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

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Chapter 20 - Photoelectric Effect and Wave-Particle Duality Exercise 365

Question 1

Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.

Solution 1

Given data:-

λ1 = 400 nm

λ2 = 780 nm

We know,

h = 6.63×10-34Js

c = 3×108 m/s

Energy of photon is given by,

E=hv 

ν=c / λ 

 E =hc / λ 

Energy E1 of a photon of wavelength λ1 

E1=hc / λ1 

=6.63×10-34 ×3×108 / 400×10-9 

=5×10-19 J 

Energy E2 of a photon of wavelength λ2 

E2 =6.63×10-34 ×3×108 / 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. 

Question 2

Calculate the momentum of a photon of light of wavelength 500 nm. 

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 

Question 3

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process. 

Solution 3

Given data:

λ1= 500 nm

λ2= 700 nm

c = 3×108 m/s

h = 6.63×10-34 Js

Energy of absorbed photon is-

E1=hc /λ1 

=hc / 500×10-9 

Energy of emitted photon,

E2=hc /λ2 

=hc / 700×10-9 

Energy absorbed by the atom in the process:

E1 - E2 = hc(1 /λ1 - 1/ λ2) 

=6.63×3×10-19(1/5 - 1/7) 

=1.136×10-19 J 

Question 4

Calculate the number of photons emitted per second by a 10 W sodium vapour lamp. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590 nm 

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 ×3×108 / 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 × 1019

Question 5

When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W m-2 of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5 × 1011 m.

 

(a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun.

(b) How many photons are there in each cubic metre near the earth's surface at any instant?

(c) How many photons does the sun emit per second?

Solution 5

Given Data:

Intensity of light is I = 1.4 × 103 W/m2

λ = 500 nm

d= 1.5×1011 m

Intensity is given as:

I=Power/ Area

 =1.4×103 W/m2 

Let n be the number of photons emitted per second.

P=nhc / λ,

Number of photons/m2 = nhc / λ×A 

=nhcλ×1 = I

 n=I×λ / hc 

 =1.4×103 ×500×10-9 / 6.63×10-34 ×3×108 

 =3.5×1021

(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/m3 

=P λ / 4πr2hc2 

=1.4×103×500×10-9 / 6.63×10-34 ×9×1016

=1.2×1013 

(c) Number of photons emitted = (Number of photons / s-m2)× Area

=(3.5×1021) ×4l2

=3.5×1021×4× (3.14) × (1.5×1011)2 

=9.9×1044 

Question 6

A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 × 1019. Calculate the force exerted by the light beam on the mirror. 

Solution 6

Given Data:

λ=663×10-9 m

θ=60° 

Number of photons per second, n=1×1019 

Momentum of photon is

p=h /λ 

Force exerted on the wall is

F=n× (pcosθ- (-pcosθ))

=2npcosθ 

=2×1×1019 ×10-27 ×0.5 

=1.0×10-8 N

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

Question 7

A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.

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

Question 8

A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light, as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror.

Figure 42-E1

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

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 ×10×3×108 ×10 /2×3 

=100 MW  

Question 9

A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber. 

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 / 3×108  

=2×10-7 N 

Pressure = Force / Area 

=2×10-7 / 4×3.14× (0.2)2  

=4×10-7 N/m2 

Question 10

A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of large aperture. The intensity of the light is 0.5 W cm-2. If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere. 

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 / 3×108 

=5.2×10-9 N 

Question 11

Consider the situation described in the previous problem. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing. 

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 2r2sinθdθ.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

 

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=2H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Dualitycosθ 

=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle DualityIΔt(ΔAcos2θ) 

Therefore, the force will be

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality = H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle DualityIΔAcos2θ 

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

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Dualitycosθ = H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle DualityIΔAcos3θ 

Now, force action on the ring,

dF=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle DualityI(2πr2sinθdθ) cos3θ 

The force on the entire sphere is given by integral of

F=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

=(πr2I/c) 

Question 12

Show that it is not possible for a photon to be completely absorbed by a free electron. 

