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=0.73eV`,`W=1.82eV`
Ionization enrgy of H atom`=13.6eV`
a. `hv=W+K_k=1.82eV+0.73eV=2.55eV`
b. The electronic energy levels of H atoms are given by
`E_n=-(Rhc)/(n^2)=-(13.6)/(n^2)eV`
for `n=1`,`E_1=-13.6eV`
for `n=2`,`E_2=-3.4eV`
For `n=3`,`E_3=-1.51eV`
For `n=4`,`E_4=-0.85eV`
Clearly,`E_4-E_2=-0.85eV-(-3.4eV)=2.55eV`
i.e., Quantum numbers involved in the photons of energy `2.55eV` are 2 and 4. The transition is specified by
`n_1=4ton_2=2`.
c. The angular momentum of electron in H atom,
`J=n(h)/(2pi)`
`n=4`,`J_1=4(h)/(2pi)=(2h)/(pi)`
For `n=2`,`J_2=2(h)/(2pi)=(h)/(pi)`
Therefore, change in angular mumantum, `triangleJ=J_1-J_2=(2h)/(pi)-(h)/(pi)=(h)/(pi)`
d. According to conservation of momentum, `(hv)/(c)+mv=0`
[since momentum of photoelectron is negligible]
`v=-(hv)/(cm)`
`=-(2.55eV)/((3xx10^8ms^-1)xx(1.67xx10^-27kg))`
`=-(2.55eVxx1.6xx10^-19)/(3xx10^8xx1.67xx10^27)=-0.814ms^-1`
Recoil speed of H atom`=0.814ms^-1`