Question

Light of wavelength $2475A^{\circ }$ 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. Work function of the barium is (Given $\frac{e}{m}=1.7\times {10}^{11})$

• A. 1.8 eV
• B. 2.1 eV
• C. 4.5 eV
• D. 3.3 eV

Answer 1

The correct option is C

4.5 eV

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

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

$=8\times {10}^{-20}J=0.5eV$

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

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