10-19

= 6.64 x 10-19 J

Threshold frequency of the metal

Therefore, Φ0 = hυ0

υ= Φ0/h = 6.64 x 10-19 /6.6 x 10-34

= 1.006 x 1015 Hz

Wavelength of the red light, λr = 6328 x 10-10 m

Frequency of the red light, υr = c/λr = 3 x 108/6328 x 10-10

= 4.74 x 1014 Hz

υ0r

Photocell will not respond to the red light produced by the He-Ne laser.

Question 11.27: Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Answer:

Wavelength of the monochromatic radiation, λ = 640.2 nm = 640.2 × 10−9 m

Stopping potential of the neon lamp, V0 = 0.54 V

Charge on an electron, e = 1.6 × 10−19 C

Planck’s constant, h = 6.6 × 10−34 Js

From photoelectric effect we have the photon-electron relation as eV0 = hν – Φ0

Work function of the metal, Φ0 = hν – eV0

= (hc/λ) – eV0

=6.6×1034×3×108640.2×1091.6×1019×0.54=\frac{6.6\times 10^{-34}\times 3\times 10^{^{8}}}{640.2\times 10^{-9}}-1.6 \times 10^{-19}\times 0.54

= 3.093 x 10-19 – 0.864 x 10-19

= 2.229 x 10-19 J

Wavelength of the radiation emitted by the iron source, λ’ = 427.2 nm = 427.2 x 10-9 m

Let V0‘ be the new stopping potential

Therefore, eV0‘ = (hc/λ’) – Φ0

=6.6×1034×3×108427.2×1092.229×1019=\frac{6.6\times 10^{-34}\times 3\times 10^{^{8}}}{427.2\times 10^{-9}}-2.229 \times 10^{-19}

= 4.63 x 10-19 – 2.229 x 10-19

= 2.401 x 10-19 J

V0‘ = 2.401 x 10-19 J/ 1.6 x 10-19 J

= 1.5 eV

The new stopping potential = 1.50 eV

Question 11.28: A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650 Å,

λ2 = 4047 Å,

λ3 = 4358 Å,

λ4 = 5461 Å,

λ5 = 6907 Å,
The stopping voltages, respectively, were measured to be: V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V. Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]

Answer

From photoelectric effect we have the photon-electron relation as eV0 = hν – Φ0

Work function of the metal, Φ0 = hν – eV0

Φ= (hc/λ) – eV0

V0=heνϕ0eV_{0} = \frac{h}{e}\nu -\frac{\phi _{0}}{e}——-(1)

Here,

V0 = Stopping potential
h = Planck’s constant
e = Charge on an electron
ν = Frequency of radiation
Φ= Work function of a material

Stopping proportional is directly proportional to the frequency.
Frequency, ν = Speed of light (c)/Wavelength (λ)

Using this equation we can find the frequency of various lines

ν1 = c/λ1 = 3 x 108/3650 x 10-10

= 8.219 x 1014 Hz

ν2 = c/λ2= 3 x 108/4047 x 10-10

= 7.412 x 1014 Hz

ν3 = c/λ3= 3 x 108/4358 x 10-10

= 6.88 x 1014 Hz

ν4= c/λ4= 3 x 108/5461 x 10-10

= 5.493 x 1014 Hz

ν5= c/λ5= 3 x 108/6907 x 10-10

= 4.343 x 1014 Hz

 

Frequency 8.219 7.412 6.884 5.493 4.343
Stopping potential 1.28 0.95 0.74 0.16 o

 

The above values can be plotted in a graph

NCERT solutions Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter Question 28

The graph is a straight line. It intersects the y-axis at 5 x 1014 Hz. This is the threshold frequency. The point D is the frequency less than the threshold frequency.

Slope of the straight line = AB/CB = (1.28 – 0.16)/(8.214 – 5.493) x 1014

From equation (1), the slope is written as

h/e = (1.28 – 0.16)/(8.214 – 5.493) x 1014

h = (1.12 x 1.6 x 10-19)/(2.726 x 1014)

= 6.573 x 10-34 Js

The work function of the metal is,

Φ0 = hν0

= (6.573 x 10-34 x 5 x 1014)

= 3.286 x 10-19 J

= 3.286 x 10-19 /1.6 x 10-19

Φ0= 2.054 eV

Question 11.29: The work function for the following metals is given:  Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Answer:

Wavelength λ = 3300 Å

Speed of light = 3 x 10m/s

Planck’s constant = 6.63 x 10-34 Js

Energy of the photon of the incident light

E = hc/λ = (6.63 x 10-34 x 3 x 108)/3300x 10-10

⇒ 6.018 x 10-19 J

⇒  (6.018 x 10-19 J)/1.6 x 10-19

= 3.7 eV

The energy of the incident radiation is greater than the work function of Na and K. It is lesser for Mo and Ni. Therefore, Mo and Ni will not show photoelectric effect.

