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 photoelectric emission is nearly:

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

Answer 1

The correct option is A $2V$

From photoelectric equation,$K.E.=hv-h{v}_{th}=e{V}_0(V_0=\text{cutoff voltage})$

$V_0=\frac{h}{e}(8.2\times {10}^{14}-3.3\times {10}^{14})$

$=\frac{6.6\times {10}^{-34}\times 4.9\times {10}^{14}}{1.6\times {10}^{-19}}\approx 2V$.