#### Page No 407:

#### Question 11.1:

Find the

**(a)** maximum
frequency, and

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

#### Answer:

Potential
of the electrons, *V* = 30 kV = 3 ×
10^{4} V

Hence,
energy of the electrons, *E *=
3 ×
10^{4} eV

Where,

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

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

The energy of the electrons is given by the relation:

*E *= *h**ν*

Where,

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

Hence, the maximum frequency of X-rays produced is

**(b)**The
minimum wavelength produced by the X-rays is given as:

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

#### Question 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)** maximum
kinetic energy of the emitted electrons,

**(b) ** Stopping potential, and

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

#### Answer:

Work function of caesium metal,

Frequency of light,

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

Where,

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

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

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

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

**(c)**Maximum
speed of the emitted photoelectrons = *v*

Hence, the relation for kinetic energy can be written as:

Where,

*m* = Mass of an electron = 9.1 ×
10^{−31} kg

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

#### 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?

#### Answer:

Photoelectric
cut-off voltage, *V*_{0} = 1.5 V

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

Where,

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

Therefore,
the maximum kinetic energy of the
photoelectrons emitted in the given experiment is 2.4 ×
10^{−19} J.

#### 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 uniform cross-section which is
less than the target area), and

**(c) ** How fast
does a hydrogen atom have to travel in order to have the same
momentum as that of the photon?

#### Answer:

Wavelength
of the monochromatic light, *λ* = 632.8 nm = 632.8 ×
10^{−9} m

Power
emitted by the laser, *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
energy of each photon is given as:

The momentum of each photon is given as:

**(b)**Number
of photons arriving per second, at a target irradiated by the beam = *n*

Assume that the beam has a uniform cross-section that is less than the target area.

Hence, the equation for power can be written as:

**(c)**Momentum
of the hydrogen atom is the same as the momentum of the photon,

Momentum is given as:

Where,

*v* = Speed of the hydrogen atom

#### Question 11.5:

The
energy flux of sunlight reaching the surface of the earth is 1.388 ×
10^{3} W/m^{2}.
How many photons (nearly) per square metre are incident on the Earth
per second? Assume that the photons in the sunlight have an average
wavelength of 550 nm.

#### Answer:

Energy
flux of sunlight reaching the surface of earth, *Φ* = 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,

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

Hence, the equation for power can be written as:

Therefore, every second, photons are incident per square metre on earth.

#### Question 11.6:

In
an experiment on 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.

#### Answer:

The
slope of the cut-off voltage (*V*)
versus frequency (ν)
of an incident light is given as:

Where,

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

*h* = Planck’s constant

Therefore, the value of Planck’s constant is

#### Question 11.7:

A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

#### Answer:

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

Wavelength
of the emitted sodium light, *λ* = 589 nm = 589 ×
10^{−9} m

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

Speed
of light,* c *=
3 ×
10^{8} m/s

**(a)**The
energy per photon associated with the sodium light is given as:

**(b)**Number
of photons delivered to the sphere = *n*

The equation for power can be written as:

Therefore, every second, photons are delivered to the sphere.

#### Page No 408:

#### Question 11.8:

The
threshold frequency for a certain metal is 3.3 × 10^{14 }Hz. If light of frequency 8.2 ×
10^{14} Hz is incident on the metal, predict the cutoff voltage for the
photoelectric emission.

#### Answer:

Threshold frequency of the metal,

Frequency of light incident on the metal,

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 =

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

Therefore, the cut-off voltage for the photoelectric emission is

#### 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?

#### Answer:

No

Work function of the metal,

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

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

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

Speed
of light, *c* = 3 ×
10^{8} m/s

The energy of the incident photon is given as:

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 × 10^{14} Hz is incident on a metal surface. Electrons with a maximum speed of
6.0 × 10^{5} m/s are ejected from the surface. What is the threshold frequency for
photoemission of electrons?

