Question

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

Answer 1

Formula used: $E=h(v-v_0)=e{V}_0$
Given, threshold frequency of the metal, $v_0=3.3\times {10}^{14}\text{Hz}$
Frequency of light incident on the metal, $v=8.2\times {10}^{14}\text{Hz}$ If the cut - off voltage for the photoelectric emission from the metal is $V_0$, then the equation for the cut-off energy is given by
$E=h(v-v_0)=e{V}_0$
Here,
Charge on an electron, $e=1.6\times {10}^{-19}C$
Plancks's constant, $h=6.626\times {10}^{-34}Js$
Therefore, the cut - off voltage for the photoeelctric emission is
$V_0=\frac{h(v-v_0)}{e}$
$V_0$
$=\frac{6.626\times {10}^{-34}\times (8.2\times {10}^{14}-3.3\times {10}^{14}}{1.6\times {10}^{-19}}$
$V_0=2.029V$
Final answer : $2.029V$