The minimum energy required for the emission of photoelectron from the surface fo a metal is ` 4. 94 xx 10 ^(19) J`. Calcualte the critical reequency and the corresponding wavelength of the critical frequency and the corresponding wavelngth of the photon requred to ejerct the electron. `( h = 6. 6 xx 10^(-234) J esc^(-1))`.

Explanation

Minimum energy required for the emission of photoelectrons ` = 4. 95 xx 10^(-19) J`
Let the critical frequency or threshold frequency of the photon to eject the electon be `v sec ^(-1)` .
We know , ` E = hv or v = E/h = ( 4.95 xx 10^(-19))/( 6.5 xx 10^(34))`
` = 7. 5 xx 10^(14) s^(-1)`
Also ` lambda = c/ v = ( 3 xx 10^8 )/( 7 . 5 xx 10^(14))= 4xx 10^(-7) m` .