Q. Light of two different frequencies whose photons have energies $1 \, eV$ and $2.5 \, eV$ respectively illuminate a metallic surface whose work function is $0.5 \, eV$ successively. The ratio of maximum speeds of both emitted electrons will be :
NTA AbhyasNTA Abhyas 2022
Solution:
From the Einstein's photoelectric equation, $KE_{max}=E_{ph}-\phi_{0}$ , where $E_{ph},\phi_{0}$ denotes energy of photon and work function of material.
$KE_{1}=1eV-0.5eV=0.5eV$
$KE_{2}=2.5eV-0.5eV=2eV$
$\frac{K E_{1}}{K E_{2}}=\frac{0 . 5 eV}{2 eV}=\frac{1}{4}$
$\frac{v_{1}}{v_{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$KE_{1}=1eV-0.5eV=0.5eV$
$KE_{2}=2.5eV-0.5eV=2eV$
$\frac{K E_{1}}{K E_{2}}=\frac{0 . 5 eV}{2 eV}=\frac{1}{4}$
$\frac{v_{1}}{v_{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$