Question

The equilibrium separation of H atoms in the H2 molecule is 0.074 nm, calculate the energies and wavelengths of photons for the rotational transitions (a) L = 1 to L = 0, (b) L = 2 to L = 1, and (c) L =3 to L = 2

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asked 2021-03-22
The equilibrium separation of H atoms in the H2 molecule is 0.074 nm, calculate the energies and wavelengths of photons for the rotational transitions (a) L = 1 to L = 0, (b) L = 2 to L = 1, and (c) L =3 to L = 2

Answers (1)

2021-03-24

in order to find the energy differences you need to find the rotational quantity:
\(\frac{hbar^2}{m\cdot Req^2}=\frac{c^2hbar^2}{Req^2mc^2}=\frac{hc^2}{Req^24\pi^2mc^2}\)
\(\displaystyle{\frac{{{\left({1240}{e}{v}\cdot{n}{m}\right)}}}{{{\left({0.5}\cdot{1.008}{u}\cdot{931.5}{M}{e}\frac{{v}}{{u}}\right)}{\left({0.072}{n}{m}\right)}^{{24}}\pi^{{2}}}}}={0.0152}{e}{V}\)
Now, the corresponding wavelengths can be found via the following equation:
\(\displaystyle\lambda={\frac{{{h}{e}}}{{\triangle{E}}}}\)
\(\displaystyle{L}={1}\to{L}={0}\triangle{E}={0.0152}{e}{V}\lambda={81.6}\mu{m}\)
\(\displaystyle{L}={2}\to{L}={1}\triangle{E}={0.0304}{e}{V}\lambda={40.8}\mu{m}\)
\(\displaystyle{L}={3}\to{L}={2}\triangle{E}={0.0456}{e}{V}\lambda={27.2}\mu{m}\)
You can also note that the wavelength is halved each time atransition occurs. Accordingly, the energy is \(\displaystyle\triangle{E},{2}\triangle{E},{3}\triangle{E},{e}{t}{c}\)

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