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 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 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

`E_(K) = 0.73 e V, W = 1.82 e V`
Lonization energy of `H` atom `= 13.6 e V`
a. `hv = W + E_(K) = 1.82 e V = + 0.73 e V = 2.55 e V`
b. The electronic energy levels of `H` -atom are given by
`E_(n) = - (Rhc)/(n^(2)) = - (13.6)/(n^(2)) e V`
`For n = 1, E_(1) = - 13.6 e V`
`For n = 2, E_(2) = - 3.4 e V`
`For n = 3, E_(3) = - 1.51 e V`
`For n = 4, E_(4) = - 0.85 e V`
Clearly `E_(4) - E_(2) = - 0.85 e V - (-3.4 e V) = 2.55 e V`
i.e. Quantum of electron in `H` atom
`J = n (h)/(2 pi)`
For `n = 4 J_(1) = 4 (h)/(2 pi) = (2 h)/(pi)`
For `n = 2 J_(2) = 2 (h)/(2 pi) = (2 h)/(pi)`
Therefore change angular momentum, `Delta J = J_(1) - J_(2) = (2h)/( pi) - (h)/(pi) = (h)/(pi)`
d. According to conservation of momentum,
`(hc)/(c) + mv = 0`
`:. v = - (hc)/(cm) = - (2.55)/((3 xx 10^(8)) xx (1.67 xx 10^(-27))`
`= - (2.55 xx 1.6 xx 10^(-19))/(3 xx 10^(8) xx 1.67 xx 10^(-27))`
`= - 0.814 m s^(-1)`
Therefore , recoil speed of `H` atom `= 0.814 m s^(-1)`