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JEE Advanced 2016 Paper 2
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© examsnet.com
Question : 1
Total: 54
The electrostatic energy of Z protons uniformly distributed through out a spherical nucleus of radius R is given by
E
=
3
5
Z
(
Z
−
1
)
e
2
4
π
ε
0
R
The measured masses of neutron
1
1
H
,
7
15
N
and
8
15
O
are
1.008665
u,
1.007825
u,
15.000109
u and
15.003065
u respectively .Given that the radii of both the
7
15
N
and
8
15
O
nuclei are ame,1u
=
931.5
MeV/
c
2
(c is the speed of light) and
e
2
(
4
π
ε
0
)
=
1.44
MeV fm. Assuming that the difference between the binding energies of
7
15
N
and
8
15
O
is purely due to the electrostatic energy ,the radius of eithr of the nuclei is
(
1
f
m
=
10
−
15
m
)
2.85
fm
3.03
fm
3.42
fm
3.80
fm
Validate
Solution:
Electrostatic energy
=
B
E
N
−
B
E
0
=
[
[
7
M
H
+
8
M
n
−
M
N
]
−
[
8
M
H
+
7
M
n
−
M
0
]
]
×
c
2
=
[
−
M
H
+
M
n
+
M
O
−
M
N
]
C
2
=
[
−
1.007825
+
1.008665
+
15.003065
−
15.000109
]
×
931.5
=
+
3.5359
MeV
ΔE
=
3
5
×
1.44
×
8
×
7
R
−
3
5
×
1.44
×
7
×
6
R
=
3.5359
R
=
3
×
1.44
×
14
5
×
3.5359
=
3.42
fm
© examsnet.com
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