# Class 12-science H C VERMA Solutions Physics Chapter 20 - Photoelectric Effect and Wave-Particle Duality

## Photoelectric Effect and Wave-Particle Duality Exercise 365

### Solution 1

Given data:-

λ_{1} = 400 nm

λ_{2} = 780 nm

We know,

*h* = 6.63×10^{-}^{34}Js

*c *= 3×10^{8} m/s

Energy of photon is given by,

*E*=*hv*

*ν*=*c / *λ

∴ *E* =*hc** / *λ

Energy *E*_{1} of a photon of wavelength λ_{1}

*E*_{1}=*hc** / **λ*_{1}

=6.63×10^{-}^{34} ×3×10^{8} / 400×10^{-9}

=5×10^{-19} J

Energy *E*_{2 }of a photon of wavelength λ_{2}

*E*_{2 }=6.63×10^{-}^{34
}×3×10^{8} / 780×10^{-9}

=2.55×10^{-19} J

Thus, The range of energy is between 2.55 × 10^{-19} J and 5 × 10^{-19} J.

### Solution 2

Given data:

λ= 500 nm

*h = *6.63×10^{-}^{34} Js

Momentum of a photon of light is,

*p*=*h /*λ

=6.63×10^{-}^{34}
/ 500×10^{-9}

=1.33×10^{-27} kg-*m*/s* *

### Solution 3

Given data:

λ_{1}= 500 nm

λ_{2}= 700 nm

*c = *3×*108* m/s

*h* = 6.63×10^{-}^{34} Js

Energy of absorbed photon is-

*E*_{1}=*hc** /*λ_{1}

=hc / 500×10^{-9}

Energy of emitted photon,

*E*_{2}=*hc** /*λ_{2}

=hc / 700×10^{-9}

Energy absorbed by the atom in the process:

*E*_{1 }- *E*_{2 }= *hc*(1* /*λ_{1 }- 1/ λ_{2})

=6.63×3×10^{-19}(1/5 - 1/7)

=1.136×10-19 J

### Solution 4

Given data:

*P* = 10 W

λ = 590 nm

Energy consumed per second = 10 J

Energy converted into light = 60 %

∴ Energy converted into light = 60 / 100 ×10 =6 J

Energy needed to emit one photon from the sodium is:

*E*_{ }=*hc /*λ

=6.63×10^{-}^{34} ×3×10^{8} / 590×10^{-9}

=19.89/590 10^{-17} J

Number of photons emitted are:

*N *= 6 / E

=6 / (19.89/590 10^{-17})

*N* = 1.77 × 10^{19}

### Solution 5

Given Data:

Intensity of light is *I *=
1.4 × 10^{3} W/m^{2}

λ = 500 nm

d= 1.5×10^{11} m

Intensity is given as:

I=Power/ Area

=1.4×10^{3} W/m^{2}

Let *n* be
the number of photons emitted per second.

∴ P=nhc / λ,

Number of photons/m^{2} = nhc / λ×A

=nhcλ×1 = I

∴ n=I×λ / hc

=1.4×10^{3} ×500×10^{-9 }/ 6.63×10^{-34 }×3×10^{8}

=3.5×10^{21}

(b) Consider number of two parts at a distance *r* and *r* + *dr* from the source.

Let *dt*
be the small time interval

Total number of photons emitted in this time interval,

N=ndt

= (P λ / hc × A) dr / c

These points will be between two spherical shells of
radius *r* and *r* + *dr*.
It will be the distance of the 1st point from the sources.

No. of photons/m^{3}

=P λ / 4πr^{2}hc^{2}

=1.4×10^{3}×500×10^{-9 }/ 6.63×10^{-34 }×9×10^{16}

=1.2×10^{13}

(c) Number of photons emitted = (Number of photons / s-m^{2})× Area

=(3.5×10^{21}) ×4⊓l^{2}

=3.5×10^{21}×4× (3.14) × (1.5×10^{11})^{2}

=9.9×10^{44}

### Solution 6

Given Data:

λ=663×10^{-9 }m

θ=60°

Number of photons per second, *n*=1×10^{19}

Momentum of photon is

*p*=*h /*λ

Force exerted on the wall is

*F*=*n*× (*p*cosθ- (-*p*cosθ))

