Use $R_{H} = 1.097 \times 1 0^{7} m^{- 1}$ . The $n = 3 \to n = 2$ line of a certain hydrogen-like ion is observed at $72.9 nm$ . Find the atomic number $Z$ of the ion (nearest integer).

Numerical answer type, tolerance 0.1.

Answer: 3

1/ λ = Z^{2} R_{H} (1/4 - 1/9) = 5 Z^{2} R_{H} /36

, so

Z^{2} = 36/ (5 R_{H} λ) = 9.003

. Hence

Z \approx 3.00

, i.e.

Z = 3

(Li2+).