Question

Light of wavelength $2475A^{\circ }$ is incident on Barium. Photoelectrons emitted describe a circle of radius $100cm$ in a magnetic field of flux density $\frac{1}{\sqrt{17}}\times {10}^{-5}T$. Work function of the barium (in $eV$) is (Given $\frac{e}{m}=1.7\times {10}^{11}CK{g}^{-1})$

Answer 1

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

$\Rightarrow K=\frac{q^2{B}^2{r}^2}{2m}=\left(\frac{e}{m}\right)\frac{e{B}^2{r}^2}{2}$

$=\frac{1}{2}\times 1.7\times {10}^{11}\times 1.6\times {10}^{-19}\times {(\frac{1}{\sqrt{17}}\times {10}^{-5})}^2\times (1{)}^2$

$=8\times {10}^{-20}J=0.5eV$$\because (1eV=1.6\times {10}^{-19}J)$

By using $E=W_0+K_{max}$

$E=\left(\frac{12375}{2475}eV\right)$

$\Rightarrow W_0=E-K_{max}=\left(\frac{12375}{2475}\right)eV-0.5eV=4.5eV$