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JEE Main 20 July 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 11
Total: 90
A nucleus of mass
M
emits
γ
-ray photon of frequency '
v
′
.
The loss of internal energy by the nucleus is :
[Take 'c' as the speed of electromagnetic wave]
[20 Jul 2021 Shift 1]
hv
0
h
v
[
1
−
h
v
2
M
c
2
]
h
v
[
1
+
h
v
2
M
c
2
]
Validate
Solution:
Energy of
γ
ray
[
E
γ
]
=
h
v
Momentum of
γ
ray
[
P
γ
]
=
h
λ
=
h
v
C
Total momentum is conserved.
→
P
γ
+
→
P
N
u
=
0
Where
→
P
N
u
=
Momentum of decayed nuclei
⇒
P
γ
=
P
N
u
⇒
h
v
C
=
P
N
u
⇒
K.E. of nuclei
=
1
2
M
v
2
=
(
P
N
u
)
2
2
M
=
1
2
M
[
h
v
C
]
2
Loss in internal energy
=
E
γ
+
K
.
E
N
u
=
h
v
+
1
2
M
[
h
v
C
]
2
=
h
v
[
1
+
h
v
2
M
C
2
]
© examsnet.com
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