=2*np*cosθ

=2×1×10^{19 }×10^{-27 }×0.5

=1.0×10^{-8 }N

Force exerted by the light beam on the mirror is 1.0×10^{-8 }N

### Solution 7

Given Data:* *

*P *= 10 watt

We know that,

λ=*h
/ p*

⇒ *p*=*h / *λ* *

On dividing both sides by t,

we get:

p / t=h / λ t ….....(1)

Energy is given as

E=hc / λ

⇒ E / t=hc / λ t

Let *P* be the power. Then,

*P*=*E / t*

=hc / λ t* *

*P
*=*pc
/ t*

⇒ *P/c *=p/t* *

F=p/t

=P/c

F = 7/10(absorbed) + 2×3/10(reflected)

F=7/10× P/c + 2×3/10 × P/c

*F
*= 4.33×10^{-8} N

### Solution 8

Given that:

*m* = 20 g = 20 × 10^{-3} kg

The weight of the mirror will be balanced if the force exerted by the photons will be equal to the weight of the mirror.

Now,

p=h /* *λ

On dividing both sides by t,

P/t=h/* *λ t .....(1)

E = hc / λ

⇒ E/t = hc/ λ t

*P* be the power. Then,

P=Et=hc / λ t

P=pc/t

⇒ P/c = p/t

F=P/t = P/c

Thus, rate of change of momentum = Power/*c*

As the light gets reflected normally,

Force exerted = 2×(Rate of change of momentum) = 2 × Power/*c*

30%of (2×Power/c)=mg

⇒ Power = 20×10^{-3} ×10×3×10^{8} ×10 /2×3

=100 MW

### Solution 9

Given that,

*P*= 100 W

*R* = 20 cm = 0.2 m

It is given that 60% of the energy supplied to the bulb is converted to light.

Therefore, power of light emitted by the bulb, *P *= 60 W

F=P/c

F=60 / 3×10^{8}

=2×10^{-7} N

Pressure = Force / Area

=2×10^{-7} / 4×3.14× (0.2)^{2}

=4×10-7 N/m2

### Solution 10

Given that,

*r = *1 cm

*I* = 0.5 Wcm^{-2}

Let *A* be
the effective area of the sphere perpendicular to the light beam.

So, force exerted by the light beam on sphere is given by,

F=P/c=AI/c

*F *= ⊓× (1)^{2} ×0.5 / 3×10^{8}

=5.2×10^{-9 }N

### Solution 11

Consider az sphere of centre O and radius OP. As shown in the figure, the
radius OP of the sphere is making an angle θ with OZ. Let us rotate the radius
about OZ to get another circle on the sphere. The part of the sphere between
the circle is a ring of area 2⊓r^{2}sinθdθ.

Consider a small part of area ΔA of the ring at point P.

Energy of the light falling on this part in time Δt,

ΔU=IΔt(ΔAcosθ)

As the light is reflected by the sphere along PR, the change in momentum,

Δp=2cosθ

=IΔt(ΔAcos^{2}θ)

Therefore, the force will be

= IΔAcos^{2}θ

The component of force on ΔA, along ZO, is given by

cosθ = IΔAcos^{3}θ

Now, force action on the ring,

dF=I(2πr2sinθdθ) cos^{3}θ

The force on the entire sphere is given by integral of

F=

=(πr^{2}I/c)

### Solution 12

We can apply the principle of conservation of energy to this collision of electron with a photon.

So, using principle of conservation of energy:

pc + m_{e}c^{2} =√( p^{2} c^{2 }+ m_{e}^{2}c^{4}) …………(1)

Where,

*m*_{e }=
rest mass of electron

*pc* =
energy of the photon

Squaring on both side of equation (i),

(pc + m_{e}c^{2})^{2} = ( p^{2} c^{2 }+ m_{e}^{2}c^{4})

⇒pc + m_{e}c^{2} +2pc×m_{e}c^{2} = (p^{2} c^{2 }+ m_{e}^{2}c^{4})

⇒2pc×m_{e}c^{2} = 0 or pc=0 …..(since, m_{e }and_{}c are non-zero)

This is giving null energy of photon which is not possible.

