Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 nm falls on a metal having work function 2.5e V.

Explanation

Energy of photoelection : ` rArr 1/ 2 mv^(2) = hc / lambda - hv_1`
` = 4.14 xx 10^(-15) xx 3 xx 10^(8) / 4 xx 10^(-7) - 2 .5 eV`
` = 0.605eV`
We know , KE ` = P^(2) / 2m`
`rArr P^(2) = 2 m xx KE `
P^(2) = 2 xx 9.1 xx 10^(-31) `
` xx 0.605 xx 1.6 xx 10^(-19) `
`P = 4.197 xx 10^(-25) kg m / s ` .