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 $1.83\, eV$. 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. $\left(B=1 T , \pi^{2}=10\right.$, 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

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] $
$=13.6 \times \frac{3}{16}=2.55 \,eV$
Energy of incident photon $=$ Work function $(\phi)+(K E)_{\max }$
Deflection $=180^{\circ}-60^{\circ}=120^{\circ}$ (From geometry)
Now, $\frac{m v^{2}}{R}=q v B $
$\Rightarrow$ Radius, $R=m v / q B$
Circumference of part inside the magnetic field,
$=\frac{\pi R \theta}{180^{\circ}}=\frac{120^{\circ}}{180^{\circ}} \pi R=\frac{2}{3} \pi R=\frac{2}{3} \pi \frac{m v}{q B}$
$=\frac{2 \pi \sqrt{2 m K E_{\max }}}{3 q B}=6 \times 10^{-6} m$