Question

What is the energy and frequency of a photon emitted during a transition from n=5 state to n=2 state in the hydrogen atom?

Answer 1

According to the Rydberg formula:
$wavenumber=R(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2})$
Here R = 109677 cm^-1
n₂ = 5 and n₁ = 2
$$\begin{gathered} wavenumber=109677(\frac{1}{2^2}-\frac{1}{5^2}) \\ wavenumber=109677(\frac{1}{4}-\frac{1}{25}) \\ wavenumber=109677\times \frac{21}{100} \\ wavenumber=23032.17c{m}^{-1} \\ \lambda =\frac{1}{wavenumber}\lambda =\frac{1}{23032.17}\lambda =434.2\times 10 \end{gathered}$$