Physics - Modern Physics Question with Solution | TestHub

Let$E_1$and$E_2$be the KE of photoelectrons for incident light of frequency$v$and$2v$respectively.

Then$hv=E_1+\mathrm{\varphi }$and$h2v=E_2+{\mathrm{\varphi }}_0$

So,$2\left(E_1+{\mathrm{\varphi }}_0\right)=E_2+{\mathrm{\varphi }}_0$or$E_2=2E_1+{\mathrm{\varphi }}_0$

It means the KE of photoelectron becomes more than double