Answer
a) $31.13$ $V$
b) $5631.25$ $V$
Work Step by Step
All symbols have the usual meaning.
$e=1.6 \times 10^{-19}$ $C$
Energy of electron when accelerated through a potential difference of $\Delta V$ (say) $=e\Delta V$
a) Wavelength of the electron $\lambda=0.220$ $nm$
de-Broglie relation:
$\lambda=\dfrac{h}{p} \Longrightarrow p\simeq 3.012 \times 10^{-24}$ $kg \hspace{2mm} m/s$
Mass of the electron $=m_e=9.11 \times 10^{-31}$ $kg$
Energy of the electron $=\dfrac{p^2}{2m_e} \simeq 4.98 \times 10^{-18}$ $J$
Potential difference $=V_o$
$eV_o=4.98 \times 10^{-18} \Longrightarrow V_o \simeq31.13$ $V $
b) Frequency of x-ray $=\dfrac{c}{\lambda}\simeq1.36 \times 10^{18} $ $Hz$
Energy of the photon $=h\dfrac{c}{\lambda}\simeq 9.01 \times 10^{ -16}$ $J$
Let $V'=$ potential difference required.
$eV'\simeq 9.01 \times 10^{-16} \Longrightarrow V'\simeq 5631.25$ $V$
