Question

The threshold frequency for certain metal is $3.3\times {10}^{14}Hz$. If light of $8.2\times {10}^{14}Hz$ frequency is incident on the metal, then the cut-off voltage of the photoelectric current will be :

• A. $3V$
• B. $4V$
• C. $2V$
• D. $2.5V$

Answer 1

Answer: (C)

In photoelectric effect, energy is conserved.

In photoelectric effect, Einstein's equation is given by

Photon energy $=$ kinetic energy of electron $+$ work function

i.e. $hv=e{V}_s+h{v}_0$

where, $V_s$ is the stopping potential and $v_0$ is the threshold frequency.

$\therefore V_s=\frac{h}{e}(v-v_0)$ .....(i)

Given, $v_0=3.3\times {10}^{14}Hz$, $v=8.2\times {10}^{14}Hz$

$h=6.6\times {10}^{-34}J-s$, $e=1.6\times {10}^{-19}C$

Substituting these values in equation (i), we get

$V_s=\frac{6.6\times {10}^{-34}(8.2\times {10}^{14}-3.3\times {10}^{14})}{1.6\times {10}^{-19}}$

$V_s=\frac{6.6\times 4.9}{1.6}\times {10}^{-1}=2V$

$\Rightarrow V_s=2V$