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JEE Advanced 2016 Paper 1
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© examsnet.com
Question : 16
Total: 54
The isotope
5
12
B
having a mass 12.014 u undergoes β-decay to
6
12
C
,
6
12
C
has an excited state of the nucleus
(
6
12
C
*
)
at
4.041
MeV above ground state.If
5
12
B
decays to
(
6
12
C
*
)
,the maximum kinectic energy of the β-particles in units of MeV is
(
1
u
=
931.5
MeV/c
2
,where c is the speed of the light in vacuum ),
Your Answer:
Validate
Solution:
5
12
B
→
6
12
C
+
−
1
0
e
+
−
v
Mass defect
=
(
12.014
−
12
)
u
∴Released energy
=
13.041
MeV
Energy used for excitation of
6
12
C
=
4.041
MeV
∴Energy converted to KE of electron
=
13.041
−
4.041
=
9
MeV
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