# Light from a discharge 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 photoelectric emission,(b) the quantum numbers of the two levels involved in the emission of these photons,(c) the change in the angular momentum of the electron in the above transition, and(d) the recoil speed of the emitting atom assuming it to be at rest before the transition.(Ionization potential of hydrogen is 13.6 eV)

Kevin Nash
7 years ago
Hello Student,
. (a) The energy of photon causing photoelectric emission
= work function of sodium metal + KE of the fastest photoelectron
= 1.82 + 0.73 = 2.55 eV
(b) We know that E = - 13.6 / n2 eV / atom for hydrogen atom.
Let electron jump from n2 to n1 then
$_{E_n}__2$- $_{E_n}__1$= - 13.6 / n22 – ( 13. 6 / n21)
⇒ 2.55 = 13. 6 (1/ n21 – 1/ n22)
By hit and trial we get n2 = 4 and n1 = 2
[angular momentum mvr = nh/ 2π]
(c) Change in angular momentum
= n1 h / 2 π – n2h / 2 π = h / 2 π (2 – 4 ) = h / 2 π x (-2) = - h / x
(d) The momentum of emitted photon can be found by de Broglie relationship
λ = h / p ⇒ p = h / λ = hc / c = E / c ∴ p = 2.55 x 1.6 x 10-19 / 3 x 108
Note : The atom was initially at rest the recoil momentum of the atom will be same as emitted photon (according to the conservation of angular momentum).
Let m be the mass and v be the recoil velocity of hydrogen atom then
m x v= 2.55 x 1.6 x 10-19 / 3 x 108
⇒ v = 2.55 x 1.6 x 10-19 / 3 x 108 x 1.67 x 10-27 = 8.14 m/s
Thanks
Kevin Nash