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Physics
Physics
Problem
Find the binding energy per nucleon of
20
Ca
40
_{20}\text{Ca}^{40}
20
Ca
40
nucleus. Given,
m
N
m_N
m
N
(
20
Ca
40
_{20}\text{Ca}^{40}
20
Ca
40
) = 39.962589 u,
m
n
m_n
m
n
= 1.008665 u and
m
p
m_p
m
p
= 1.007825 u. (Take,
1
amu
=
931
c
2
MeV
1 \text{ amu} = \frac{931}{c^2} \text{ MeV}
1
amu
=
c
2
931
MeV
)
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