When photon of energy $hv$ falls on a photosensitive metallic surface (work function $hv_0$) electrons are emitted from the metallic surface. It is possible to say that:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The ejected electrons have a distribution of KE, the most energetic ones having KE equal to $(hv-hv_0)$
According to photoelectric equation,
$(KE)_{max}=hv-\phi$
where,
$v$ = frequency of incident light
$\phi$ = Work function