Light form a dicharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find (a) the energy of the photons causing the photoelectrons emission.
(b) the quatum numbers of the two levels involved in the emission of these photons.
(c ) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and
(d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (lonization potential of hydrogen is 13.6 eV.)

Explanation

(a) From Einstein's equation of photoelectric effect,
Energy of photons causing the photoelectric emission
= Maximum kinetic energy of emitted photons + work - function
or `E = K_(max) + W = (0.73 + 1.82 ) eV`
or `E=2.55 eV`
(b) in case of a hydrogen atom, `E_1 =- 13.6 eV, E_2 = - 3.4 eV, E_3 =- 1.5 eV`
`E_4 = - 0.85 eV`
Since, `E_4 - E_2 = 2.55 eV`
Therefore, quantum numbers of the two levels involeved in the emission of these photons are 4 and `2(4 rarr 2).
(c ) Change in angular momentum in transition form 4 to will be
`DeltaL= L_2 - L_4= 2((h)/(2pi)) -4 ((h)/(2pi)) or Delta L =- (h)/(pi)`
(d) From conservation of linear momentum
|Momentum of hydrogen atom | = |Momentum of emitted photon |
or ` mv= (E)/(c ) (m=mass of hydrogen atom)
or ` v= (E)/(mc) = ((2.55xx1.6xx10^(-19) J)/(1.67xx10^(-27) kg)(3.0xx10^8 m//s))`
`v = 0.814 m/s.`