Question

When light of wavelength $310\text{nm}$ is incident on a material. The electron emitted with the maximum kinetic energy. If emitted photo electros are allowed to move through transverse magnetic field strength $B=5.2\times {10}^{-6}\text{T}$ performs circular motion, then find the radius of the circular path.

Work function of the material, $\varphi =2.5\text{eV}$.

• A. $0.25\text{m}$
• B. $1.5\text{m}$
• C. $2.5\text{m}$
• D. $0.79\text{m}$

Answer 1

The correct option is D $0.79\text{m}$

Given:
$\lambda =310\text{nm};B=5.2\times {10}^{-6}\text{T}$
$\varphi =2.5\text{eV}$

When an electron undergoes circular motion in a transverse magnetic field $\left(B\right)$ with a speed $V_{max}$, the radius of the circular path is given by,

$r=\frac{m{V}_{max}}{eB}$

To find $V_{max}$, Using Einstein's photoelectric equation,

$K.E_{max}=E-\varphi$

$\Rightarrow \frac{1}{2}m{V}_{ma{x}^2}=\frac{hc}{\lambda }-\varphi$

$\Rightarrow \frac{1}{2}\times 9.1\times {10}^{-31}\left(V_{max}^2\right)=(\frac{1240}{310}-2.5)\times 1.6\times {10}^{-19}$

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

$\Rightarrow V_{max}^2=5.2747\times {10}^{11}$

$\Rightarrow V_{max}=7.26\times {10}^5{\text{ms}}^{-1}$

$\Rightarrow r=\frac{m{V}_{max}}{Be}=\frac{9.1\times {10}^{-31}\times 7.26\times {10}^5}{5.2\times {10}^{-6}\times 1.6\times {10}^{-19}}$

$\Rightarrow r=0.79\text{m}$

Hence, option $\left(\text{D}\right)$ is correct.