Question

When a certain metal is irradiated with a radiation of frequency $3.2\times {10}^{16}{S}^{-1}$, the photo electrons emitted had twice the kinetic energy as did the phototelectrons emitted when the same metal was irradiated with radiation of frequency $2\times {10}^{16}{S}^{-1}$. Calculate threshold frequency of metal.

Answer 1

Taking,

$X=h{\vartheta }_0$

$Y=\frac{1}{2}m{v}^2$

$X+2Y=6.626\times {10}^{-34}\times 3.2\times {10}^{16}$......(1)

$X+Y=6.626\times {10}^{-34}\times 2\times {10}^{16}.....\left(2\right)$

Then by solving the above equations we get,

$X=5.30\times {10}^{-18}$

$Y=7.94\times {10}^{-18}$

Threshold frequency$=h{\vartheta }_0=5.30\times {10}^{-18}$

${\vartheta }_0=\frac{5.30\times {10}^{-18}}{6.626\times {10}^{-34}}$

$=8\times {10}^{15}{s}^{-1}$ Hence the solution.