(a) How much work would it take to push two protons very slowly from a separation of$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{C}$(a typical atomic distance) to$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$(a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

Consider the formula for the work required to move a charge from the point a to the point b is:

${\mathbf{W}}_{\mathbf{a}\mathbf{\to }\mathbf{b}}\mathbf{=}{\mathbf{U}}_{\mathbf{b}}\mathbf{-}{\mathbf{U}}_{\mathbf{a}}$

Here,$U_a$and$U_b$are the initial and final electric potential respectively.

Consider the expression for the initial potential as:

$U_a=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r_a}$

Consider the expression for the final potential as:

$U_b=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_2{q}_2}{r_b}$

Consider the formula for the change in the kinetic energy when the mass of proton and electron are equal and are projected in the same direction.

$$\begin{gathered} \nabla K=\frac{m{v}_2^2}{2}+\frac{m{v}_2^2}{2} \\ =m{v}_2^2 \end{gathered}$$

Then, the change in the work done is.

$\nabla U=m{v}_2^2$

Rewrite the expression in terms of the velocity as,

$v_2=\sqrt{\frac{\nabla U}{m}}$ ….. (1)

(a)

Consider the equation for the work done as:

$$\begin{gathered} W=U_b-U_a \\ W=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{1}{r_b}-\frac{1}{r_a}\right] \end{gathered}$$

Put given values

$$\begin{gathered} \mathrm{W}=\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2\left[\frac{1}{3.0\times {10}^{-15}}-\frac{1}{2\times {10}^{-10}}\right] \\ \mathrm{W}=-7.68\times {10}^{-14}\mathrm{J} \end{gathered}$$

But work done is never negative so$7.68\times {10}^{-14}\hspace{0.33em}\mathrm{J}$.

Hence, the amount done is $7.68\times {10}^{-14}\hspace{0.33em}\mathrm{J}$.

(b)

Substitute the values in the equation (1) and solve as:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{7.68\times {10}^{-14}\mathrm{J}}{1.67\times {10}^{-27}\mathrm{kg}}} \\ \mathrm{v}=6.78\times {10}^6\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, the velocity is $6.78\times {10}^6\frac{\mathrm{m}}{\mathrm{s}}$.