A photon of frequency `1.5xx10^(15)` Hz is incident on a metal surface of work function `1.672` eV. Calculate the stopping potential.

Explanation

Given : `v=1.5xx10^(15)`Hz
`w=1.672 eV = 1.672xx1.6xx10^(-19)=2.675xx10^(-19)`J
We know that, `E=w+(KE)_(max)`
where, `E=hv=6.626xx10^(-34)xx1.5xx10^(15)=9.939xx10^(-19)`J
`therefore (KE)_(max)=(9.939-2.675)x10^(-19)J = 7.264xx 10^(-19)J div 1.6 xx 10^(-19) = 4.54` eV
`therefore ` Stopping potential `= 4.54` V.