An electron in a hydrogen atom jumps from the third excited state ($n=4$) to the ground state ($n=1$). What is the wavelength $\lambda$ of the emitted photon? (Take $R=1.097\times 10^7$ m${}^{-1}$)

Solution

1. Identify the transition parameters: the ground state $n_1=1$ and the third excited state $n_2=4$.
2. Apply the Rydberg formula: $$\begin{array}{rl}\frac{1}{\lambda }& =R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\\ & =1.097\times 10^7{\text{ m}}^{-1}\left(\frac{1}{1^2}-\frac{1}{4^2}\right)\end{array}$$.
3. Simplify the expression: $\frac{1}{\lambda }=1.097\times 10^7\times \frac{15}{16}\approx 1.0284\times 10^7{\text{ m}}^{-1}$.
4. Calculate the wavelength: $$\begin{array}{rl}\lambda & =\frac{1}{1.0284\times 10^7}\approx 9.72\times 10^{-8}\text{ m}\\ & =97.2\text{ nm}\end{array}$$.
5. Hence the answer is (A).