Question

For a clean metallic surface, it is known that its threshold frequency is $3.3\times 10^{14} \, Hz$ . Calculate the cut-off voltage for photoelectric emission if the light of frequency $8.2\times 10^{14} \, Hz$ is incident on the same metallic surface. (Given $ \, h=6.63\times 10^{- 34} \, J \, s$ )

• A. $2 \, V$
• B. $4 \, V$
• C. $6 \, V$
• D. $8 \, V$

Answer 1

Answer: (A)

Given threshold frequency, $\nu_{0}=3.3\times 10^{14} \, Hz$
Frequency of incident light, $\nu=8.2\times 10^{14} \, Hz$
As $eV=h\left(\nu - \left(\nu\right)_{0}\right)$ or $V_{0}=\frac{h \left(\right. \nu - \left(\nu\right)_{0} \left.\right)}{e}$
$V_{0}=\frac{6.63 \times \left(10\right)^{- 34} \left(\right. 8.2 \times \left(10\right)^{14} - 3.3 \times \left(10\right)^{14} \left.\right)}{1.6 \times \left(10\right)^{- 19}}=2 \, V$