Question

If the frequency of the incident radiation is increased from $4\times {10}^{15}Hz$ to $8\times {10}^{15}Hz$, by how much will the stopping potential for a given photosensitive surface go up?

Answer 1

$e{V}_0=hv-h{v}_0$
$=h(v-v_0)$
$=6.6\times {10}^{-34}(8\times {10}^{15}-4\times {10}^{15})$
$e{V}_0=2.64\times {10}^{-18}$
$V_0=\frac{2.64\times {10}^{-18}}{1.6\times {10}^{-19}}$
$V_0=16.5volt$