The energy change for an electronic transition in a one-electron atom or ion H, He⁺, Li²⁺, etc.) from n _initial to n_final Is given by , $\left(-2.178 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n_{\text {final }} 2}-\frac{1}{n_{\text {initial }} 2}\right)$
where Z is the atomic number.
Which one of the following species will have the longest wavelength emission line for the transition between the n_initial = 2 and n_final = 1 levels?

Question

The energy change for an electronic transition in a one-electron atom or ion H, He⁺, Li²⁺, etc.) from n _initial to n_final Is given by ,

\left(-2.178 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n_{\text {final }} 2}-\frac{1}{n_{\text {initial }} 2}\right)

where Z is the atomic number.
Which one of the following species will have the longest wavelength emission line for the transition between the n_initial = 2 and n_final = 1 levels?

Quizplus · Accepted Answer

The wavelength of the emission line is inversely proportional to the energy change. The energy change formula shows that the energy change is directly proportional to $Z^2$. Therefore, the atom with the smallest $Z$ (atomic number) will have the smallest energy change and thus the longest wavelength. Hydrogen (H) has the smallest atomic number (Z=1) among the given options.