#### Answer:

Frequency of the incident photon,

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

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

Mass
of an electron, *m* = 9.1 ×
10^{−31} kg

For
threshold frequency *ν*_{0},
the relation for kinetic energy is written as:

Therefore, the threshold frequency for the photoemission of electrons is

#### 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.

#### Answer:

Wavelength of light produced by the argon laser, λ = 488 nm

=
488 × 10^{−9} m

Stopping
potential of the photoelectrons, *V*_{0} = 0.38 V

1eV
= 1.6 × 10^{−19} J

∴ *V*_{0} =

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

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

Speed
of light, *c* = 3 × 10 m/s

From
Einstein’s photoelectric effect, we have the relation involving
the work function *Φ*_{0} of the material of the emitter as:

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

#### Answer:

Potential
difference, *V* = 56 V

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

The momentum of each accelerated electron is given as:

*p* = *mv*

= 9.1 × 10^{−31} × 4.44 × 10^{6}

= 4.04 × 10^{−24} kg m s^{−1}

Therefore,
the momentum of each electron is 4.04 × 10^{−24} kg m s^{−1}_{.}

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

_{ }

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 kinetic energy of 120 eV.

#### Answer:

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
the electron, we can write the relation for kinetic energy as:

Where,

*v* = Speed of the electron

Momentum
of the electron, *p* = *mv*

= 9.1
× 10^{−31} × 6.496 × 10^{6}

= 5.91
× 10^{−24} kg m s^{−1}

Therefore,
the momentum of the electron is 5.91 × 10^{−24} kg m s^{−1}_{.}

**(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:

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.

#### Answer:

Wavelength
of light of a sodium line, *λ* = 589 nm = 589 × 10^{−9} m

Mass
of an electron, *m*_{e}= 9.1 × 10^{−31} kg

Mass
of a neutron, *m*_{n}= 1.66 × 10^{−27} kg

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

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

We have the relation for de Broglie wavelength as:

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

Hence,
the kinetic energy of the electron is 6.9 × 10^{−25} J or 4.31 μeV.

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

Hence,
the kinetic energy of the neutron is 3.78 × 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?

#### Answer:

**(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 × 10^{−34} Js

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

**(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:

_{ }

**(c)**Mass
of the dust particle, *m* = 1 × 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:

#### 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 electron.

#### Answer:

Wavelength of an electron

= 1 × 10^{−9} m

Planck’s
constant, *h* = 6.63 × 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.

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

Where,

Speed
of light, *c* = 3 × 10^{8} m/s

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:

Where,

*m* = Mass of the electron = 9.1 × 10^{−31} kg

*p *= 6.63 ×
10^{−25} kg m s^{−1}

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 × 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.

#### Answer:

**(a)** De
Broglie wavelength of the neutron, *λ* = 1.40 × 10^{−10} m

Mass
of a neutron, *m*_{n}_{ }= 1.66 × 10^{−27} kg

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

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

… (1)

De
Broglie wavelength (*λ*)
and velocity (*v*)
are related as:

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

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

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

Boltzmann
constant, *k* = 1.38 × 10^{−23} kg m^{2} s^{−2} K^{−1}

Average kinetic energy of the neutron:

The relation for the de Broglie wavelength is given as:

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).

#### Answer:

The
momentum of a photon having energy (*h**ν*)
is given as:

Where,

λ = Wavelength of the electromagnetic radiation

*c* = Speed of light

*h* = Planck’s constant

De Broglie wavelength of the photon is given as:

Where,

*m* = Mass of the photon

*v* = Velocity of the photon

Hence,
it can be inferred 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)

#### Answer:

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

∴*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^{−1}

We have the expression that relates mean kinetic energy of the nitrogen molecule with the root mean square speed as:

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

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

#### Page No 409:

#### 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 × 10^{11} C kg^{−1}.

**(b) ** Use the same
formula you employ in (a) to obtain electron speed for an collector
potential of 10 MV. Do you see what is wrong? In what way is the
formula to be modified?

