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

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

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

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.

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

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* *

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.

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

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

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}

When the
sun is directly overhead, the surface of the earth receives 1.4 × 10^{3} 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 × 10^{11} 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?

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}

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 × 10^{19}. Calculate the force exerted by the light beam on the
mirror.

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

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.

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

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

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

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.

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

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.

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

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.

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)

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

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.

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.

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

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

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

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 × 10^{14} Hz,
what will be the stopping potential?

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

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.

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}

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.

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

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.

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

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.

(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

The
electric field associated with a monochromatic beam is 1.2 × 10^{15} 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.

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

The
electric field associated with a light wave is given byE=E_{0} sin (1.57×10^{7 }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.

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.

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

The
electric field at a point associated with a light wave isE= (100
Vm^{-1}) sin [(3.0×10^{15} s^{-1})t] sin [(6.0 ×10^{15} 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?

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

A
monochromatic light source of intensity 5 mW emits
8 × 10^{15} 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.

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

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

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

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.

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.

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 10^{6} photons is able to eject a
photoelectron, find the photocurrent in the circuit.

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

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?

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.

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

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

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

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.

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.

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

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.

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

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.

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

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?

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}

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.

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)

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.

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.

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