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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 32
Total: 54
The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion.
Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be
E
b
p
and the binding energy of a neutron be
E
b
n
in the nucleus.
Which of the following statement(s) is(are) correct?
E
b
p
−
E
b
n
is proportional to
Z
(
Z
−
1
)
where
Z
is the atomic number of the nucleus.
E
b
p
−
E
b
n
is proportional to
A
−
1
3
where
A
is the mass number of the nucleus.
E
b
p
−
E
b
n
is positive.
E
b
p
increases if the nucleus undergoes a beta decay emitting a positron.
Validate
Solution:
Binding energy of proton \& neutron due to nuclear force is same. So difference in binding energy is only due to electrostatic P.E. and it is positive
E
0
p
−
E
0
n
=
electrostatic P.E.
=
Z
×
P.E. of one proton
=
Z
×
1
4
π
ε
0
(
Z
−
1
)
e
2
R
Where
R
=
R
0
A
1
∕
3
=
1
4
π
ε
0
Z
(
Z
−
1
)
e
2
R
0
A
1
3
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