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VITEEE 2016 Solved Paper
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© examsnet.com
Question : 2
Total: 125
In photoelectric effect, initially when energy of electrons emitted is
E
0
, de-Broglie wavelength associated with them is
λ
0
. Now, energy is doubled then associated de-Broglie wavelength λ' is
λ' =
λ
0
√
2
λ' =
√
2
λ
0
λ' =
λ
0
λ' =
λ
0
2
Validate
Solution:
de-Broglie wavelength is given by
λ =
h
p
, where h = Planck’s constant and p = momentum
Also, energy (E) and momentum are related as
E =
p
2
2
m
∴ p =
√
2
m
E
∴ λ =
h
√
2
m
E
×
1
√
E
as h and m are constants
Hence ,
λ
0
λ
′
=
√
E
′
E
=
√
2
E
E
=
√
2
∴
λ
′
=
λ
0
√
2
© examsnet.com
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