A photon of wavelength $4 \times 10^{-7} \mathrm{~m}$ strikes on metal surface,

Solution:

(i) Energy (E) of a photon $=h v=\frac{h c}{\lambda}$

Where, $\mathrm{h}=$ Planck's constant $=6.626 \times 10^{-34} \mathrm{Js}$

$\mathrm{c}=$ velocity of light in vacuum $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

$\lambda=$ wavelength of photon $=4 \times 10^{-7} \mathrm{~m}$

Substituting the values in the given expression of $E$ :

$E=\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 10^{-7}}=4.9695 \times 10^{-19} \mathrm{~J}$

Hence, the energy of the photon is $4.97 \times 10^{-19} \mathrm{~J}$.

(ii) The kinetic energy of emission $E_{\mathrm{k}}$ is given by

$=h v-h v_{0}$

$=(E-W) \mathrm{eV}$

$=\left(\frac{4.9695 \times 10^{-19}}{1.6020 \times 10^{-19}}\right) \mathrm{eV}-2.13 \mathrm{eV}$

$=(3.1020-2.13) \mathrm{eV}$

$=0.9720 \mathrm{eV}$

Hence, the kinetic energy of emission is $0.97 \mathrm{eV}$.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

$\frac{1}{2} m v^{2}=h v-h v_{0}$

$\Rightarrow v=\sqrt{\frac{2\left(\mathrm{~h} v-\mathrm{h} v_{0}\right)}{m}}$

Where $\left(h v-h v_{0}\right)$ is the kinetic energy of emission in Joules and ' $m^{\prime}$ is the mass of the

photoelectron. Substituting the values in the given expression of $v$ :

$v=\sqrt{\frac{2 \times\left(0.9720 \times 1.6020 \times 10^{-19}\right) \mathrm{J}}{9.10939 \times 10^{-31} \mathrm{~kg}}}$

$=\sqrt{0.3418 \times 10^{12} \mathrm{~m}^{2} \mathrm{~s}^{-2}}$

$v=5.84 \times 10^{5} \mathrm{~ms}^{-1}$

Hence, the velocity of the photoelectron is $5.84 \times 10^{5} \mathrm{~ms}^{-1}$.