Answer
a) $0.4\;\rm V$
b) $3.75\times10^5\;\rm m/s$
c) $1.94\;\rm nm$
Work Step by Step
a)
We know that the energy of the electron is given by
$$E=W_0+KE$$
Thus, its kinetic energy is given by
$$KW=E-W_0= eV$$
So that the needed voltage is given by
$$V=\dfrac{E-W_0}{e}$$
Recalling that $E=hf=\dfrac{hc}{\lambda}$;
$$V=\dfrac{\dfrac{hc}{\lambda}-W_0}{e}$$
Plugging the known;
$$V=\dfrac{\dfrac{6.626\times10^{-34}\times 3\times10^8}{ 464\times10^{-9}}-(2.28\times1.6\times10^{-19})}{1.6\times10^{-19}}$$
$$V=0.397\;\rm V\approx \color{red}{\bf 0.4}\;\rm V$$
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b)
First, we need to find the electron speed by using the normal kinetic energy law and then compare it to the speed of light ($0.1C$), if it was too small to be compared with it so be it, if not, we will need to use the relativistic kinetic energy law.
$$KE=\frac{1}{2}m_ev^2$$
Solving for $v$;
$$v=\sqrt{\dfrac{2KE}{m_e}}=\sqrt{\dfrac{2eV}{m_e}}$$
Plugging the known;
$$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times0.4 }{9.11\times 10^{-31}}}=\color{red}{\bf 3.75\times10^5}\;\rm m/s$$
which is about $0.00125c$, so our result is acceptable.
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c)
We know that the wavelength is given by
$$\lambda=\dfrac{h}{p}=\dfrac{h}{m_ev}$$
Plugging the known;
$$\lambda= \dfrac{6.626\times10^{-34}}{9.11\times10^{-31}\times 3.75\times10^5}=\color{red}{\bf1.94}\;\rm nm$$