Question

In a single isolated atom of $He ^{+}$ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in $eV$ ) will be:

Answer 1

Since electron is coming from $3rd$ excited state to ground state by emitting only two photrons, it can be $4 \rightarrow 3$ and $3 \rightarrow 1$
Hence, higher energy value will be
$\Delta E_{3 \rightarrow 1}=(13.6-1.5) \times 2^{2}=48.4\, eV$