The work function of caesium is $2.14 \ eV$. Find
$(a)$ the threshold frequency for caesium,and
$(b)$ the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.

(N/A) For the cut-off or threshold frequency,the energy $h \nu_{0}$ of the incident radiation must be equal to the work function $\phi_{0}$,so that
$\nu_{0} = \frac{\phi_{0}}{h} = \frac{2.14 \ eV}{6.63 \times 10^{-34} \ J \cdot s}$
$= \frac{2.14 \times 1.6 \times 10^{-19} \ J}{6.63 \times 10^{-34} \ J \cdot s} = 5.16 \times 10^{14} \ Hz$
Thus,for frequencies less than this threshold frequency,no photoelectrons are ejected.
$(b)$ Photocurrent reduces to zero when the maximum kinetic energy of the emitted photoelectrons equals the potential energy $e V_{0}$ provided by the retarding potential $V_{0}$. Einstein's photoelectric equation is
$e V_{0} = h \nu - \phi_{0} = \frac{h c}{\lambda} - \phi_{0}$
$\lambda = \frac{h c}{e V_{0} + \phi_{0}}$
$= \frac{(6.63 \times 10^{-34} \ J \cdot s) \times (3 \times 10^{8} \ m/s)}{(0.60 \ eV + 2.14 \ eV) \times 1.6 \times 10^{-19} \ J/eV}$
$= \frac{19.89 \times 10^{-26} \ J \cdot m}{4.384 \times 10^{-19} \ J} \approx 454 \ nm$