Question

The maximum kinetic energy of the photo-electrons is found to be $6.63\times {10}^{-19}J$ when the metal is irradiated with a radiation of frequency $3\times {10}^{15}Hz$. The threshold frequency of the metal is?

Answer 1

Given that:-

Kinetic energy $=6.63\times {10}^{-19}J$

Frequency of radiation $\left(\nu \right)=3\times {10}^{15}Hz$

Planck's constant $\left(h\right)=6.63\times {10}^{-34}Js$

Therefore,

Absorbed energy $=h\nu =6.63\times {10}^{-34}\times 3\times {10}^{15}=6.63\times 3\times {10}^{-19}$

As we know that,

Absorbed energy $=$ Threshold energy $+$ Kinetic energy

Threshold energy $=$ Absorbed energy $-$ Kinetic energy

$\therefore h{\nu }^0=h\nu -K.E.$

Here, ${\nu }^0$ is the frequency of metal.

$\Rightarrow {\nu }^0=\frac{6.63\times 3\times {10}^{-19}-6.63\times {10}^{-19}}{6.63\times {10}^{-34}}$

$\Rightarrow {\nu }^0=\frac{(6.63\times {10}^{-19})\times (3-1)}{6.63\times {10}^{-34}}=2\times {10}^{15}Hz$

Hence the threshold frequency of the metal will be $2\times {10}^{15}Hz$.