Question

The threshold frequency for a photo-sensitine 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 photo-electric emission is nearly

• A. $2V$
• B. $3V$
• C. $5V$
• D. $1V$

Answer 1

The correct option is A $2V$

Here, $V_0=\frac{E-V}{e}=\frac{h(v-v_0)}{e}$
$=\frac{6.62\times {10}^{-34}(8.2\times {10}^{14}-3.3\times {10}^{14})}{1.6\times {10}^{-19}}$
$=\frac{6.62\times {10}^{-34}}{1.6}\times 4.9\times {10}^{33}$
$=\frac{6.62\times 4.9\times {10}^{-1}}{1.6}$
$V_0=2$ volt