UV radiation of 6.2 eV falls on aluminium surface of work function 4.2 eV. The stopping potential for the surface is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 V
Using Photoelectric equation,
$eV_s=K.E.max=E_{photon}-\phi$
$⇒eV_s=6.2-4.2$
$⇒eV_s=2eV$
$⇒V_s=\frac{2}{e}=\frac{2}{1}=2V$