In the arrangement shown if Fig. `y=1.0mm`,`d=0.24mm` and `D=1.2m`. The work function of the material of the emitter if 2.2 eV. Find the stopping potential V needed to stop photocurrent (in`xx10^(-1)`V).

Explanation

Given: Fringe width,
`y=1.0mmxx2=2.0mm`
`d=0.24mm`,`W_0=2.2eV`,`D=1.2m`
`y=(lamdaD)/(d)` or `lamda=(yd)/(D)`
`=(2xx10^(-3)xx0.24xx10^(-3))/(1.2)=4xx10^(-7)`m
`E=(hc)/(lamda)=(4.14xx10^(-15)xx3xx10^(8))/(4xx10^(-7))=3.105 eV`
Stopping potential
`eV_0=3.105-2.2=0.905 eV`
`V_0=(0.905)/(1.6xx10^(-19))xx1.6xx10^(-19)V=0.905V`