Radiation emitted from the n = 2 to n = 1 transition of hydrogen...

The energy of the photon emitted during the transition of a hydrogen atom from $n = 2$ to $n = 1$ is given by:

$E = 13.6 \times 1^2 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) = 13.6 \times \dfrac{3}{4} \text{ eV}$

This energy is absorbed by a helium ion ($He^+$) with atomic number $Z = 2$. Let the transition be from state $n_1$ to $n_2$. The energy required for this transition is:

$E' = 13.6 \times 2^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) = 13.6 \times 4 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)<br><br>Equating the two energies,$E = E'$:<br><br>$13.6 $\times \dfrac{3}{4} =$13.6 $\times$4 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)$<br><br>$\dfrac{3}{16} = \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}$<br><br>The helium ions are initially in$n_1 = 1$or$n_1 = 2$states.<br><br>If$n_1 = 1$:<br><br>$\dfrac{3}{16} = 1 - \dfrac{1}{n_2^2} \Rightarrow \dfrac{1}{n_2^2} = \dfrac{13}{16}$<br><br>This does not yield an integer value for$n_2$, so this transition is not possible.<br><br>If$n_1 = 2$:<br><br>$\dfrac{3}{16} = \dfrac{1}{2^2} - \dfrac{1}{n_2^2} = \dfrac{1}{4} - \dfrac{1}{n_2^2}$<br><br>$\dfrac{1}{n_2^2} = \dfrac{1}{4} - \dfrac{3}{16} = \dfrac{1}{16}$<br><br>$n_2^2 = 16 \Rightarrow n_2 = 4$<br><br>Thus, the possible transition for the helium ion is from$n = 2$to$n = 4$.$