Question

Light from a discharge tube containing hydrogen atoms falls on a piece of sodium due to the transition of electron from $4^{\text{th }}$ orbit to $2^{\text{nd }}$ orbit. Work function of sodium is $1.83eV$. The fastest moving photoelectron is allowed to enter in a magnetic field, which is perpendicular to the direction of motion of photoelectron as shown in the figure. Find the distance (in $\mu m$ ) covered by the electron in the magnetic field. $(B=1T,{\pi }^2=10$, Mass of electron $=9\times 10^{-31}kg$ and $R=$ Radius of the path that the most energetic electron takes in the presence of applied magnetic field)

Answer 1

Answer: 6

Energy of incident photon
$=13.6\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=13.6\left[\frac{1}{4}-\frac{1}{16}\right]$
$=13.6\times \frac{3}{16}=2.55eV$
Energy of incident photon $=$ Work function $(\varphi )+(KE{)}_{\max }$
Deflection $=180^{\circ }-60^{\circ }=120^{\circ }$ (From geometry)
Now, $\frac{m{v}^2}{R}=qvB$
$\Rightarrow$ Radius, $R=mv/qB$
Circumference of part inside the magnetic field,
$=\frac{\pi R\theta }{180^{\circ }}=\frac{120^{\circ }}{180^{\circ }}\pi R=\frac{2}{3}\pi R=\frac{2}{3}\pi \frac{mv}{qB}$
$=\frac{2\pi \sqrt{2mK{E}_{\max }}}{3qB}=6\times 10^{-6}m$