Looking for classes? Ksquare Career Institute, Bengaluru →
The threshold frequency for a photosensitive metal is $3.3\times10^{14}$ Hz. If light of frequency $8.2\times10^{14}$ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
A
5 V
B
1 V
C
2 V
D
3 V
Detailed Solution
Einstein's equation: $eV_0 = h\nu - h\nu_0 = h(\nu - \nu_0)$
$\nu - \nu_0 = (8.2 - 3.3)\times10^{14} = 4.9\times10^{14}$ Hz
$V_0 = \frac{h(\nu - \nu_0)}{e} = \frac{6.63\times10^{-34}\times4.9\times10^{14}}{1.6\times10^{-19}}$
$V_0 = \frac{3.25\times10^{-19}}{1.6\times10^{-19}} = 2.03$ V
$V_0 \approx 2$ V
$\nu - \nu_0 = (8.2 - 3.3)\times10^{14} = 4.9\times10^{14}$ Hz
$V_0 = \frac{h(\nu - \nu_0)}{e} = \frac{6.63\times10^{-34}\times4.9\times10^{14}}{1.6\times10^{-19}}$
$V_0 = \frac{3.25\times10^{-19}}{1.6\times10^{-19}} = 2.03$ V
$V_0 \approx 2$ V