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WBJEE 2016 Physics Solved Paper
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© examsnet.com
Question : 38
Total: 40
The distance between a light source and photoelectric cell is d. If the distance is decreased to d/2 then
The emission of electron per second will be four times
Maximum kinetic energy of photoelectrons will be four times
Stopping potential will remain same
The emission of electrons per second will be doubled
Validate
Solution:
I
∝
1
r
2
and
I
∝
N
(number of photons per second)
∴
N
∝
1
r
2
,
∴ number of ejected electron become 4 times
K
E
max
=
h
υ
−
Φ
since
υ
remains unchanged hence,
K
E
max
as well as stopping potential remains unchanged
K
E
max
=
e
V
s
© examsnet.com
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