91Ó°ÊÓ

The proton charge is $q_p=+e$

The proton mass is $m_p=1.67×{10}^{−27}\mathrm{kg}$

Right after the release, the force is maximum and therefore the acceleration is also maximum,

Acceleration $a_p=\frac{F}{m_0}$

Now, $F=k\frac{\left|q{q}^{\mathrm{′}}\right|}{r^2}$

And r is the distance which is equal to0.225 nm.

So, solve for F,

$\begin{array}{r}\mathrm{F}=\left(8.99×{10}^9\mathrm{N}×{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(2\right){\left(1.60×{10}^{−19}\mathrm{C}\right)}^2}{{\left(0.225×{10}^{−9}\mathrm{m}\right)}^2}\\ =9.09×{10}^{−9}\mathrm{N}\end{array}$

$\begin{array}{r}{\mathrm{a}}_{\mathrm{p}}=\frac{9.09×{10}^{−9}\mathrm{N}}{1.67×{10}^{−27}\mathrm{kg}}\\ =5.44×{10}^{18}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Similarly, for the alpha particle,

The charge is $q_{\mathrm{Î\pm }}=+2e$

The mass is $m_{\mathrm{Î\pm }}=4m_p=6.68×{10}^{−27}\mathrm{kg}$

Acceleration $a_{\mathrm{Î\pm }}=\frac{F}{m_{\mathrm{Î\pm }}}$

$\begin{array}{r}{\mathrm{a}}_{\mathrm{Î\pm }}=\frac{9.09×{10}^{−9}\mathrm{N}}{6.68×{10}^{−27}\mathrm{kg}}\\ =1.36×{10}^{18}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus the acceleration of the proton and alpha particle respectively is $5.44×{10}^{18}\mathrm{m}/{\mathrm{s}}^2$ and $1.36×{10}^{18}\mathrm{m}/{\mathrm{s}}^2$.

Speed is calculated by the law of conservation of energy,

${\mathrm{U}}_1+{\mathrm{K}}_1={\mathrm{U}}_2+{\mathrm{K}}_2$

Now, K₁ = 0 and U₂ = 0.

So, U₁ = K₂

Solve for U₁,

localid="1668229301464" $\begin{array}{r}{U}_1=\left(8.99×{10}^9\mathrm{N}×{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(2\right)\left(1.60×{10}^{−19}\mathrm{C}\right)}{0.225×{10}^{−9}\mathrm{m}}\\ =2.05×{10}^{−18}\mathrm{J}\end{array}$

So this is the total kinetic energy of the system, localid="1668229340261" $\begin{array}{r}{K}_2=\frac{1}{2}{m}_p{v}_p^2+\frac{1}{2}{m}_{\mathrm{Î\pm }}{v}_{\mathrm{Î\pm }}^2\\ =2.05×{10}^{−18}\mathrm{J}\end{array}$

This amount is divided between the proton and the alpha particle, Therefore, use the momentum conservation principle to solve for the speed of the proton.

$\begin{array}{r}{p}_1=p_2\\ m_p{v}_p−m_{\mathrm{Î\pm }}{v}_{\mathrm{Î\pm }}=0\\ v_a=\left(\frac{m_p}{m_{\mathrm{Î\pm }}}\right)v_p\end{array}$

So, kinetic energy is,

$\begin{array}{r}{K}_2=\frac{1}{2}{m}_p{\mathrm{v}}_p^2+\frac{1}{2}{m}_{\mathrm{Î\pm }}{\mathrm{v}}_{\mathrm{Î\pm }}^2\\ =\frac{1}{2}{m}_p{\mathrm{v}}_p^2\left(1+\frac{m_p}{m_{\mathrm{Î\pm }}}\right)\end{array}$

Solve for v_p,

localid="1668229807440" $\begin{array}{r}{V}_p=\sqrt{\frac{2K_2}{m_p\left(1+\left(\frac{m_p}{m_a}\right)\right)}}\\ =\sqrt{\frac{2\left(2.05×{10}^{−18}\mathrm{J}\right)}{\left(1.67×{10}^{−27}\mathrm{kg}\right)\left(1+\frac{1}{4}\right)}}\\ =4.43×{10}^4\mathrm{m}/\mathrm{s}\end{array}$

Similarly, solve for v_α,

localid="1668229670474" $\begin{array}{r}{v}_a=\left(\frac{m_p}{m_a}\right)v_p\\ =\frac{1}{2}\frac{1}{2}\left(4.43×{10}^4\mathrm{m}/\mathrm{s}\right)\\ =1.11×{10}^4\mathrm{m}/\mathrm{s}\end{array}$

Thus the speed of the proton and alpha particle respectively is localid="1668229814402" $4.43×{10}^4\mathrm{m}/\mathrm{s}$and $1.11×{10}^4\mathrm{m}/\mathrm{s}$.