Question

The threshold frequency of a certain metal is $3.3\times 10^{14}Hz$ If light of frequency $8.2\times 10^{24}Hz$ is incident on the metal, then the cut off voltage for photoelectric emission is (Given $h=6.63\times 10^{-34}Js)$

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

Answer 1

Answer: (A)

Given threshold frequency, ${\upsilon }_0=3.3\times 10^{14}Hz$
Frequency of incident light, $\upsilon =8.2\times 10^{14}Hz$
As $e{V}_0=h\left(\upsilon -{\upsilon }_0\right)$ or $V_0=\frac{h\left(\upsilon -{\upsilon }_0\right)}{e}$
$V_0=\frac{6.63\times 10^{-34}\left(8.2\times 10^{14}-3.3\times 10^{14}\right)}{1.6\times 10^{-19}}=2V$