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JEE Advanced 2018 Paper 2
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© examsnet.com
Question : 13
Total: 54
In a photoelectric experiment a parallel beam of monochromatic light with power of 200W is incident on a perfectly absorbing cathode of work function
6.2
e
V
The frequency of light is just above the thres hold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force
F
=
n
×
10
−
4
N
due to the impact of the electrons. The value of n is_______ . Mass of the electron
m
e
=
9
×
10
−
31
k
g
and
1.0
e
V
=
1.6
×
10
−
19
J
Your Answer:
Validate
Solution:
power = nhv n= number of protons per second
Since KE=0
h
c
=
ϕ
200
=
n
[
6.25
×
16
×
10
−
19
j
o
u
l
e
]
n
=
200
1.6
×
10
−
19
×
6.25
As photon just above threshold frequency
K
E
max
is zero and they are accelerated by potential difference of 500V.
K
E
f
=
q
Δ
V
P
2
2
m
=
q
Δ
V
⇒
P
=
√
2
m
Δ
V
Since efficiency is 100%, number of electrons = number of photons per second As photon is completely absorbed force exerted = nmv
=
200
6.25
×
1.6
×
10
−
19
×
√
2
(
9
×
10
−
31
)
×
1.6
×
10
−
19
×
500
=
3
×
200
×
10
−
25
×
√
1600
6.25
×
1.6
×
10
−
19
=
2
×
40
6.25
×
1.6
×
10
−
4
×
3
=
24
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