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 + mec2 =( p2 c2 + me2c4) …………(1) 

Where,

me = rest mass of electron

pc = energy of the photon

Squaring on both side of equation (i),

(pc + mec2)2 = ( p2 c2 + me2c4)

pc + mec2 +2pc×mec2 = (p2 c2 + me2c4)

2pc×mec2 = 0 or pc=0  …..(since, me andc are non-zero) 

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

Question 13

Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible. 

Solution 13

Given Data:-

r = 1 m

Electric potential energy is given by

E1= kq2r = kq2,

where k is 1/4⊓ɛ0 

Energy of photon is given by

E2 = hc / λ,

Given that, E1 = E2

 kq2 =hc / λ 

 λ =hc / kq2 

For λ to be maximum q should be minimum.

q = e =1.6×10-19 C 

λ = hc / kq2 

= 6.63×3×10-34 ×108  / 9×102 × (1.6)2×10-38 

= 863 m 

Next smaller wave length,

λ =6.63×3×10-34 ×108 / 9×102 ×4× (1.6)2×10-38 

=215.74 m 

Question 14

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 eV 

Solution 14

Given Data :-

λ = 350 nm

ϕ = 1.9 eV

From Einstein's photoelectric equation,

E= φ + Kinetic energy of electron

K.E. =E - φ 

Maximum kinetic energy of electrons,

Emax =hc/λ - ϕEmax =6.63×10-34×3×108  / 350×10-9×1.6×10-19 - 1.9Emax 

Emax =1.6 eV 

Question 15

The work function of a metal is 2.5 × 10-19 J.

 

(a) Find the threshold frequency for photoelectric emission.

(b) If the metal is exposed to a light beam of frequency 6.0 × 1014 Hz, what will be the stopping potential? 

Solution 15

Given Data:-

W0 = 2.5 × 10-19 J

v = 6.0 × 1014 Hz

(a) Work function of a metal,

W0 = hv0,

v0 = W0/h 

v0 = 2.5×10-19 / 6.63×10-34

=3.8×1014 Hz 

(b) Einstein's photoelectric equation gives

 ev0 =hv - W0,

 v0=(hv - W0)/e 

= (6.63×10-34×6×1014 - 2.5×10-19) / 1.6×10-19  

=0.91 V 

Question 16

The work function of a photoelectric material is 4.0 eV.

 

(a) What is the threshold wavelength?

(b) Find the wavelength of light for which the stopping potential is 2.5 V.

Solution 16

Given Data:

φ = 4 eV = 4×1.6×10-19 J

V0 = 2.5 V

 

(a) Work function of a photoelectric material,

φ = hc/λ0 

λ 0 = hc/φ 

λ 0=6.63×10-34 ×3×108 / 4×1.6×10-19

λ 0 = 3.1×10-7 m

(b) From Einstein's photoelectric equation,

E=φ+eV0 

On substituting the respective values,

hc / λ = 4×1.6×10-19 +1.6×10-19×2.5

λ =1.9125×10-7

Question 17

Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 eV. 

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 ×3×108 /4×10-7 ×1.6×10-19) - 2.5 eV 

=0.605 eV 

And K.E. =p2/2m,

p2=2m×K.E.

p2=2×9.1×10-31×0.605×1.6×10-19 

p=4.197×10-25 kg-m/s 

Question 18

When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal. 

Solution 18

Given Data:

λ = 400 nm

V0 = 1.1 V

Using Einstein's photoelectric equation,

hc/ λ =hc / λ 0 + eV0,

Here, λ 0 = threshold wavelength

V0 = stopping potential

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

6.63×10-34 ×3×108 /400×10-9 = (6.63×10-34 ×3×108 / λ0) +1.6×10-19 ×1.1

λ0 =6.196×10-7 m

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

Question 19

In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:

 

Wavelength (nm):

350

400

450

500

550

Stopping potential (V):

1.45

1.00

0.66

0.38

0.16

 

Plot the stopping potential against inverse of wavelength (1/λ) on a graph paper and find

 

(a) Planck's constant

(b) The work function of the emitter and

(c) The threshold wavelength.

