Answer
a) $0.06$ $V$
b) $3.98 \times 10^{-17}$ $J$
c) $2.05 \times 10^{-5}$ $m$
Work Step by Step
All the symbols have usual meaning.
$e=1.6 \times 10^{-19}$ $C$
a) Wavelength $=\lambda=5$ $nm$
Momentum of the electron $\hspace{1mm}p=\dfrac{h}{\lambda}\simeq 1.33 \times 10^{-25}$ $kg \hspace{1mm}m/s$
Energy of electron $=\dfrac{p^2}{2m_e}\simeq 9.71 \times 10^{-21}$ $J$
$V_o=$ potential difference required.
$eV_o\simeq 9.71 \times 10^{-21}\Longrightarrow V_o\simeq 0.06$ $V$
b) Energy of the photon $=h\dfrac{c}{\lambda} \simeq 3.98 \times 10^{-17}$ $J$
c) Energy of photon now $=9.71 \times 10^{-21}$ $J$
$\lambda'=$ Wavelength of the photon
$h\dfrac{c}{\lambda'}=9.71 \times 10^{-21} \Longrightarrow \lambda' \simeq 2.05 \times 10^{-5}$ $m$
