Question

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

• A. $5\text{V}$
• B. $2\text{V}$
• C. $3\text{V}$
• D. $1\text{V}$

Answer 1

The correct option is B $2\text{V}$

Given:
${\nu }_{th}=3.3\times {10}^{14}\text{Hz}$.
$\nu =8.2\times {10}^{14}\text{Hz}$

we know that,
$K.E.=hv-h{v}_{th}=e{V}_0$

Where, ($V_0=$ cut-off voltage)

$\Rightarrow V_0=\frac{h}{e}(8.2\times {11}^{14}-3.3\times {10}^{14})$

$\Rightarrow V_0=\frac{6.6\times {10}^{-34}\times 4.9\times {10}^{14}}{1.6\times {10}^{-19}}\approx 2\text{V}$

Hence, option $\left(A\right)$ is correct.