#### Answer:

**(a)**Potential
difference across the evacuated tube, *V* = 500 V

Specific
charge of an electron, *e/m* = 1.76 ×
10^{11} C
kg^{−1 }

The speed of each emitted electron is given by the relation for kinetic energy as:

Therefore, the speed of each emitted electron is

**(b)**Potential
of the anode, *V* = 10 MV = 10 ×
10^{6} V

The speed of each electron is given as:

This
result is wrong because nothing can move faster than light. In the
above formula, the expression (*mv*^{2}/2)
for energy can only be used in the non-relativistic limit, i.e., for *v* << *c*.

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:

*E* = *mc*^{2}

Where,

*m *= Relativistic mass

*m*_{0} = Mass of the particle at rest

Kinetic energy is given as:

*K* = *mc*^{2} − *m*_{0}*c*^{2}

#### Question 11.21:

**(a)** A
monoenergetic electron beam with electron speed of 5.20 × 10^{6} 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 × 10^{11} C kg^{−1}.

**(b)** Is the
formula you employ in (a) valid for calculating 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:

**(a)**Speed
of an electron, *v* = 5.20 ×
10^{6} m/s

Magnetic
field experienced by the electron, *B* = 1.30 ×
10^{−4} T

Specific
charge of an electron, *e/m* = 1.76 ×
10^{11} C
kg^{−1}

Where,

*e* = Charge on the electron = 1.6 ×
10^{−19} C

*m* = Mass of the electron = 9.1 ×
10^{−31} kg^{−1}

The force exerted on the electron is given as:

θ = Angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam.

The
beam traces a circular path of radius, *r*.
It is the magnetic field, due to its bending nature, that provides
the centripetal force for the beam.

Hence, equation (1) reduces to:

Therefore, the radius of the circular path is 22.7 cm.

**(b) **Energy
of the electron beam, *E *=
20 MeV

The energy of the electron is given as:

This
result is incorrect because nothing can move faster than light. In
the above formula, the expression (*mv*^{2}/2)
for energy can only be used in the non-relativistic limit, i.e., for *v* << *c*

When very high speeds are concerned, the relativistic domain comes into consideration.

In the relativistic domain, mass is given as:

Where,

= Mass of the particle at rest

Hence, the radius of the circular path is given as:

#### 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 an anode, *V *=
100 V

Magnetic
field experienced by the electrons,* B* = 2.83 ×
10^{−4} T

Radius
of the circular orbit *r* = 12.0 cm = 12.0 × 10^{−2} m

Mass
of each electron = *m*

Charge
on each electron = *e*

Velocity
of each electron = *v*

The energy of each electron is equal to its kinetic energy, i.e.,

It is the magnetic field, due to its bending nature, that provides the centripetal force for the beam. Hence, we can write:

Centripetal force = Magnetic force

Putting
the value of *v* in equation (1), we get:

Therefore,
the specific charge ratio (*e/m*)
is

#### 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 an X-ray tube,

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

Speed
of light, *c *=
3 ×
10^{8} m/s

The maximum energy of a photon is given as:

Therefore, the maximum energy of an X-ray photon is 27.6 keV.

**(b) **Accelerating
voltage provides energy to the electrons for producing X-rays. To get
an X-ray of 27.6 keV, the incident electrons must possess at least
27.6 keV of kinetic electric energy. Hence, an 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 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 = 10^{9} eV)

#### Answer:

Total
energy of two *γ*-rays:

*E* = 10. 2 BeV

=
10.2 × 10^{9} eV

=
10.2 ×
10^{9} ×
1.6 ×
10^{−10} J

Hence,
the energy of each *γ*-ray:

Planck’s constant,

Speed of light,

Energy is related to wavelength as:

Therefore,
the wavelength associated with each *γ*-ray
is

#### 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 radiowaves 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 cm^{2},
and the average frequency of white light to be about 6 × 10^{14} Hz.

#### Answer:

**(a)** Power
of the medium wave transmitter, *P* = 10 kW = 10^{4 }W
= 10^{4} J/s

Hence,
energy emitted by the transmitter per second, *E* = 10^{4}

Wavelength
of the radio wave, *λ* = 500 m

The energy of the wave is given as:

Where,

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

*c* = Speed of light = 3 × 10^{8} m/s

Let *n* be the
number of photons emitted by the transmitter.

