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 `2.2 eV`. Find the stopping potential `V` needed to stop the photocurrent.

Options

A. `0.9V`
B. `0.5V`
C. `0.4V`
D. `0.1V`

Answer

A

Explanation

Fringe width `(beta) = (lamdaD)/d`
Here `beta = 2y implies 2y =(lamdaD)/d`
`lamda=(2yd)/D=(2xx1xx10^(-3)xx0.24 xx10^(-3))/(1.2)`
`=4 xx10^(-7)m=400Å`
`E(eV)=(12375)/(lamdaÅ)= (12375)/(4000) =3.1eV`
Now , `eV_0=E-phi_0`
`impliesV_0= (3.1- 2.2)/e eV=0.9V`