### Solution 13

Given Data:-

*r* = 1 m

Electric potential energy is given by

*E _{1}*= kq

^{2}r = kq

^{2},

where *k *is 1/4⊓ɛ_{0}

Energy of photon is given by

*E*_{2} = hc / λ,

Given that, *E*_{1} = *E*_{2}

∴ kq^{2 }=hc / λ

⇒ λ =hc / kq^{2}

For λ to be maximum *q* should be minimum.

q = e =1.6×10^{-19} C

λ = hc / kq^{2}

= 6.63×3×10^{-34} ×10^{8} / 9×10^{2} × (1.6)^{2}×10^{-38}

= 863 m

Next smaller wave length,

λ =6.63×3×10^{-34} ×10^{8} / 9×10^{2} ×4× (1.6)^{2}×10^{-38}

=215.74 m

### Solution 14

Given Data :-

*λ** *= 350 nm

*ϕ* = 1.9 *e*V

From Einstein's photoelectric equation,

E= φ + Kinetic energy of electron

⇒K.E. =E - φ

Maximum kinetic energy of electrons,

E_{max} =hc/λ - ϕE_{max} =6.63×10^{-34}×3×10^{8} / 350×10^{-9}×1.6×10^{-19} - 1.9E_{max}

E_{max}_{ }=1.6 eV

### Solution 15

Given Data:-

*W _{0}* = 2.5 × 10

^{-19}J

*v = *6.0 × 10^{14} Hz

(a) Work function of a metal,

*W*_{0} = *hv*_{0},

∴ *v*_{0 }= *W*_{0}/h

⇒* v*_{0 }= 2.5×10^{-19} / 6.63×10^{-34}

=3.8×10^{14} Hz

(b) Einstein's photoelectric equation gives

e*v*_{0 }=hv -* W*_{0},

∴ *v*_{0}=(hv -* W*_{0})/e

= (6.63×10^{-34}×6×10^{14} - 2.5×10^{-19}) / 1.6×10^{-19}

=0.91 V

### Solution 16

Given Data:

φ = 4 eV = 4×1.6×10^{-19} J

V_{0} = 2.5 V

(a) Work function of a photoelectric material,

φ = hc/λ_{0}

∴ λ_{ 0 }= hc/φ

λ_{ 0}=6.63×10^{-34} ×3×10^{8} / 4×1.6×10^{-19}

λ_{ 0 }= 3.1×10^{-7} m

(b) From Einstein's photoelectric equation,

E=φ+eV_{0}

On substituting the respective values,

hc / λ = 4×1.6×10^{-19} +1.6×10^{-19}×2.5

⇒ λ =1.9125×10^{-7}

### Solution 17

Given Data:-

λ = 400 nm

φ = 2.5 eV

Using Einstein's photoelectric equation,

Kinetic energy=hc/ λ - φ

∴K.E.=(6.63×10^{-34}^{ }×3×10^{8}^{ }/4×10^{-7}^{ }×1.6×10^{-19}) - 2.5 eV

=0.605 eV

And K.E. =p^{2}/2m,

∴p^{2}=2m×K.E.

⇒ p^{2}=2×9.1×10^{-31}×0.605×1.6×10^{-19}

⇒p=4.197×10^{-25}^{ }kg-m/s

### Solution 18

Given Data:

λ = 400 nm

V_{0} = 1.1 V

Using Einstein's photoelectric equation,

hc/ λ =hc / λ_{ 0} + eV_{0},

Here, λ_{ 0} = threshold
wavelength

V_{0} = stopping potential

On substituting the respective values in the above formula, we get:

6.63×10^{-34}^{ }×3×10^{8}^{ }/400×10^{-9}^{ }= (6.63×10^{-34}^{ }×3×10^{8}^{ }/ λ_{0}) +1.6×10^{-19}^{ }×1.1

λ_{0} =6.196×10^{-7}^{ }m

Threshold
wavelength for the metal is 6.196×10^{-7}^{ }m

### Solution 19

(a)

