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KEAM 2014 Physics and Chemistry Paper
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© examsnet.com
Question : 20
Total: 120
An electron of mass
m
e
and a proton of mass
m
p
are accelerated through the same potential. Then the ratio of their de Broglie wavelengths is
1
√
m
e
m
p
m
e
m
p
m
p
m
e
√
m
p
m
e
Validate
Solution:
Given,
m
e
=
mass of the electron
m
p
=
mass of proton
By de-Broglie wavelength of matter waves
λ
=
h
p
(
∵
p
=
m
v
)
∴
λ
=
h
m
v
If
K
be the kinetic energy of the electron, then
K
=
1
2
m
v
2
or
v
=
√
2
K
∕
m
Now,
λ
=
h
m
√
2
K
/
m
λ
=
h
√
2
K
m
So,
λ
∝
1
√
m
λ
e
=
1
√
m
e
....(i)
λ
p
=
1
√
m
p
.....(ii)
From Eqs. (i) and (ii), we get
λ
e
λ
p
=
1
/
√
m
e
1
/
√
m
p
λ
e
λ
p
=
√
m
p
m
e
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