Answer
a) $51.3\;\rm V$
b) $8.68\times10^4 \;\rm V$
Work Step by Step
a) According to de Broglie, the momentum of a particle is given by
$$p=\dfrac{h}{\lambda}$$
and we know that the momentum is given by $p=mv$;
$$mv=\dfrac{h}{\lambda}\tag 1$$
Now its final speed is
$$ v =\dfrac{h}{\lambda m } \tag 2$$
And we know that kinetic energy is given by
$$KE=eV=\frac{1}{2}mv^2$$
and hence,
$$V=\dfrac{KE}{e}=\dfrac{mv^2}{2e} \tag A$$
$$V =\dfrac{m}{2e} \;v^2 $$
Plugging from (2);
$$V =\dfrac{m}{2e} \;\left(\dfrac{h}{\lambda m }\right)^2 $$
Therefore, the potential difference is given by
$$V = \dfrac{h^2}{2e\lambda^2 m } \tag 3$$
Plugging the known for the proton case;
$$V = \dfrac{(6.626\times 10^{-34})^2}{2\times (1.6\times10^{-19})(4\times10^{-12})^2 (1.67\times 10^{-27}) } $$
$$V =\color{red}{\bf 51.3}\;\rm V$$
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b) In the electron case, if we used (2) to find its velocity, we will see that it is about
$$v=\dfrac{h}{\lambda m_e}=\dfrac{6.626\times 10^{-34}}{4\times 10^{-12}\times9.11\times 10^{-31}}$$
$$v=1.8\times10^8\;\rm m/s\approx 0.61c$$
which is 61$\%$ speed of light, so we can not use the same equations above since its speed is that way.
Now we need to find its energy
$$E=\sqrt{(pc)^2+(m_ec^2)^2}$$
Recall that
$$E=KE+m_ec^2$$
and hence,
$$KE=E-m_ec^2$$
Thus,
$$KE=\sqrt{(pc)^2+(m_ec^2)^2}-m_ec^2$$
Plugging that into (A) above;
$$V=\dfrac{\sqrt{(pc)^2+(m_ec^2)^2}-m_ec^2}{e}$$
$$V=\dfrac{\sqrt{(pc)^2+(m_ec^2)^2}-m_ec^2}{e}$$
Plugging from (1);
$$V=\dfrac{\sqrt{\left(\dfrac{hc}{\lambda}\right)^2+(m_ec^2)^2}-m_ec^2}{e}$$
Plugging the known;
$$V=\dfrac{\sqrt{\left(\dfrac{6.626\times 10^{-34}\times3\times10^8}{4\times10^{-12}}\right)^2+(9.11\times10^{-31}\times [3\times10^8]^2)^2}- 9.11\times10^{-31}\times [3\times10^8]^2}{1.6\times10^{-19}}$$
$$V=86779.2 \;\rm V=\color{red}{\bf 8.68\times10^4 }\;\rm V$$
