Discussion
In a hydrogen atom, the transition of an electron from the energy level,\(n_{2} = \infty\) and \(n_{1} = 3\) is the
Answer: A
The transition gives the shortest wavelength of the Paschen series.
The Paschen series contains the transitions ending on \(n_{1} = 3\). The photon energy is largest, and so the wavelength shortest, when the electron falls from the highest level, \(n_{2} = \infty\) (the series limit):
\[ \begin{aligned} \frac{1}{\lambda} \\ &= R\left(\frac{1}{3^{2}} - \frac{1}{\infty^{2}}\right) \\ &= \frac{R}{9} \end{aligned} \]
The longest Paschen wavelength comes from \(4 \to 3\). The Balmer series ends on \(n = 2\) and the Brackett series on \(n = 4\), so they do not contain this transition.
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