(i) Energy (E) of a photon = hν = hc/$\mathrm{\lambda }$
Where, h = Planck’s constant = 6.626 × ${10}^{-34}$ Js
c = velocity of light in vacuum = 3 × ${10}^8$ m/s
λ = wavelength of photon = 4 × ${10}^{-7}$ m
Substituting the values in the given expression of E:
$\mathrm{E} = \frac{\left(6.626 \mathrm{x} {10}^{-34}\right)\left(3 \mathrm{x} {10}^8\right)}{4 \mathrm{x} {10}^{-7}} = 4.9695 \mathrm{x} {10}^{-19} \mathrm{J}$
Hence, the energy of the photon is 4.97 × ${10}^{-19}$ J.

(ii) The kinetic energy of emission Ek is given by
$= \mathrm{hv} - \mathrm{hc}$
$= \left(\mathrm{E} - \mathrm{W}\right)\mathrm{eV}$
$= \left(\frac{4.9695 \mathrm{x} {10}^{-19}}{1.6020 \mathrm{x} {10}^{-19}}\right)\mathrm{eV} - 2.13 \mathrm{eV}$
= (3.1020 – 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (ν) can be calculated by the expression,
$\frac{1}{2}{\mathrm{mv}}^2 = \mathrm{hv} - {\mathrm{hc}}_0$
$\Rightarrow \mathrm{v} = \sqrt{\frac{2\left(\mathrm{hv} - \mathrm{hc}\right)}{\mathrm{m}}}$
Where $\left(\mathrm{hv} - {\mathrm{hc}}_0\right)$ is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:
$\mathrm{v} = \sqrt{\frac{2 \mathrm{x} \left(0.90720 \mathrm{x} 1.6020 \mathrm{x} {10}^{19}\right) \mathrm{J}}{9.10939 \mathrm{x} {10}^{-11} \mathrm{kg}}}$
$= \sqrt{0.3418 \mathrm{x} {10}^{12} {\mathrm{m}}^2{\mathrm{s}}^{-2}}$
v = 5.84 × 105 m/s
Hence, the velocity of the photoelectron is 5.84 × 105 m/s.