∴*nE*_{1} = *E*

The
energy (*E*_{1})
of a radio photon is very less, but the number of photons (*n*)
emitted per second in a radio wave is very large.

The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.

**(b)** Intensity
of light perceived by the human eye, *I* = 10^{−10} W m^{−2}

Area
of a pupil, *A* = 0.4 cm^{2 }=
0.4 × 10^{−4} m^{2}

Frequency
of white light, *ν*=
6 × 10^{14} Hz

The energy emitted by a photon is given as:

*E* = *h*ν

Where,

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

∴*E* = 6.6 × 10^{−34} × 6 × 10^{14}

= 3.96 × 10^{−19} J

Let *n* be the
total number of photons falling per second, per unit area of the
pupil.

The
total energy per unit for *n* falling photons is given as:

*E* = *n* ×
3.96 × 10^{−19} J s^{−1} m^{−2}

The energy per unit area per second is the intensity of light.

∴*E
= I*

*n* × 3.96 × 10^{−19} = 10^{−10}

= 2.52 × 10^{8} m^{2} s^{−1}

The total number of photons entering the pupil per second is given as:

*n*_{A }= *n
× A*

= 2.52 × 10^{8 }× 0.4 × 10^{−4}

= 1.008 × 10^{4} s^{−1}

This
number is not as large as the one found in problem **(a)**,
but it is large enough for the human eye to never see the individual
photons.

#### Page No 410:

#### 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 a high intensity (∼10^{5} 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, *V*_{0} = 1.3 V

Planck’s
constant, *h* = 6.6 × 10^{−34} J

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

Work function of the metal =

Frequency
of light = *ν*

We have the photo-energy relation from the photoelectric effect as:

= *h**ν* − *eV*_{0}

Let *ν*_{0} be the threshold frequency of the metal.

∴= *h**ν*_{0}

Wavelength
of red light, = 6328 × 10^{−10} m

∴Frequency of red light,

Since *ν*_{0}> *ν*_{r},
the photocell will not respond to the red light produced by the
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, *V*_{0} = 0.54 V

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

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

Let be the work function and *ν* be the frequency of emitted light.

We have the photo-energy relation from the photoelectric effect as:

*eV*_{0} = *h**ν* −

Wavelength
of the radiation emitted from an iron source, *λ*‘
= 427.2 nm

*= *427.2 × 10^{−9} m

Let be the new stopping potential. Hence, photo-energy is given as:

Hence, the new stopping potential is 1.50 eV.

#### Question 11.28:

A mercury lamp is a convenient source for studying 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:

*λ*_{1} = 3650 Å, *λ*_{2}=
4047 Å, *λ*_{3}=
4358 Å, *λ*_{4}=
5461 Å, *λ*_{5}=
6907 Å,

The stopping voltages, respectively, were measured to be:

*V*_{01} = 1.28 V, *V*_{02} = 0.95 V, *V*_{03} = 0.74 V, *V*_{04} = 0.16 V, *V*_{05} = 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:

Einstein’s photoelectric equation is given as:

*eV*_{0} = *h**ν*−

Where,

*V*_{0} = Stopping potential

*h* = Planck’s constant

*e* = Charge on an electron

ν = Frequency of radiation

= Work function of a material

It
can be concluded from equation (1) that potential *V*_{0} is directly proportional to frequency *ν*.

Frequency is also given by the relation:

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

The given quantities can be listed in tabular form as:

**Frequency × 10**^{14}**Hz**8.219

7.412

6.884

5.493

4.343

**Stopping potential****V**_{0}1.28

0.95

0.74

0.16

0

The
following figure shows a graph between *ν*and *V*_{0.}

It
can be observed that the obtained curve is a straight line. It
intersects the *ν*-axis
at 5 × 10^{14} Hz, which is the threshold frequency (*ν*_{0})
of the material. Point D corresponds to a frequency less than the
threshold frequency. Hence, there is no photoelectric emission for
the *λ*_{5} line, and therefore, no stopping voltage is required to stop the
current.

