Question

In the arrangement shown if figure (42-E4), y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photo current.

Answer 1

Given : Fringe width,

$y=1.0mm\times 2=2.0mm$

$d=0.24mm,$

$W_0=2.2ev$

$D=1.2m$

$y=\frac{\lambda D}{d}$

$\text{or}\lambda =\frac{yd}{D}$

$=\frac{2\times {10}^{-3}\times 0.24\times {10}^{-3}}{1.2}$

$=4\times {10}^{-7}m$

$E=\frac{hc}{\lambda }$

$=\frac{4.14\times {10}^{-15}\times 3\times {10}^8}{1\times {10}^{-7}}$

$=3.105ev$

$\text{Stopping potential}$

$e{V}_0=3.105-2.2=0.905eV$

$V_0=\frac{0.905}{1.6\times {10}^{-19}}\times 1.6\times {10}^{-19}V$

$=0.905V$