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WBJEE 2019 Physics Solved Paper
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© examsnet.com
Question : 3
Total: 40
A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton
(
λ
p
)
and that of electron
(
λ
e
)
?
(
λ
p
)
=
2000
(
λ
e
)
(
λ
p
)
=
λ
e
2000
(
λ
p
)
=
20
√
5
λ
e
(
λ
p
)
=
λ
e
20
√
5
Validate
Solution:
As we know that de-Broglie wavelength is given as
λ
=
h
p
=
h
m
v
.....(i)
where,
h = Planck constant
p = momentum of particle
v = velocity of particle and
m = mass of the particle.
Eq. (i) can be written as,
λ
=
h
2
m
(
K
E
)
=
h
√
2
m
q
v
[
∵
K
E
=
q
v
]
where, KE = Kinetic energy of particle
Hence,
λ
∝
1
√
m
Now,
λ
proton
λ
electron
=
√
m
electron
m
proton
Given, mass of proton,
m
proton
=
2000
m
electron
λ
proton
λ
electron
=
√
1
2000
λ
proton
λ
electron
=
1
20
√
5
⇒
λ
p
=
λ
e
20
√
5
© examsnet.com
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