When *λ* = 350, *V _{s}* = 1.45

and when *λ* = 400, *V _{s}* = 1

∴ hc/350=w +1.45 ...(1)

And hc /400= w +1 .....(2)

Subtracting (2) from (1), we get:

*h* = 4.2 × 10^{-15} eVs

(b) Now, work function,

w= (12240 / 350) -1.45

=2.15ev

(c) w= nc/* **λ** *

λ_{threshold}_{ }= hc/w

λ_{threshold}_{ }=576.8 nm

### Solution 20

Given Data:-

*E* = 1.2 × 10^{15} times per
second

*v* = 1.2×10^{15} / 2

=0.6×10^{15} Hz

φ = 2.0 eV

Using Einstein's photoelectric equation,

K=hv - φ_{0}

⇒ K=6.63×10^{-34 }×0.6×10^{15}* */1.6×10^{-19} - 2

=0.486 eV

### Solution 21

Given Data:

E= E_{0}sin[(1.57×10^{7 }m^{-1 })(x-ct)]

φ = 1.9 eV

Comparing the given equation with the standard equation,

E= E_{0}sin(kx-wt), we get:

ω=1.57×10^{7 }×c

For frequency,

v=1.57×10^{7 }×3×10^{8 } / 2⊓ Hz

Using Einstein's photoelectric equation,

e V_{0 }=hv - φ

Substituting the values, we get

e V_{0 }=6.63×10^{-34 }×(1.57×3×10^{15} / 2π) ×1.6×10^{-19 }- 1.9eV

⇒ V_{0}=1.205×1.6×10^{-19}/ 1.6×10^{-19}

=1.205 V

The value of the stopping potential is 1.205 V.

## Photoelectric Effect and Wave-Particle Duality Exercise 366

### Solution 22

Given Data:

E=(100 Vm^{-1})sin [(3.0×10^{15} s^{-1})t ] sin [(6.0 ×10^{15} s^{-1})t]

=100×12cos [(9×10^{15 }s^{-1})t]-cos[(3×10^{15 }s^{-1})t]

Angular frequency ω are 9 × 10^{15} and 3 × 10^{15} .

φ = 2 eV

Maximum frequency,

v=ω_{max} /2π = 9×10^{15 }/ 2π Hz

Using Einstein's photoelectric equation,

*K* = *hv* - φ

⇒*K = *6.63×10^{-34 }× (9×10^{15} / 2⊓) × (1/1.6×10^{-19 }) - 2 eV

⇒K = 3.938 eV

### Solution 23

Given Data:

Intensity I = 5 mW

Number of photons
emitted per second are n =
8 × 10^{15}

V_{0} = 2 V

E = hv = I/n = 5×10^{-3 }/ 8×10^{15}

Using Einstein's photoelectric equation

W_{0 }= hv - eV_{0}

Substituting the values, we get:

W_{0 }=5×10^{-3}/^{}8×10^{15 } - 1.6×10^{-19 }×2

=3.05×10^{-19 }J

=1.906 eV

### Solution 24

We have to take two cases.Case(I) :- When stopping potential,

V_{0}=1.656 volts

v=5×10^{14 }Hz

Case(II):- When stopping potential,

V_{0}=0

v=1×10^{14} Hz

Using Einstein's equation,

eV_{0}=hv - W_{0}

Using the values of above two cases,

we get:

1.656e = 5h×10^{14 }- W_{0} …....(1)

0 = 5h×10^{14 }- 5×W_{0} …...(2)

Subtracting equation(2) from (1), we get:

W_{0}=
1.656 eV / 4

=0.414 eV

Using the value
of W_{0} in
equation (2), we get:

⇒ 5W_{0 }= 5h×10^{14}

h/e= 4.414×10^{-15} Vs

### Solution 25

Given Data:-

W_{0} = 0.6 eV

Now, work function,

W_{0 }= hcλ,

∴λ=hc / W_{0}

=6.63×10^{-34 }×3×10^{8} / 0.6×1.6×10^{-19}

=2071 nm

The maximum wavelength of light that can be recorded by the film is 2071 nm.

