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JEE Main Physics Class 12 Atoms and Nuclei Part 2 Questions
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© examsnet.com
Question : 95
Total: 100
Given the masses of various atomic particles
m
p
=
1.0072
u
m
n
=
1.0087
u
,
m
e
=
0.000548
u
,
m
v
−
=
0
,
m
d
=
2.0141
u
,
where
p
equiv
proton,
n
≡
neutron,
e
≡
electron,
v
≡
antineutrino and
d
≡
deuteron. Which of the following process is allowed by momentum and energy conservation?
[Sep. 06,2020 (II)]
n
+
n
→
deuterium atom (electron bound to the nucleus)
p
→
n
+
e
+
+
v
n
+
p
→
d
+
γ
e
+
+
e
−
→
γ
Validate
Solution:
For the momentum and energy conservation, mass defect
(
Δ
m
)
should be positive. Since some energy is lost in every process.
(
m
p
+
m
n
)
>
m
d
© examsnet.com
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