Question

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

• A. 4.9 V
• B. 3.0 V
• C. 2.0 V
• D. 1.0 V

Answer 1

Answer: (C)

In photoelectric effect energy is conserved.
In photoelectric effect Einstein's equation is given by.
Photon energy $=KE $ of electron + work function i.e.,
$ hv=e{{V}_{s}}+h{{v}_{0}} $ where $ {{V}_{s}} $ is 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 the values in Eq. (i), we get
$ {{V}_{s}}=\frac{6.6\times {{10}^{-34}}\times (8.2\times {{10}^{14}}-3.3\times {{10}^{14}})}{1.6\times {{10}^{-19}}} $
$=\frac{6.6\times 4.9}{1.6}\times {{10}^{-1}}$
$=2.0V $