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 cutoff voltage for the photoelectric emission is nearly :

• A. 1V
• B. 2V
• C. 3V
• D. 5V

Answer 1

Answer: (B)

2V

Given $v_0=3.3\times {10}^{14}$, $v=8.2\times {10}^{14}$, $V_0=?$

Using Einstein's photoelectric equation, $e{V}_0=hv-h{v}_0$

$\Rightarrow 1.6\times {10}^{-19}{V}_0=6.626\times {10}^{-34}(8.2-3.3)\times {10}^{14}$ $\Rightarrow V_0=2.029\approx 2V$