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:

The ejected electrons have a distribution of KE, the most energetic ones having KE equal to $(hv-hv_0)$

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