If the laser is brought nearer and placed 50 cm away, then the intensity of the radiation will increase. The energy of the radiation will not be affected. Therefore, the result will be the same. However, the photoelectrons from Na and K will increase in proportion to intensity.

Question 11.30: Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

Answer:

Intensity of the light = 10–5 W m–2

Surface area of the sodium photocell, A = 2 cm2

Incident power of the light, P = I x A

= 10-5 x 2 x 10-4

= 2 x 10-9 W

Work function of the metal, Φ0 = 2 eV

= 2 x 1.6 x 10-19 

= 3.2 x 10-19 J

The number of layers of sodium that absorbs the incident energy, n = 5

Atomic area of the sodium atom, Ae is 10-20 m2

Hence, the number of conduction electrons in n layers is given as:

n’ = n x (A/Ae)

= 5 x [(2 x 10-4)/10-20] = 1017

The incident power is absorbed by all the electrons continuously. The amount of energy absorbed per electron per second is

E = P/n’

= (2 x 10-9)/1017

= 2 x 10-26 J/s

The time for photoelectric emission

t = Φ0/E

= (3.2 x 10-19)/(2 x 10-26) = 1.6 x 107 s ≈ 0.507 years

The time required for the photoelectric emission is almost half a year. This is not practical. Therefore, the wave function is in disagreement with the given experiment.

Question 11. 31: Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of interatomic spacing in the lattice) (me =9.11 × 10–31 kg).

Answer:

For Electrons, Kinetic energy, K.E= (1/2) mev2

= (mev)2/2m

K.E = p2/2me

⇒ p=2meK.Ep = \sqrt{2m_{e}K.E}

 

λ=hp=h2meK.E\lambda = \frac{h}{p}= \frac{h}{\sqrt{2m_{e}K.E}}

 

λ2=h22meK.E\lambda^{2} = \frac{h^{2}}{{2m_{e}K.E}}

 

K.E=h22meλ2K.E= \frac{h^{2}}{{2m_{e}\lambda^{2}}}

 

K.E=(6.64×1034)22×9.1×1031×(1010)2K.E= \frac{(6.64 \times 10^{-34})^{2}}{{2\times 9.1 \times 10^{-31}\times (10^{-10})^{2}}}

K.E = 2.4 x 10-17 J

K.E = 2.4×10171.6×1019\frac{2.4\times 10^{-17}}{1.6\times 10^{-19}}

= 149.375 eV

For photon of X-rays, Energy, E = hc/λe

= (6.63 x 10-34 x 3 x 108)/ (10-10 x 1.6 x 10-19)

= 12.375 x 103 eV

= 12.375 keV

The energy of the photons of X-rays is more than the energy of the electron.

Question 11.32: (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in question 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 × 10–27 kg).

(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Answer:

(a) Kinetic energy of the neutron = 150 eV

= 150 x 1.6 x 10-19 

= 2.4 x 10-17 J

Mass of the neutron, mn = 1.675 × 10–27 kg

The kinetic energy of the neutron is given by the relation

K.E = (1/2) mev2

K.E = p2/2me

⇒ p=2meK.Ep = \sqrt{2m_{e}K.E}

 

λ=hp=h2meK.E\lambda = \frac{h}{p}= \frac{h}{\sqrt{2m_{e}K.E}}

Wavelength and mass are inversely proportional. Wavelength decreases with increase in the mass and vice versa.

λ=(6.63×1034)2.24×1017×1.675×1027\lambda = \frac{(6.63 \times 10^{-34})}{{\sqrt{2.24 \times 10^{-17}\times 1.675\times 10^{-27}}}}

= 2.327 x 10-12 m

In question 11.31 it is given inter-atomic spacing of the crystal is about 1 Å, i.e., 10-10 m. The interatomic spacing is about 100 times greater. Therefore, a neutron of kinetic energy is 150 eV is not good for a diffraction experiment.

