Question

The threshold frequency for certain metal is $3.3\times 10^{14}$ $Hz$ . If the 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. $4.9V$
• B. $3.0 \, V$
• C. $2.0 \, V$
• D. $1V$

Answer 1

Answer: (C)

From relation
$ \, \, \, eV_{s}=h\left(v - v_{0}\right)$
or $V_{s}=$ threshold or cut off voltage
$=\frac{h}{e}\left(v - v_{0}\right)$
$= \, \frac{6.6 \times \left(10\right)^{- 34}}{1.6 \times \left(10\right)^{- 19}} \, \left(\right.8.2-3.3\left.\right)\times \left(10\right)^{14}$
$ \, = \, \frac{6 .6 \times 4 .9 \times 10^{- 1}}{1 .6} \, = \, 2V$