Question

Calculate the neutron separation energy from the following data
$m\left( {{}_{20}^{40}Ca} \right) = 39.962591u$
$m\left( {{}_{20}^{41}Ca} \right) = 40.9622780u$
${m_u} = 1.00865u$
$1u = 931.5Mev/{c^2}$
A. $7.57Mev$
B. $8.36Mev$
C. $9.12Mev$
D. $9.56Mev$

Answer 1

Hints:First we need to calculate the mass defect then multiplied to ${c^2}$ ($c$ is the velocity of light) then the result will be multiplied to $931.5Mev/{c^2}$.

Formula Used:
$E = \Delta m.{c^2}$, where $\Delta m$ is the mass defect.

Complete step by step answer: According to the question we can say that,
${}_{20}^{41}Ca + E \to {}_{20}^{40}Ca + {}_0^1n$
In this equation calcium’s isotope having mass number 41 is getting energy to separate one of its neutrons and we need to find that energy. But for that energy we must be known to the mass defect such that we can apply mass energy equivalence formula.
$\Delta m \to $(Final mass – Initial mass)
$\Delta m \to \left( {m\left( {{}_{20}^{40}Ca} \right) + m\left( {{}_0^1n} \right) - m\left( {{}_{20}^{41}Ca} \right)} \right)$
As value given,
$m\left( {{}_{20}^{40}Ca} \right) = 39.962591u$
$m\left( {{}_{20}^{41}Ca} \right) = 40.9622780u$
${m_u} = 1.00865u$
Putting all the values,
$\Rightarrow \Delta m \to \left( {39.962591 + 1.00865 - 40.962278} \right)$
$ \to 0.008963u$
$\Rightarrow E = \Delta m{c^2}$
$E = 0.008963{c^2}$
As per the question $1u$ mass taken $931.5Mev/{c^2}$
$1u = 931.5Mev/{c^2}$
$\Rightarrow 1u.{c^2} = 931.5Mev$
$\Rightarrow E = 0.008963 \times 931.5$
$\Rightarrow E = 8.36Mev$

Note: In this type of problem, we should not calculate the value of ${c^2}$ as because conversion is necessary. So calculating the value makes the problem complicated.