Light of wavelength 2475$\stackrel{0}{\mathrm{A}}$  is incident on Barium. Photoelectrons emitted describe a circle of radius 100 cm by a magnetic field of flux density $\frac{1}{\sqrt{17}}\times {10}^{-5}$Tesla. The workfunction of Barium is (Given $\frac{e}{m}=1.7\times {10}^{11}$ C/Kg)

 

Radius of circular path described by the charged particle in the given magnetic field is

$\begin{array}{l}r=\frac{\sqrt{2m\left(KE\right)}}{qB}\\ KE=\frac{q^2{B}^2{r}^2}{2m}=\frac{e}{m}\times \frac{e{B}^2{r}^2}{2}\\ =1.7\times {10}^{11}\times \frac{1}{2}\times 1.6\times {10}^{-19}\times {\left(\frac{1}{\sqrt{17}}\times {10}^{-5}\right)}^2\times 1^2\\ =8\times {10}^{-20}J\\ =0.5eV\end{array}$

By using the relation, 

$W_0=E-K{E}_{\max }=\frac{12400}{2475}-0.5=4.5eV$