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Standard XII
Physics
Question
A
α
- particle is accelerated through a potential difference of V volts from rest. The de-Broglie wavelength associated with it is (in angstrom):
0.101
√
V
0.983
√
V
√
150
V
0.286
√
V
A
0.101
√
V
B
0.983
√
V
C
0.286
√
V
D
√
150
V
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Solution
Verified by Toppr
Kinetic Energy gained by the
α
-particle by potential difference
V
is,
p
2
2
m
α
=
q
α
V
=
2
e
V
,
and
m
α
=
4
m
p
p
2
=
=
16
m
p
e
V
p
=
4
√
m
p
e
V
=
4
√
(
1.67
×
10
−
27
)
k
g
×
(
1.6
×
106
−
19
)
C
×
V
=
4
×
1.634
×
10
−
23
√
V
λ
=
h
p
, here
h
=
6.62
×
10
−
34
J
−
s
e
c
is Plank's constant.
λ
=
h
p
=
6.62
×
10
−
34
4
×
1.634
×
10
−
23
√
V
=
0.101
√
V
∘
A
Option 'C' is correct.
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