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 $h=6.63\times10^{-34}\,J\,s)$

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

Answer 1

Answer: (A)

Given threshold frequency, $\upsilon_{0}=3.3\times10^{14}\,Hz$
Frequency of incident light, $\upsilon=8.2\times10^{14}\,Hz$
As $eV_{0}=h\left(\upsilon-\upsilon_{0}\right)$ or $V_{0}=\frac{h\left(\upsilon-\upsilon_{0}\right)}{e}$
$V_{0}=\frac{6.63\times10^{-34}\left(8.2\times10^{14}-3.3\times10^{14}\right)}{1.6\times10^{-19}}=2V$