Slope of the straight line =

From equation (1), the slope can be written as:

The work function of the metal is given as:

= *h*ν_{0}

=
6.573 × 10^{−34} × 5 × 10^{14}

=
3.286 × 10^{−19} J

= 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 a 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:

Mo and Ni will not show photoelectric emission in both cases

Wavelength
for a radiation, *λ* = 3300 Å = 3300 × 10^{−10} m

Speed
of light, *c* = 3 × 10^{8} m/s

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

The energy of incident radiation is given as:

It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted 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 cm^{2}.
Assuming that the top 5 layers of sodium absorb the incident energy,
estimate 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 incident light, *I* = 10^{−5} W m^{−2}

Surface
area of a sodium photocell, *A* = 2 cm^{2} = 2 × 10^{−4} m^{2}

Incident
power of the light, *P = I × A*

= 10^{−5} × 2 × 10^{−4}

= 2 × 10^{−9} W

Work function of the metal, = 2 eV

= 2 × 1.6 ×
10^{−19}

= 3.2 × 10^{−19} J

Number
of layers of sodium that absorbs the incident energy, *n* = 5

We
know that the effective atomic area of a sodium atom, *A*_{e} is 10^{−20} m^{2}_{.}

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

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

Time required for photoelectric emission:

The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture 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 inter-atomic
spacing in the lattice) (*m*_{e}= 9.11 × 10^{−31} kg).

#### Answer:

An X-ray probe has a greater energy than an electron probe for the same wavelength.

Wavelength
of light emitted from the probe, *λ* = 1 Å = 10^{−10} m

Mass
of an electron, *m*_{e} = 9.11 × 10^{−31} kg

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

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

The kinetic energy of the electron is given as:

Where,

*v* = Velocity of the electron

*m*_{e}*v* = Momentum (*p*)
of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

Energy of a photon,

Hence, a photon has a greater energy than an electron for the same wavelength.

#### Question 11.32:

**(a)** Obtain the
de Broglie wavelength of a neutron of kinetic energy 150 eV. As you
have seen in Exercise 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. (*m*_{n}= 1.675 × 10^{−27} kg)

**(b)** Obtain the
de Broglie wavelength associated with thermal neutrons at room
temperature (27 ºC). 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)** De Broglie
wavelength =;
neutron is not suitable for the diffraction experiment

Kinetic
energy of the neutron, *K* = 150 eV

= 150 × 1.6 × 10^{−19}

= 2.4 × 10^{−17} J

Mass
of a neutron, *m*_{n} = 1.675 × 10^{−27} kg

The kinetic energy of the neutron is given by the relation:

Where,

*v *= Velocity of the neutron

*m*_{n}*v* = Momentum of the neutron

De-Broglie wavelength of the neutron is given as:

It
is given in the previous problem that the inter-atomic spacing of a
crystal is about 1 Å, i.e., 10^{−10} m. Hence, the inter-atomic spacing is about a hundred times greater.
Hence, a neutron beam of energy
150 eV is not suitable for
diffraction experiments.

**(b) **De
Broglie wavelength =

Room
temperature, *T* = 27°C = 27 + 273 = 300 K

The average kinetic energy of the neutron is given as:

Where,

*k* = Boltzmann constant = 1.38 × 10^{−23} J mol^{−1} K^{−1}

The wavelength of the neutron is given as:

This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high-energy neutron beam should first be thermalised, before using it for diffraction.

#### Page No 411:

#### 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, *V* = 50 kV = 50 × 10^{3} V

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

Mass
of an electron, *m*_{e} = 9.11 × 10^{−31} kg

Wavelength
of yellow light = 5.9 × 10^{−7} m

The kinetic energy of the electron is given as:

*E* = *eV*

=
1.6 × 10^{−19} × 50 × 10^{3}

=
8 × 10^{−15} J

De Broglie wavelength is given by the relation:

This
wavelength is nearly 10^{5 }times
less than the wavelength of yellow light.

