Photocurrent recorded in the micro
ammeter in an experimental set-up of
photoelectric effect vanishes when the
retarding potential is more than 0.8 V if
the wavelength of incident radiation is
4950 Å. If the source of incident radiation
is changed, the stopping potential turns
out to be 1.2 V. Find the work function of
the cathode material and the wavelength
of the second source.
Answers
Answer:
What would you like to ask?
12th
Physics
Dual Nature of Radiation and Matter
Photons and Photoelectric Effect
When a surface is irradiate...
PHYSICS
avatar
Asked on December 27, 2019 by
Preksha Kanojia
When a surface is irradiated with light of wavelength 4950 , a photocurrent appears which vanishes if a retarding potential greater than 0.6 V is applied across the phototube. When different source of light is used, it is found that the critical retarding potential is changed to 1.1 V. Find the work function of the emitted surface and the wavelength of the second source.
MEDIUM
Share
Study later
ANSWER
Given, wavelength of light used, λ=4950
A
˙
Stopping potential, $$V_{o} =0.6V$$
According to Einstein potential equation we have;
2
1
mv
2
=eV
o
=hv−V
o
........ (1)
ie, eV
o
=
λ
hc
−ϕ
o
Where, ϕ
o
= work function
V
o
= stopping potential
For the first source
λ
1
=4950
A
˙
=4950×10
−10
V
o
=0.6v
Putting these values in equation (1), we get
1.6×10
−19
×0.6=
495×10
−9
6.6×10
−34×3×10
8
−ϕ
o
ϕ
o
=3.04×10
−19
J
=
1.6×10
−19
3.04×10
−19
eV
=1.9eV
Let λ
2
be the wavelength of the second source.
Stopping potential V
o
=1.1 v (given)
Therefore,
1.6×10
−19
×1.1=
λ
2
6.6×10
−34×3×10
8
−3.04×10
−19
J
1.76×10
−19
=
λ
2
19.8×10
−27
−3.04×10
−19
λ
2
19.8×10
−26
=4.8×10
−19
∴λ
2
=
4.8×10
−19
19.8×10
−26
=4.125×10
−7
m
=4125
A
˙