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 $h=6.63\times 10^{-34}Js$ )

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

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 $V_0=\frac{h(\nu -{\left(\nu \right)}_0)}{e}$
$V_0=\frac{6.63\times {\left(10\right)}^{-34}(8.2\times {\left(10\right)}^{14}-3.3\times {\left(10\right)}^{14})}{1.6\times {\left(10\right)}^{-19}}=2V$