Question

If an electron emitted from photo-electric emissions revolves in a circle when exposed to a uniform transverse magnetic field of strength $B=5.2\times {10}^{-7}\text{T}$ with a maximum radius of $5\text{m}$, find the wavelength of incident energy if work function is $2.5\text{eV}$ $(\text{Use}hc=12400\text{J}\dot{A})$

• A. $4000\dot{A}$
• B. $5000\dot{A}$
• C. $3100\dot{A}$
• D. $4500\dot{A}$

Answer 1

The correct option is A $4000\dot{A}$

Given:
$B=5.2\times {10}^{-7}\text{T}$
$r=5\text{m}$
$\varphi =2.5\text{eV}$

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

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

$\Rightarrow V_{max}=\frac{B.e.r}{m}$

$\Rightarrow V_{max}=\frac{(5.2\times {10}^{-7})(1.6\times {10}^{-19})\times \left(5\right)}{9.1\times {10}^{-31}}$

$\Rightarrow V_{max}=4.57\times {10}^5{\text{ms}}^{-1}$

Now, according to Einstein's photoelectric equation,

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

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

$\Rightarrow \frac{1}{2}\times 9.1\times {10}^{-31}\times {(4.57\times {10}^5)}^2=(\frac{12400\text{J}\dot{A}}{\lambda \left(in\dot{A}\right)}-2.5)\times 1.6\times {10}^{-19}$

$\Rightarrow \lambda =4000\dot{\text{A}}$

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

Why this question? Key Concept: When an electron undergoes circular motion with speed $V_{max}$ in magnetic field $\left(B\right)$, then the radius of the circle is given by, $r=\frac{m{V}_{max}}{eB}$