Question

The threshold frequency for a photosensitive metal is $3.3 \times 10^{14}\, Hz. $ If light of frequency $8.2 \times 10^{14}\, Hz. $ is incident on this metal, the cut-off voltage for the photoelectron emission is nearly

• A. 1 V
• B. 2 V
• C. 3 V
• D. 5 V

Answer 1

Answer: (B)

$V _{0}=\frac{ E _{ Ph }- W }{ e }=\frac{ h \left( v - v _{0}\right)}{ e }$
$=\frac{6.62 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}$
$=\frac{6.62 \times 10^{-34}}{1.6} \times 4.9 \times 10^{14+19}$
$=\frac{6.62 \times 4.9}{1.6} \times 10^{-1}=2$ volt