### Solution 26

Given Data:-

*λ** *= 400 nm

*P *= 5 W

*E* = hc/ *λ*

=(1242 / 400) eV

and *no. of photons* = P/E

=5×400 / 1.6×10^{-}^{19}×1242

Number of electrons = 1 electron per 10^{6} photons

Number of photoelectrons emitted are:

n' = 5×400 / 1.6×1242×10^{-}^{19 }×10^{6}

*I* = n' × Charge on electron

I = ( 5×400 / 1.6×1242×10^{-}^{19}×10^{6 }) ×1.6×10^{-}^{19}

=1.6 μA

### Solution 27

Given Data:-

*r* = 4.8 cm

*λ* = 200 nm

*E* = 1.0 × 10^{-7} J

According to condition one photon out of every ten thousand is able to eject a photoelectron.

Energy of one photon,

E'=hc* / **λ*

E' =6.63×10^{-}^{34}×3×10^{8} / 2×10^{-}^{7}

=9.945×10^{-}^{19}

Number of photons,

n* *= E / E

= 1×10^{-}^{7} / 9.945×10^{-}^{19}

= 1×10^{11}

Number of photoelectrons = 1×10^{11} / 10^{4}= 1×10^{7}

The amount of positive charge developed due to the outgoing electrons is given by,

q = Number of photoelectrons × charge on a electron

=1×10^{7}×1.6×10^{-}^{19} =1.6×10^{-}^{12} C

Potential developed at the centre as well as on surface,

*V*=Kq / r, ……(*K* = 14⊓ε_{0})

∴ V* *= 9×10^{9}×1.6×10^{-}^{12} / 4.8×10^{-}^{2}

=0.3 V

So, 0.3 V is potential at the centre of the sphere.

### Solution 28

Given Data:-

d = 10 cm

φ* *= 2.39 eV

λ_{1} = 400 nm

λ_{2} = 600 nm

We know that,
Magnetic field *B* will
be minimum if energy is maximum.

And for maximum
energy, wavelength *λ* should
be minimum.

Using Einstein's photoelectric equation:

*E
*= *hc** /* λ* **- *φ

∴ *E*=1242 / 400
- 2.39

=0.715 eV

We know that, the beam of ejected electrons will gwt bent by the magnetic field. There will be no current, if the electrons do not reach the other plates.

When a charged particle is sent perpendicular to a magnetic field, it moves along a circle of radius,

r = mv / qB,

We wanted that no
current flows in the circuit. So, radius of the circle should be equal
to *r = d,*

Using *mv *= √(2*mE)*

⇒ *r*=√)

⇒0.1=√(2×9.1×10^{-31}×1.6×10^{-19}×0.715 / 1.6×10^{-19}×B)

⇒*B*=2.85×10^{-5} T

### Solution 29

Given Data:-

y = 1 mm × 2 = 2 mm

*W*_{0} = 2.2 eV

*D* = 1.2 m

*d* = 0.24 mm

*y*=λD / *d*

∴ λ=2×10^{-3}×0.24×10^{-3} / 1.2

= 4×10^{-7} m

E=hc / λ

= 4.14×10^{-15}×3×10^{8} / 4×10^{-7}

= 3.105 eV

Using Einstein's photoelectric equation, we have,

eV_{0}=E - W_{0}

∴ e*V*_{0 }= 3.105 -
2.2=0.905 eV

*V*_{0}=0.905 / 1.6×10^{-19}×1.6×10^{-19} V

= 0.905 V

The stopping
potential *V* needed to stop the
photocurrent is 0.905 V.

### Solution 30

Given Data:-

*φ*= 4.5 eV,

*λ** *= 200 nm

Using Einstein's photoelectric equation, we have,

K = E - φ* *

= hc /λ - φ

∴ *K*=1242 / 200 - 4.5

= 1.71 eV

At least 1.7 eV is required to stop the electron and thus minimum kinetic energy is 2 eV.

