Work function of nickel is 5.01 eV. When ultraviolet radiation of wavelength $200\mathop {\text{A}}\limits^0 $ is incident on it, electrons are emitted. What will be the maximum velocity of emitted electrons?
$
  {\text{A}}{\text{. 3}} \times {\text{1}}{{\text{0}}^8}m/s \\
  {\text{B}}{\text{. 4}}{\text{.48}} \times {\text{1}}{{\text{0}}^6}m/s \\
  {\text{C}}{\text{. 10}}{\text{.36}} \times {\text{1}}{{\text{0}}^5}m/s \\
  {\text{D}}{\text{. 8}}{\text{.54}} \times {\text{1}}{{\text{0}}^6}m/s \\
$

Hint: In photoelectric effect, when the energy of the incident radiation exceeds the work function of the metal, the electrons are emitted from the surface of the metal and kinetic energy of the electrons is equal to the difference between energy of radiation and the work function of metal.

Formula used:
The formula for photoelectric effect is given as

$h\nu = W + K{\text{ }}...{\text{(i)}}$

where $h\nu $ is the energy of incident radiation; h is called the Planck’s constant which has value $h = 6.62 \times {10^{ - 34}}Js$ and $\nu $ is the frequency of radiation which is related to wavelength $\lambda $ as follows:

$\nu = \dfrac{c}{\lambda }$

Here c is the velocity of light given as $c = 3 \times {10^8}m/s$.

W signifies the work function of the metal while K is the kinetic energy imparted to the electron given as

$K = \dfrac{1}{2}m{{\text{v}}^2}$

where v is the velocity of the electron.

Also $1eV = 1.6 \times {10^{ - 19}}J$

Complete step-by-step answer:
In photoelectric effect, we have a radiation of certain wavelength or frequency incident on a metal surface. Work function of a metal is defined as the minimum amount of energy required to pull out an electron from that metal. When the energy of the radiation is greater than the work function of metal, it is observed that electrons are emitted from the metal surface and their kinetic energy is equal to the difference between energy of photon and the work function of metal.

$\therefore \dfrac{1}{2}m{{\text{v}}^2} = \dfrac{{hc}}{\lambda } - W{\text{ }}...{\text{(ii)}}$

We have the following values given to us and we need to calculate the velocity of the electron which has mass $m = 9.1 \times {10^{ - 31}}kg$

$
  W = 5.01eV = 5.01 \times 1.6 \times {10^{ - 19}}J \\
  \lambda = 200\mathop {\text{A}}\limits^0 = 2 \times {10^{ - 8}}m \\
$

Using these values in equation (ii),

$
  \dfrac{1}{2}m{{\text{v}}^2} = \dfrac{{hc}}{\lambda } - W \\
   = \dfrac{{\left( {6.62 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{2 \times {{10}^{ - 8}}}} - \left( {5.01 \times 1.6 \times {{10}^{ - 19}}} \right) \\
   = 99.3 \times {10^{ - 19}} - 8.016 \times {10^{ - 19}} \\
   = 91.284 \times {10^{ - 19}}J \\
   \Rightarrow {{\text{v}}^2} = \dfrac{{2 \times 91.284 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}} = 20.062 \times {10^{12}} \\
$

Taking square root on both sides, we get

$
  {\text{v}} = \sqrt {20.062 \times {{10}^{12}}} \\
   = 4.479 \times {10^6}m/s \\
   \simeq 4.48 \times {10^6}m/s \\
$

Therefore, the correct answer is option B.

Note:
 1. Higher the frequency and lower the wavelength of light, higher is the energy of that light.
 2. Photoelectric effect shows the particle nature of light photons transmitting energy to the metal in a chunk called quanta.