Question

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

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

Answer 1

The correct option is C $4.5eV$

As we know,

Radius of circular path describe by a charged particle in a magnetic field is given by $r=\frac{\sqrt{2mK}}{qB}$; where $K=$ kinetic energy of electron $\Rightarrow K=\frac{q^2{B}^2{r}^2}{2m}=\left(\frac{e}{m}\right)\frac{e{B}^2{r}^2}{2}$
$=\frac{1}{2}\times 1.7\times {10}^{11}\times 1.6\times {10}^{-19}\times {\left(\frac{1}{\sqrt{17}\times {10}^{-5}}\right)}^2\times (1{)}^2$
$8\times {10}^{-20}J=0.5eV$
by using $E=W_0+K_{max}$
$\Rightarrow W_0=E-K_{max}=\left(\frac{12375}{2475}\right)eV-0.5eV=4.5eV$