Light quanta with energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electron flies out.

Explanation

According to Einstein's photoelectric equation
`E=(1)/(2)mv_(max)^(2) = hv-phi = 4.9-4.5 = 0.4 eV`
If E be the energy of each ejected photo electron momentum of electrons is P = `sqrt(2mE)`
We know that change of momentum is impulse. Here the whole momentum of electron is gained when it is ejected out thus impulse on surface is
Impulse= `sqrt(2mE)`
Substituting the values, we get
Maximum impulse = `sqrt(2xx9.1xx10^(-31)xx0.4xx1.6xx10^(-19)) = 3.45xx10^(-25) kg m//sec`