Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium . The kinetic energyof the fastest photonelectrons emittede from from sodium is `0.73 eV` . The work function for sodium is `1.82 eCV` find
(a) the energy of the photons causing the photoelectric emission ,
(b) the quantum number of the two level inveloved in the emission of there photons,
(c ) the change in the angular momentum of the electron in the hydrogen atom in the above transition and,
(d)the recoll speed of emitted atom assuming it to be at rest before the transition.
(lonization potential of hydrogen is `13.6 eV`)

Answer

A, B, C, D

Explanation

The energy of photon causing photoelectric emission = work function of sodium metal `= K.E. ` of the festest photoelectron
`= 1.82 + 0.73 = 2.55 eV`
(b) we know that `E_(0) = (-13.6)/(n^(2)) (eV)/(atom)` for atom
Let electron jump from n_(2) to n_(1)` then
E_(n_2) - E_(n_1) = (13.6)/(n_(2)^(2)) - ((13.6)/(n_(2)^(2)))`
`rArr 2.55 = 13.6 ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
By hit and trial we get `n_(2) = 4 and n_(1) = 2`
[anhular momentum `m v r = (nh)/(2 pi)]`
Change in angular momentum
`= (m_(1)h)/(2 pi) - (m_(2)h)/(2 pi) = (h)/(2 pi) (2 - 4) = (h)/(2 pi) xx (-2) = (h)/(pi) `
The momentum of emitted photon can be found by de Broge relationship
`lambda(h)/(p) rArr P = (h)/(lambda) = (hv)/(c) =(E)/(c) :. (2.55 xx 1.6 xx 10^(-19))/(3 xx 10^(8))`