Question

Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo-electrons emitted from sodium is $0.73eV$. The work function for sodium is $1.82eV$. Find
(a) the energy of the photons causing 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 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 $13.6volt$ and the mass of the hydrogen atom is $1.67\times {10}^{-27}kg,1eV=1.6\times {10}^{-19}J$).

Answer 1

(a) According to Einsteins photo-electric equation, the maximum kinetic energy $E_K$ of the emitted electrons is given by
$E_K^{max}=hv-W$,
where $hv$ is the energy of photons causing the photo-electric emission and $W$ is the work-function of the emitting surface.
Given that, $E_K^{max}=0.73eV$ and $W=1.82eV$
$\therefore hv=E_K^{max}+W=0.73eV+1082eV=2.55eV$
(b) These photons (whose energy is $2.55eV$)are hydrogen atoms.
$As{(I.E.)}_H=13.6eV$, Hence $E_1^H=-{(I.E.)}_H=-13.6eV$
The energy of higher levels is given by
$E_n^H=\frac{E_1^H}{n^2}$
Hence, $E_2^H=\frac{13.6}{4}=-3.4eV,E_3^H=-\frac{13.6}{9}=-1.5eV$
and $E_4^H=-\frac{13.6}{16}=-0.85eV$
The energy of the emitted photon is $2.55eV$.
Now $E_4^H-E_2^H=-0.85eV-\left(3.4eV\right)=2.55eV$
Thus the quantum number of two levels involved in the emission of photon of energy $22.55eV$ are $4$ and $2$.
(c) The electron transition causing the emission of photon of energy $2.55eV$ is from $n=4$ level to $n=2$ level. Now, according to Bohrs $2^{nd}$ postulate, the angular momentum of electron in the hydrogen atom is $\left(nh/2\pi \right)$. Thus, the change in angular momentum in the above transition is
$\Delta L=\frac{4h}{2\pi }-\frac{2h}{\pi }=\frac{h}{\pi }$
(d) The momentum of the photon emitted from the hydrogen atom
$p_{ph}=\frac{hv}{c}=\frac{2.55\times (1.6\times {10}^{-19})J}{3\times {10}^{8}m/s}$
$=1.36\times {10}^{-27}Kg.m/s$
According to the law of conservation of momentum, the recoil momentum of a hydrogen atom will be equal and opposite to the momentum of the emitted photon. $({\overline{P}}_{ph}+{\overrightarrow{P}}_A=0or{\overrightarrow{P}}_A-{\overrightarrow{P}}_{ph})$
Hence the recoil speed of the atom is
$v=\frac{\left|Momentum\right|}{mass}=\frac{\left|{\overrightarrow{p}}_A\right|}{m_A}=\frac{\left|{\overrightarrow{p}}_{ph}\right|}{m_A}$
$=\frac{1.36\times {10}^{-27}Kg-m/s}{1.67\times {10}^{-27}Kg}=0.814m/s$.