Question

In the arrangement shown in figure, $y =1.0\, mm , d =0.24 \,mm$ and $D =1.2 \,m$. The work function of the material of the emitter is $2.18\, eV$. The stopping potential needed to stop the photocurrent is ________volts. (Take $h =4.14 \times 10^{-15} eVs$ )

Answer 1

The distance between bright band and dark band is given by,
$y =\frac{\lambda D }{2 d } $
$\therefore \lambda =\frac{2 yd }{ D }$
$=\frac{2 \times 10^{-3} \times 0.24 \times 10^{-3}}{1.2} $
$=4 \times 10^{-7} m$
$E =\frac{ hc }{\lambda} $
$=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{4 \times 10^{-7}} $
$=3.105\, eV$
Stopping potential,
$eV _{0}=3.105-2.18=0.925\, eV $
$ V _{0}=0.925\, V \approx 0.92\, V$