According to given condition that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy

= (2+1.7) eV

= 3.7 eV

### Solution 31

Given Data:-

σ = 1.0 × 10^{-9} Cm^{-2}

φ = 1.9 eV

*λ*= 400 nm

*d* = 20 cm

*V* = *E* × *d,*

∴ V=σ/ε_{0}×d

=1×10^{-9}×20 / 8.85×10^{-12}×100

=22.598 V=22.6 V

Using Einstein's photoelectric equation, we have:

e*V*_{o}=hv - W_{0}

=hc /λ - W

V_{0}=4.14×10^{-15}×3×10^{8} /4×10^{-7} - 1.9

=1.205 V

As *V*_{0} is much less than '*V*', the minimum energy required to reach
the charged plate must be equal to 22.7*e*V.

For maximum KE, 'V' must have an accelerating value.

Hence maximum kinetic energy,

K.E.=V_{0}+V

=1.205+22.6

=23.8005 eV

### Solution 32

We know that,

E = σ / ε_{0}

=1×10^{-}^{9} / 8.85×10^{-}^{12}

=113 V/m

Acceleration is given by

a=

*a *= 1.6×10^{-}^{19}×113 / 9.1×10^{-}^{31}

=19.87×10^{12}

*T *= √(2y/*a)*

=√(2×20×10^{-}^{2} / 19.87×10^{12})

=1.41×10^{-}^{7} s

In previous problem we got KE = 1.2eV

Using Einstein'*s* photoelectric equation, we have,

K.E.=*hc /*√ - W

=1.2 eV=1.2×1.6×10^{-}^{19} J

∴Velocity is *v*=√)

=√(2×1.2×1.6×10^{-}^{19} / 4.1×10^{-}^{31})

=0.665×10^{-}^{6}^{ }m/s

∴ Horizontal displacement,

*S *= v×t* *

=0.665×10^{-}^{6} ×1.4×10^{-}^{7}* *

=0.092 m

### Solution 33

Given Data:-

*φ** *= 1.9 eV

*λ** *= 250 nm

Energy of a photon,

*E*=*hc** / *λ

∴*E *= = 4.96 eV

Using Einstein's photoelectric equation, we have,

K=E - φ* *

⇒ *K*= *hc /**λ** *- φ

=4.96 eV - 1.9 eV

=3.06 eV.

For non-positive velocity of each photo electron, the velocity of a photoelectron should be equal to minimum velocity of the plate.∴ Velocity is given by,

*v*=√(2*K/m)*

∴ v= √(2×3.06×1.6×10^{-}^{19} / 9.1×10^{-}^{31})

=1.04×10^{6} ms^{-}^{1}

### Solution 34

Using Einstein's photoelectric equation,

e*V*_{0}=*hc** /*λ* *- φ

⇒V_{0}=(*h*c
/* *λ - φ)(1/e*)*

The particle will move in a circle when the stopping potential is equal to the potential due to the singly charged ion at that point so that the particle gets the required centripetal force for its circular motion.

⇒Ke / 2d=(hc /* *λ - φ)(1/e)

⇒hc / *λ* =Ke^{2} /2d +
φ = (Ke^{2}+2dφ) /2d

⇒* *λ =(hc)(2d) /(ke^{2}+2dφ)

⇒* *λ = 8⊓∊_{0}hcd /(e^{2}+8⊓∊_{0}dφ)

### Solution 35

Given Data:-

λ= 400 nm

ϕ = 2.2 eV

E=hc / λ

∴*E*= = 3.1 eV

This energy will be supplied to the electrons.

Energy lost is

= 3.1 eV × 10%

= 0.31 eV

Now, energy of the electron after the first collision = 3.1 - 0.31 = 2.79 eV

Energy lost in the second collision

= 2.79 eV× 10%

= 0.279 eV

Total energy lost in two collisions

= 0.31 + 0.279 = 0.589 eV

Using Einstein's photoelectric equation,

K = E - φ - energy lost in collisions

= (3.1 - 2.2 - 0.589) eV

= 0.31 eV

(b) Similarly for the third collision,

The energy lost = (2.79 - 0.279) eV × 10%

= 0.2511 eV

Energy after the third collision = 2.790 - 0.2511 = 2.5389

Energy lost in the fourth collision = 2.5389 × 10%

Energy after the fourth collision = 2.5389 - 0.25389 = 2.28501

After the fifth collision, the energy of the electron becomes less than the work function of the metal.

Therefore, the electron can suffer maximum four collisions before it becomes unable to come out of the metal.