Solution 19

(a)

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

 

When λ = 350, Vs = 1.45

and when λ = 400, Vs = 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  

λthreshold = hc/w

λthreshold =576.8 nm 

Question 20

The electric field associated with a monochromatic beam is 1.2 × 1015 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV. 

Solution 20

Given Data:-

E = 1.2 × 1015 times per second

v = 1.2×1015 / 2

=0.6×1015 Hz 

φ = 2.0 eV

Using Einstein's photoelectric equation,

 K=hv - φ0 

 K=6.63×10-34 ×0.6×1015 /1.6×10-19 - 2

=0.486 eV 

Question 21

The electric field associated with a light wave is given byE=E0 sin (1.57×107 m-1)(x-ct). Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV. 

Solution 21

Given Data:

E= E0sin[(1.57×107 m-1 )(x-ct)] 

φ = 1.9 eV

Comparing the given equation with the standard equation,

E= E0sin(kx-wt), we get:

ω=1.57×107 ×c 

For frequency,

 v=1.57×107 ×3×108  / 2 Hz 

Using Einstein's photoelectric equation,

 e V0 =hv - φ 

Substituting the values, we get

 e V0 =6.63×10-34 ×(1.57×3×1015 / 2π) ×1.6×10-19 - 1.9eV

V0=1.205×1.6×10-19/ 1.6×10-19 

=1.205 V 

The value of the stopping potential is 1.205 V.

Chapter 20 - Photoelectric Effect and Wave-Particle Duality Exercise 366

Question 22

The electric field at a point associated with a light wave isE= (100 Vm-1­­­­­­­­) sin [(3.0×1015 s-1)t] sin [(6.0 ×1015 s-1)t ]. If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons? 

Solution 22

Given Data:

E=(100 Vm-1­­­­­­­­)sin [(3.0×1015 s-1)t ] sin [(6.0 ×1015 s-1)t]

=100×12cos [(9×1015 s-1)t]-cos[(3×1015 s-1)t] 

Angular frequency ω are 9 × 1015 and 3 × 1015 .

φ = 2 eV

Maximum frequency,

v=ωmax /2π = 9×1015 / 2π Hz

Using Einstein's photoelectric equation,

K = hv - φ 

K = 6.63×10-34 × (9×1015 / 2) × (1/1.6×10-19 ) - 2 eV

K = 3.938 eV 

Question 23

A monochromatic light source of intensity 5 mW emits 8 × 1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal. 

Solution 23

Given Data:

Intensity I = 5 mW

Number of photons emitted per second are n = 8 × 1015

V0 = 2 V

E = hv = I/n = 5×10-3 / 8×1015 

Using Einstein's photoelectric equation

W­0 = hv - eV0

Substituting the values, we get:

W­0 =5×10-3/8×1015  - 1.6×10-19 ×2

=3.05×10-19 J 

=1.906 eV

Question 24

The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find

(a) The ratio h/e and

(b) The work function.

Figure 42-E2

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

Solution 24

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

V0=1.656 volts

v=5×1014 Hz

Case(II):- When stopping potential,

V0=0

v=1×1014 Hz

Using Einstein's equation,

eV0=hv - W0

Using the values of above two cases,

we get:

1.656e = 5h×1014 - W0  …....(1)

0 = 5h×1014 - 5×W0  …...(2)

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

W0= 1.656 eV / 4

=0.414 eV

Using the value of W0 in equation (2), we get:

5W0 = 5h×1014

h/e= 4.414×10-15 Vs

Question 25

A photographic film is coated with a silver bromide layer. When light falls on this film, silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film. 

Solution 25

Given Data:-

W0 = 0.6 eV

Now, work function,

W0 = hcλ,

λ=hc / W0

=6.63×10-34 ×3×108 / 0.6×1.6×10-19 

=2071 nm

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

Question 26

In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit.

