It is given that the work function (${\mathrm{W}}_0$) for the cesium atom is 1.9 eV.

(a) From the ${\mathrm{W}}_0 = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_0}$ expression, we get:
${\mathrm{\lambda }}_0 = \frac{\mathrm{hc}}{{\mathrm{W}}_0}$
Where λ0 = threshold
wavelength h = Planck’s
constant c = velocity of
radiation
Substituting the values in the given expression of (${\mathrm{\lambda }}_0$):
Hence, the threshold wavelength is 653 nm.

(b) From the expression, ${\mathrm{W}}_0 = {\mathrm{hv}}_0$, we get:
${\mathrm{v}}_0 = \frac{{\mathrm{W}}_0}{\mathrm{h}}$
Where ${\mathrm{v}}_0$ = threshold
frequency h = Planck’s
constant
Substituting the values in the given expression of ${\mathrm{v}}_0$:
${\mathrm{v}}_0 = \frac{1.9 \mathrm{x} 1.602 \mathrm{x} {10}^{-19} \mathrm{J}}{6.626 \mathrm{x} {10}^{-34} \mathrm{Js}}$
(1 eV = 1.602 × 10–19 J) ${\mathrm{v}}_0$ = 4.593 × 1014 ${\mathrm{s}}^{-1}$
Hence, the threshold frequency of radiation (${\mathrm{v}}_0$) is 4.593 × 1014 ${\mathrm{s}}^{-1}$.

(c) According to the question:
The wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ${\mathrm{v}}_0$)

$= \mathrm{hc}\left(\frac{1}{\mathrm{\lambda }} - \frac{1}{{\mathrm{\lambda }}_0}\right)$
$= \left(6.626 \mathrm{x} {10}^{-34} \mathrm{Js}\right)\left(3.0 \mathrm{x} {10}^8 \mathrm{m}/\mathrm{s}\right)\left(\frac{{\mathrm{\lambda }}_0 - \mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)$
$= \left(1.9878 \mathrm{x} {10}^{-26} \mathrm{Jm}\right)\left[\frac{\left(653 \mathrm{x} 500\right){10}^{-9} \mathrm{m}}{\left(653\right)\left(500\right){10}^{-18} {\mathrm{m}}^2}\right]$
$= \frac{\left(1.9878 \mathrm{x} {10}^{-26}\right)\left(153 \mathrm{x} {10}^9\right)}{\left(653\right)\left(500\right)}$

= 9.3149 × 10–20 J

The kinetic energy of the ejected photoelectron = 9.3149 × 10–20J

$\mathrm{since} \mathrm{K}.\mathrm{E}. = \frac{1}{2}{\mathrm{mv}}^2 = 9.3149 \mathrm{x} {10}^{-20} \mathrm{J}$
$\mathrm{v} = \sqrt{\frac{2\left(9.3149 \mathrm{x} {10}^{-20} \mathrm{J}\right)}{9.10939 \mathrm{x} {10}^{-31} \mathrm{kg}}}$
$= \sqrt{2.0451 \mathrm{x} {10}^{11} {\mathrm{m}}^2{\mathrm{s}}^{-2}}$

v = $4.52 \times {10}^5 m/s$

Hence, the velocity of the ejected photoelectron (v) is $4.52 \times {10}^5 m/s$.