The
resolving power of a microscope is inversely proportional to the
wavelength of light used. Thus, the resolving power of an electron
microscope is nearly 10^{5} times that of an 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 early 1970’s
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 a proton or a neutron, *λ* ≈
10^{−15} m

Rest mass energy of an electron:

*m*_{0}*c*^{2} = 0.511 MeV

=
0.511 × 10^{6} × 1.6 × 10^{−19}

=
0.8176 × 10^{−13 }J

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

Speed
of light, *c* = 3 × 10^{8} m/s

The momentum of a proton or a neutron is given as:

The
relativistic relation for energy (*E*)
is given as:

Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.

#### Question 11.35:

Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

#### Answer:

De Broglie wavelength associated with He atom =

Room
temperature, *T* = 27°C = 27 + 273 = 300 K

Atmospheric
pressure, *P* = 1 atm = 1.01 × 10^{5} Pa

Atomic weight of a He atom = 4

Avogadro’s
number, N_{A} = 6.023 × 10^{23 }

Boltzmann
constant, *k* = 1.38 × 10^{−23} J mol^{−1} K^{−1}

Average
energy of a gas at temperature *T*,is given as:

De Broglie wavelength is given by the relation:

Where,

*m* = Mass of a He atom

We have the ideal gas formula:

*PV = RT*

*PV =
kNT*

Where,

*V* = Volume of the gas

*N* = Number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

#### Question 11.36:

Compute
the typical de Broglie wavelength of an electron in a metal at 27 ºC
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
indistinguishibility 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 given as:

Where,

*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^{−1} K^{−1}

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 *=

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 carry fractional charges. This is because
nuclear force increases extremely if they are pulled apart.
Therefore, fractional charges may exist in nature; observable charges
are still the integral multiple of an electrical charge.

**(b)** The
basic relations for electric field and magnetic field are

.

These
relations include *e* (electric charge), *v* (velocity), *m* (mass), *V* (potential), *r* (radius), and *B* (magnetic field). These relations give the value of velocity of an
electron as and

It
can be observed from these relations that the dynamics of an electron
is determined not by *e* and *m* separately, but by the ratio e/*m*.

**(c)** At
atmospheric pressure, the ions of gases have no chance of reaching
their respective electrons because of collision and recombination
with other gas molecules. Hence, gases are insulators at atmospheric
pressure. At low pressures, ions have a chance of reaching their
respective electrodes and constitute a current. Hence, they conduct
electricity at these pressures.

**(d)** The work
function of a metal is the minimum energy required for a conduction
electron to get out of the metal surface. All the electrons in an
atom do not have the same energy level. When a ray having some photon
energy is incident on a metal surface, the electrons come out from
different levels with different energies. Hence, these emitted
electrons show different energy distributions.

**(e)** The absolute
value of energy of a particle is arbitrary within the additive
constant. Hence, wavelength (*λ*)
is significant, but the frequency (*ν*)
associated with an electron has no direct physical significance.

Therefore,
the product *νλ*(phase speed)has no physical significance.

Group speed is given as:

This quantity has a physical meaning.

**NCERT Solutions for Class 12 Physics Chapters**

**Physics Part I**

- Chapter 1 – Electric Charges And Fields
- Chapter 2 – Electrostatic Potential And Capacitance
- Chapter 3 – Current Electricity
- Chapter 4 – Moving Charges And Magnetism
- Chapter 5 – Magnetism And Matter
- Chapter 6 – Electromagnetic Induction
- Chapter 7 – Alternating Current
- Chapter 8 – Electromagnetic Waves

**Physics Part II**

- Chapter 1 – Ray Optics And Optical Instruments
- Chapter 2 – Wave Optics
- Chapter 3 – Dual Nature Of Radiation And Matter
- Chapter 4 – Atoms
- Chapter 5 – Nuclei
- Chapter 6 – Semiconductor Electronics: Materials Devices and Simple Circuits
- Chapter 7 – Communication Systems

**NCERT Solutions for Class 12-science:**