(b) Room temperature = 270 C = 27 +  273 = 300 K

Average kinetic energy of the neutron, E = (3/2) kT

here, k = Boltzmann constant = 1.38 x 10-23 J/mol/K

The wavelength of the neutron is

λ=h2mnE\lambda =\frac{h}{\sqrt{2m_{n}E}}

 

λ=h2mn32kT\lambda =\frac{h}{\sqrt{2m_{n}\frac{3}{2}kT}}

 

λ=6.6×10343×1.675×1027×1.38×1023×300\lambda =\frac{6.6\times 10^{-34}}{\sqrt{3\times 1.675\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}

= 1.447 x 10-10 m

This wavelength is comparable to the inter-atomic spacing of the crystal. Hence a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

Question 11.33: An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Answer:

Electrons are accelerated by a voltage = 50 kV

Charge on an electron, e = 1.6 x 10-19 C

Mass of the electron, me = 9.11 x 10-31 kg

Wavelength of the yellow light = 5.9 x 10-7 m

The kinetic energy of the electron, E = eV

= (1.6 x 10-19) x (50 x 103)

= 8 x 10-15 J

De Broglie wavelength of electron is given as

λ=h2meE\lambda =\frac{h}{\sqrt{2m_{e}E}}

 

λ=6.6×10342×9.11×1031×8×1015\lambda =\frac{6.6\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 8\times 10^{-15}}}

= 5.467 x 10-12 m

The wavelength is 105 times lesser than the wavelength of the yellow light. The resolving power of the microscope and the wavelength of the light used is inversely proportional. Therefore, the resolving power of the electron microscope is 105 times greater than the optical microscope.

Question 11.34: The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10–15 m or less. This structure was first probed in the early 1970s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

Answer:

Wavelength of the proton or neutron, λ ≈ 10-15 m

Rest mass-energy of an electron :

m0c2 = 0.511 MeV

= 0.511 x 106 x 1.6 x 10-19

= 0.8176 x 10-13 J

Planck’s constant, h = 6.6 x 10-34 Js

Speed of light, c = 3 x 108 m/s

The momentum  of the proton or a neutron is given as

p = h/λ

=  6.6 x 10-34 /10-15

= 6.6 x 10-19 kg m/s

The relativistic relation for energy (E) is given as

E2 = p2c2 + m20C4

= (6.6 x 10-19 x 3 x 108)2 + (0.8176 x 10-13)2

= 392.04 x 10-22 + 0.6685 x 10-26

≈ 392.04 x 10-22

⇒ E = 19.8 x 10-11

= 19.8 x 10-11/1.6 x 10-19

= 12.375 x 108 eV

Order of energy of these electron beams is 12.375 x 108 eV

Question 11. 35: Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Answer:

Room temperature, T = 270 C = 27 + 273 = 300 K

Atmospheric pressure, P = 1 atm = 1.01 x 105 Pa

Atomic weight of He atom = 4

Avogadro’s number, NA = 6.023 x 1023

Boltzmann’s constant, k = 1.38 x 10-23 J/mol/K

Average energy of a gas at temperature T is given as

De Broglie wavelength is given as

λ=h2mE\lambda =\frac{h}{\sqrt{2mE}}

E = (3/2) kT

m = mass of the He atom

= Atomic weight/NA

= 4/(6.023 x 1023)

= 6.64 x 10-24 g

m = 6.64 x 10-27 kg

λ=h3mkT\lambda =\frac{h}{\sqrt{3mkT}}

 

λ=6.6×10343×6.64×1027×1.38×1023×300\lambda =\frac{6.6\times 10^{-34}}{\sqrt{3\times 6.64\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}

= 0.7268 x 10-10 m

We have the ideal gas formula

PV = RT

PV = kNT

V/N = kT/P

Here,

V is the volume of the gas

N is the number of moles of the gas

Mean separation between the two atoms of the gas is given as

r=[VN]1/3=[kTP]1/3r = \left [\frac{V}{N}\right ]^{1/3}=\left [ \frac{kT}{P} \right ]^{1/3}

 

r=[1.38×1023×3001.01×105]1/3r =\left [ \frac{1.38\times 10^{-23}\times 300}{1.01\times 10^{5}} \right ]^{1/3}

= 3.35 x 10-9 m

The mean separation between the atom is greater than the de Broglie wavelength.

