Question

Light of wavelength 2475$\AA$ is incident on barium. Photoelectrons emitted described a circle of radius 100 $cm$ in a magnetic field of magnitude $\frac{1}{\sqrt{17}}\times {10}^{-5}$ Tesla. Work function of the barium is ( Given: $\frac{e}{m}=1.7\times {10}^{11}Ck{g}^{-1}$

• 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 magnetic field is given by $r=\frac{\sqrt{2mK}}{qB}$; where $K$ = Kinetic energy of electron

$K=\frac{q^2{B}^2{r}^2}{2m}=\left(\frac{e}{m}\right)\frac{e{B}^2{r}^2}{2}$
$=\frac{1}{2}\times 1.7\times {10}^{11}\times 1.6\times {10}^{-19}\times (\frac{1}{\sqrt{17}}\times {10}^{-5}{)}^2\times 1^2$

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

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

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