91Ó°ÊÓ

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

The mass of the alpha particle is$m_{\text{alpha}}=6.640678×{10}^{−27}\text{ k²µ}$

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

The mass Defect is:

$\mathbf{Δ³¾}\mathbf{=}\mathbf{m}\mathbf{−}\mathbf{(}{\mathbf{m}}_{\mathbf{\text{alpha}}}\mathbf{+}{\mathbf{m}}_{\mathbf{\text{new}}}\mathbf{)}$

The kinetic energy is:

$\mathbf{K}\mathbf{=}{\mathbf{Δ³¾c}}^{\mathbf{2}}$

The total energy is:

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

(a)

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

$\mathrm{Δ³¾}=\mathrm{m}+{\mathrm{m}}_{\text{alpha}}+{\mathrm{m}}_{\text{new}}$

Substitute the given values to the formula above,

The formula used to find kinetic energy,

$\mathrm{K}={\mathrm{Δ³¾c}}^2$

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

Now, substitute the values to the formula above.

$\begin{array}{c}\mathrm{K}={\mathrm{Δ³¾c}}^2\\ =(9.322×{10}^{−30}\text{ k²µ})(3.0×{10}^8\text{″¾}/\text{s})\\ =8.3898×{10}^{−13}\text{ J}\end{array}$

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

(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:

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

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

$\begin{array}{c}{\stackrel{→}{\mathrm{p}}}_{\mathrm{Î\pm }}=−\stackrel{→}{\mathrm{p}}\\ {\mathrm{m}}_{\mathrm{Î\pm }}{\stackrel{→}{\mathrm{v}}}_{\mathrm{Î\pm }}=−\mathrm{m}\stackrel{→}{\mathrm{v}}\mathrm{n}\\ \left|\stackrel{→}{\mathrm{v}}\right|=\frac{{\mathrm{m}}_{\mathrm{Î\pm }}}{\mathrm{m}}\left|{\stackrel{→}{\mathrm{v}}}_{\mathrm{Î\pm }}\right|\end{array}$

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

$\begin{array}{c}\mathrm{K}=\frac{{\mathrm{mv}}^2}{2}\\ =\frac{\mathrm{m}}{2}â‹\dots \frac{{\mathrm{m}}_{\mathrm{Î\pm }}^2}{{\mathrm{m}}^2}{\mathrm{v}}_{\mathrm{Î\pm }}^2\\ =\frac{{\mathrm{m}}_{\mathrm{Î\pm }}}{\mathrm{m}}â‹\dots {\mathrm{K}}_{\mathrm{Î\pm }}\end{array}$

So, the energy conservation law becomes:

$\begin{array}{c}(\mathrm{M}−{\mathrm{m}}_{\mathrm{Î\pm }}−\mathrm{m}){\mathrm{c}}^2={\mathrm{K}}_{\mathrm{Î\pm }}+\frac{{\mathrm{m}}_{\mathrm{Î\pm }}}{\mathrm{m}}{\mathrm{K}}_{\mathrm{Î\pm }}\\ {\mathrm{K}}_{\mathrm{Î\pm }}=\frac{(\mathrm{M}−{\mathrm{m}}_{\mathrm{Î\pm }}−\mathrm{m}){\mathrm{c}}^2}{(1+\frac{{\mathrm{m}}_{\mathrm{Î\pm }}}{\mathrm{m}})}\\ =\frac{(3.917268â‹\dots {10}^{−25}\text{ k²µ}−6.640678â‹\dots {10}^{−27}\text{ k²µ}−3.850768â‹\dots {10}^{−25}\text{ k²µ})â‹\dots 9â‹\dots {10}^{16}{\text{″¾}}^2{\text{/s}}^2}{\left(1+\frac{6.640678â‹\dots {10}^{−27}\text{ k²µ}}{3.850768â‹\dots {10}^{−25}\text{ k²µ}}\right)}\\ =\left(\frac{8.2477â‹\dots {10}^{−13}}{1.6â‹\dots {10}^{−19}}\right)\text{ e³Õ}\\ =5.15\text{ M±ð³Õ}.\end{array}$

Therefore, the conservation of momentum is ${\mathrm{K}}_{\mathrm{Î\pm }}=5.15\text{ M±ð³Õ}$.