A nucleus whose mass is $\mathbf{3}\mathbf{.917268}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathbf{}\mathbf{\text{kg}}$ undergoes spontaneous alpha decay. The original nucleus disappears and there appear two new particles: a He-4 nucleus of mass$\mathbf{6}\mathbf{.640678}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{\text{kg}}$ (an alpha particle consisting of two protons and two neutrons) and a new nucleus of mass$\mathbf{3}\mathbf{.850768}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathbf{}\mathbf{\text{kg}}$ . (Note that the new nucleus has less mass than the original nucleus, and it has two fewer protons and two fewer neutrons.)

(a) What is the total kinetic energy of the alpha particle and the new nucleus?

(b) Use the conservation of momentum in order to determine the kinetic energy of the alpha particle and kinetic energy of the new nucleus.

The mass of the nucleus is$\mathrm{m}=3.917268\times {10}^{-25}\text{kg}$

The mass of the alpha particle is$m_{\text{alpha}}=6.640678\times {10}^{-27}\text{\hspace{0.17em}kg}$

The mass of the new nucleus is $m_{\text{new}}=3.850768\times {10}^{-25}\text{kg}$

The mass Defect is:

$\mathbf{\Delta m}\mathbf{=}\mathbf{m}\mathbf{-}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{alpha}}}\mathbf{+}{\mathbf{m}}_{\mathbf{\text{new}}}\mathbf{)}$

The kinetic energy is:

$\mathbf{K}\mathbf{=}{\mathbf{\Delta mc}}^{\mathbf{2}}$

The total energy is:

${\mathbf{Mc}}^{\mathbf{2}}\mathbf{=}{\mathbf{K}}_{\mathbf{\alpha }}\mathbf{+}\mathbf{K}\mathbf{+}{\mathbf{m}}_{\mathbf{\alpha }}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{mc}}^{\mathbf{2}}$

(a)

The expression of the mass defect in the alpha decay can be expressed as,

$\mathrm{\Delta m}=\mathrm{m}+{\mathrm{m}}_{\text{alpha}}+{\mathrm{m}}_{\text{new}}$

Substitute the given values to the formula above,

The formula used to find kinetic energy,

$\mathrm{K}={\mathrm{\Delta mc}}^2$

Where, $\mathrm{m}$ is the mass$\mathrm{c}$ is the speed of light with a value of $3.0\times {10}^8\text{m}/\text{s}$

Now, substitute the values to the formula above.

$\begin{array}{c}\mathrm{K}={\mathrm{\Delta mc}}^2\\ =(9.322\times {10}^{-30}\text{\hspace{0.17em}kg})(3.0\times {10}^8\text{\hspace{0.17em}m}/\text{s})\\ =8.3898\times {10}^{-13}\text{\hspace{0.17em}J}\end{array}$

Therefore, the total kinetic energy of the alpha particle and the new nucleus is$8.3898\times {10}^{-13}\text{J}$.

(b)

Since original nucleus is at rest, in the initial state of the system, only the energy of nucleus is there. In the final state there are total energies of the alpha particle and the new nucleus:

${\mathrm{Mc}}^2={\mathrm{K}}_{\mathrm{\alpha }}+\mathrm{K}+{\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{c}}^2+{\mathrm{mc}}^2$

Also, total momentum of the system is zero, which means that the two products of the decay have equal but opposite momentum:

$\begin{array}{c}{\overrightarrow{\mathrm{p}}}_{\mathrm{\alpha }}=-\overrightarrow{\mathrm{p}}\\ {\mathrm{m}}_{\mathrm{\alpha }}{\overrightarrow{\mathrm{v}}}_{\mathrm{\alpha }}=-\mathrm{m}\overrightarrow{\mathrm{v}}\mathrm{n}\\ \left|\overrightarrow{\mathrm{v}}\right|=\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{\mathrm{m}}\left|{\overrightarrow{\mathrm{v}}}_{\mathrm{\alpha }}\right|\end{array}$

This means that kinetic energy of new nucleus can be written as following:

$\begin{array}{c}\mathrm{K}=\frac{{\mathrm{mv}}^2}{2}\\ =\frac{\mathrm{m}}{2}\cdot \frac{{\mathrm{m}}_{\mathrm{\alpha }}^2}{{\mathrm{m}}^2}{\mathrm{v}}_{\mathrm{\alpha }}^2\\ =\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{\mathrm{m}}\cdot {\mathrm{K}}_{\mathrm{\alpha }}\end{array}$

So, the energy conservation law becomes:

$\begin{array}{c}(\mathrm{M}-{\mathrm{m}}_{\mathrm{\alpha }}-\mathrm{m}){\mathrm{c}}^2={\mathrm{K}}_{\mathrm{\alpha }}+\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{\mathrm{m}}{\mathrm{K}}_{\mathrm{\alpha }}\\ {\mathrm{K}}_{\mathrm{\alpha }}=\frac{(\mathrm{M}-{\mathrm{m}}_{\mathrm{\alpha }}-\mathrm{m}){\mathrm{c}}^2}{(1+\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{\mathrm{m}})}\\ =\frac{(3.917268\cdot {10}^{-25}\text{\hspace{0.17em}kg}-6.640678\cdot {10}^{-27}\text{\hspace{0.17em}kg}-3.850768\cdot {10}^{-25}\text{\hspace{0.17em}kg})\cdot 9\cdot {10}^{16}{\text{\hspace{0.17em}m}}^2{\text{/s}}^2}{\left(1+\frac{6.640678\cdot {10}^{-27}\text{\hspace{0.17em}kg}}{3.850768\cdot {10}^{-25}\text{\hspace{0.17em}kg}}\right)}\\ =\left(\frac{8.2477\cdot {10}^{-13}}{1.6\cdot {10}^{-19}}\right)\text{\hspace{0.17em}eV}\\ =5.15\text{\hspace{0.17em}MeV}.\end{array}$

Therefore, the conservation of momentum is ${\mathrm{K}}_{\mathrm{\alpha }}=5.15\text{\hspace{0.17em}MeV}$.