Question

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

Answer 1

Given: The threshold frequency of metal is $3.3\times {10}^{14}Hz$.

Frequency of the light incident on metal is $8.2\times {10}^{14}Hz$.

To find: The cut-off voltage for photoelectric emission.

The cut-off voltage is given by:

$e{V}_o=h(\nu -{\nu }_o)$

The cut-off voltage is $V_o$.

Here, ${\nu }_o=3.3\times {10}^{14}Hz$ and $\nu =8.2\times {10}^{14}Hz$

So,

$V_0=\frac{6.6\times {10}^{-34}(8.2\times {10}^{14}-3.3\times {10}^{14})}{1.6\times {10}^{-19}}\Rightarrow 2.012V$