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 106 photons

Number of photoelectrons emitted are:

n' = 5×400 / 1.6×1242×10-19 ×106 

I = n' × Charge on electron

I = ( 5×400 / 1.6×1242×10-19×106 ) ×1.6×10-19 

=1.6 μA 

Question 27

A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which light energy of 1.0 × 10-7 J falls on the surface. Assuming that on average, one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball, assuming zero potential at infinity. What is the potential at the centre of the ball? 

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' =6.63×10-34×3×108 / 2×10-7 

=9.945×10-19 

Number of photons,

n = E / E 

= 1×10-7 / 9.945×10-19

= 1×1011 

Number of photoelectrons = 1×1011 / 104= 1×107 

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

q = Number of photoelectrons × charge on a electron 

=1×107×1.6×10-19 =1.6×10-12 C 

Potential developed at the centre as well as on surface,

V=Kq / r, ……(K = 14ε0) 

 V = 9×109×1.6×10-12 / 4.8×10-2 

=0.3 V

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

Question 28

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.

Figure 42-E3

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

Solution 28

Given Data:-

d = 10 cm

φ = 2.39 eV

λ1 = 400 nm

λ2 = 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 = (2mE)  

r=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality)

0.1=(2×9.1×10-31×1.6×10-19×0.715 / 1.6×10-19×B)

B=2.85×10-5 T

Question 29

In the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.

Figure 42-E4

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

Solution 29

Given Data:-

y = 1 mm × 2 = 2 mm

W0 = 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 

= 4.14×10-15×3×108 / 4×10-7

= 3.105 eV

Using Einstein's photoelectric equation, we have,

eV0=E - W0

eV0 = 3.105 - 2.2=0.905 eV

V0=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.

Question 30

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector. 

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 

Question 31

A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10-9 C m-2 on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present. 

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/ε0×d 

=1×10-9×20 / 8.85×10-12×100

=22.598 V=22.6 V

Using Einstein's photoelectric equation, we have:

eVo=hv - W0

=hc /λ - W

V0=4.14×10-15×3×108 /4×10-7 - 1.9

 =1.205 V

As V0 is much less than 'V', the minimum energy required to reach the charged plate must be equal to 22.7eV.

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

Hence maximum kinetic energy,

K.E.=V0+V

=1.205+22.6

=23.8005 eV

Question 32

Consider the situation of the previous problem. Consider the faster electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate. 

Solution 32

We know that,

E = σ / ε0 

=1×10-9 / 8.85×10-12 

=113 V/m

Acceleration is given by

 a=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality 

a = 1.6×10-19×113 / 9.1×10-31 

=19.87×1012

T = (2y/a)

=(2×20×10-2 / 19.87×1012)

=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=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality)

=(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 

Question 33

A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?

Solution 33

Given Data:-

φ = 1.9 eV

λ = 250 nm

Energy of a photon,

E=hc / λ 

E =H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality = 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=(2K/m) 

 v= (2×3.06×1.6×10-19 / 9.1×10-31) 

=1.04×106 ms-1 

Question 34

A small metal plate (work function φ) is kept at a distance d from a singly-ionised, fixed ion. A monochromatic light beam is incident on the metal plate and photoelectrons are emitted. Find the maximum wavelength of the light beam, so that some of the photoelectrons may go round the ion along a circle. 

Solution 34

Using Einstein's photoelectric equation,

eV0=hc /λ - φ 

V0=(hc / 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 =Ke2 /2d + φ = (Ke2+2dφ) /2d

λ =(hc)(2d) /(ke2+2dφ)

λ = 80hcd /(e2+80dφ)

Question 35

A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV.

(a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal.

(b) Under the same assumptions, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal. 

Solution 35

Given Data:-

λ= 400 nm

ϕ = 2.2 eV

 E=hc / λ 

E=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Photoelectric Effect And Wave Particle Duality = 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.