Question

Show thatthe energy required to remove a neutron from helium-$4$ is .$20.6\text{MeV}$

Short answer

Energy required to remove one neutron from helium- $4$is $20.6\text{MeV}$.

Step-by-step solution

Given data

Mass of neutron, $m_n=1.008665\text{u}$.

Atomic mass of helium, $-4,M_{\frac{4}{2}}\text{He}=4.002603\text{u}$.

Speed of light, $c=3\times {10}^8\text{m}/\text{s}$.

Formula for Binding energy

For helium- $4$, binding energy is given by formula, $BE=(Z{m}_H+N{m}_n-M_{\frac{4}{1}x})c^2$.

Calculation for energy required to remove neutron

After removing $1$ neutron from helium-4, the mass of helium becomes:

$M_{\text{He}-3}=3.016029\text{u}$and number of neutron becomes.$N=1$

The binding energy is given as:

$\begin{array}{l}BE=(M_{\text{He}-3}+N{m}_n-M_{\frac{4}{2}}\text{He})c^2\\ BE=[3.016029\text{u}+(1\left)\right(1.008665\text{u})-4.002603\text{u}]{(3\times {10}^8\text{m}/\text{s})}^2\end{array}$

The conversion factor may be used to express $c^2$in $\text{MeV}/\text{u}$.

$\begin{array}{l}BE=[3.016029\text{u}+(1.008665\text{u})-4.002603\text{u}]\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ BE=20.6\text{MeV}\end{array}$