The Bohr Model of the Hydrogen Atom — Question 3

Question 3

A hydrogen emission line follows the difference of two negative energies

A hydrogen electron undergoes the transition $n=4\to n=2$. Neglect recoil and fine structure. Use the nonrelativistic Bohr model with an infinitely heavy nucleus unless stated otherwise: $E_n=-13.6Z^2/n^2\,\mathrm{eV}$, $r_n=a_0n^2/Z$, $a_0=.0529\,\mathrm{nm}$, and $hc=1240\,\mathrm{eV\,nm}$. Here Z is the nuclear charge number.

1. Find the initial and final electron energies and emitted photon energy.

2. Find the photon wavelength and the fractional decrease in the Bohr radius.

Question 3 - Solution

A hydrogen emission line follows the difference of two negative energies

(a) Method and calculation

$$\boxed{E_4=-.850\,\mathrm{eV},\quad E_2=-3.400\,\mathrm{eV}}.$$$$\boxed{E_\gamma=E_4-E_2=2.550\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$\boxed{\lambda=hc/E_\gamma=486.275\,\mathrm{nm}}.$$$$r_2/r_4=4/16=1/4\Rightarrow\boxed{(r_4-r_2)/r_4=3/4}.$$

Physical check

The electron energy becomes more negative during emission. The photon energy is positive because the initial energy is higher than the final energy.