Question 11. 36: Compute the typical de Broglie wavelength of an electron in metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This
indistinguishability has many fundamental implications which you will explore in more advanced Physics courses.]

Answer:

Temperature, T = 27°C = 27 + 273 = 300 K

Mean separation between two electrons, r = 2 × 10−10 m

De Broglie wavelength of an electron is

λ=h3mkT\lambda =\frac{h}{\sqrt{3mkT}}

h = Planck’s constant = 6.6 × 10−34 Js

m = Mass of an electron = 9.11 × 10−31 kg

k = Boltzmann constant = 1.38 × 10−23 J/mol/K.

λ=6.6×10343×9.11×1031×1.38×1023×300\lambda =\frac{6.6\times 10^{-34}}{\sqrt{3\times 9.11\times 10^{-31}\times 1.38\times 10^{-23}\times 300}}

≈ 6.2 x 10-9 m

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

Question 11.37: Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
E = h ν, p = h/λ
But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?

Answer:

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. This is because the nuclear forces grow stronger if they are pulled apart. Therefore, it seems that fractional charges may exist in nature. The observable charges are an integral multiple of electrical charges (e).

(b) The relation between electric field and the magnetic field,

eV = (1/2) mv2  and eBv = mv2/r

Here,

e = electric charge

v = velocity

V = potential

r = Radius

B = magnetic field

From these equations, it can be understood that the dynamics of an electron can be determined only by the ratio e/m and not by e and m separately.

(c)  At the atmospheric pressure, the ions in the gas does not have a chance of reaching their respective electrons due to collision and recombination with other molecules in the gas. At low pressures, ions have a chance to reach their respective electrons and results in the flow of current.

(d)  The minimum energy required for an electron in the conduction band to get out of the metal is called work function. These electrons occupy different energy levels, because of which for the same incident radiation, electrons come out with different energies.

(e) The absolute value of the energy of a particle is arbitrary within the addictive constant. Therefore, the wavelength(λ) is significant, but the frequency (ν) of the electron does not have direct physical significance. Therefore, product νλ has no physical significance.

 

NCERT Solutions for Class 12 Physics 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 for Class 12 is one of the best reference material to prepare for the Class 12 Physics exam. Students must prepare for this chapter well to score high marks in their examination.

Concepts involved in Chapter 11 Dual Nature of Radiation Matter

  1. Introduction
  2. Electron Emission
  3. Photoelectric Effect
    1. Hertz’s observations
    2. Hallwachs’ and Lenard’s observations
  4. Experimental Study Of Photoelectric Effect
    1. Effect of intensity of light on photocurrent
    2. Effect of potential on photoelectric current
    3. Effect of frequency of incident radiation on stopping potential
  5. Photoelectric Effect And Wave Theory Of Light Ex
  6. Einstein’s Photoelectric Equation: Energy Quantum Of Radiation
  7. Particle Nature Of Light: The Photon
  8. Wave Nature Of Matter

Some of the topics covered in this chapter’s NCERT Solutions 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 the 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.

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 in-depth to avoid difficulty in understanding higher concepts.

BYJU’S provides you with the complete NCERT Solutions for Class 12 to make you well equipped for the all-important CBSE Class 12 Board examination. BYJU’S also provides a downloadable PDF of chapter wise NCERT Solutions for Class 12 for subjects along with exemplar problems.

Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 11

What are the topics and subtopics covered under Chapter 11 of NCERT Solutions for Class 12 Physics?

The topics and subtopics covered under Chapter 11 of NCERT Solutions for Class 12 Physics are –
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

What is photoelectric effect in Chapter 11 of NCERT Solutions for Class 12 Physics?

The photoelectric effect occurs because the electrons at the surface of the metal tend to absorb energy from the incident light and use it to overcome the attractive forces that bind them to the metallic nuclei. This concept can be clearly understood by the students using the NCERT Solutions created by the faculty at BYJU’S. The solutions contain explanations provided in a comprehensive manner to boost the exam preparation of students.

Can students depend on the NCERT Solutions for Class 12 Physics Chapter 11 from BYJU’S?

The NCERT Solutions for Class 12 Physics Chapter 11 from BYJU’S are designed with the main aim of helping students to focus on the important concepts. Each and every minute detail is explained with utmost care to improve conceptual knowledge among students. The solutions also contain various shortcut techniques which can be used to remember the concepts effectively. Students can refer to the solutions while answering the textbook questions and understand the method of answering them without any difficulty.

 

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*