Explanation

(c) Radius of circular path described by a charged particle in a magnetic field is given by
$r = \frac{\sqrt{2mK}}{qB};$ where $K =$ Kinetic energy of electron$\ \Rightarrow K = \frac{q^{2}B^{2}r^{2}}{2m} = \left( \frac{e}{m} \right)\frac{eB^{2}r^{2}}{2}$
$= \frac{1}{2} \times 1.7 \times 10^{11} \times 1.6 \times 10^{- 19} \times \left\lbrack \frac{1}{\sqrt{17}} \times 10^{- 5} \right\rbrack^{2} \times (1)^{2}$
$= 8 \times 10^{- 20}J = 0.5\ eV$
By using $E = W_{0} + K_{\max}$
$\Rightarrow W_{0} = E - K_{\max} = \left( \frac{12375}{2475} \right)eV - 0.5\ eV = 4.5\ eV$