Question

Light of wavelength $2475\dot{A}$ is incident on barium. Photoelectrons emitted describe a circle of maximum radius of $100\text{cm}$ in a magnetic field of flux density $\frac{1}{\sqrt{17}}\times {10}^{-5}$$\text{T}$. Find the work function of Barium.

$(\frac{e}{m}=1.7\times {10}^{11}:\text{Take}hc=12375\text{eV}\dot{\text{A}})$

• A. $4.5\text{eV}$
• B. $1.3\text{eV}$
• C. $3.2\text{eV}$
• D. $5.6\text{eV}$

Answer 1

The correct option is A $4.5\text{eV}$

Given:
$\lambda =2475\dot{A}$
$r=100\text{cm}$
$B=\frac{1}{\sqrt{17}}\times {10}^{-5}\text{T}$

The radius of a circular path described by a charged particle of charge $e$, mass $m$ moving with a kinetic energy $K_m$ in a magnetic field $B$ is given by,

$r=\frac{\sqrt{2m{K}_m}}{qB}$

$\Rightarrow K_m=\frac{e^2{B}^2{r}^2}{2m}=\frac{1}{2}\times \left(\frac{e}{m}\right)\times e\times B^2\times r^2$

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

$\Rightarrow K_m=0.5\times 1.6\times {10}^{-19}\text{J}=\text{eV}$

By Einstein's photoelectric equation,

$E=\varphi +K_m$

$\Rightarrow \varphi =E-K_m=\frac{hc}{\lambda }-K_m$

$\Rightarrow \varphi =\frac{12375\text{eV}\dot{A}}{2475\dot{A}}-0.5\text{eV}$

$\therefore \varphi =4.5\text{eV}$

Hence, option $\left(A\right)$ is correct.