Question

In $H{e}^{+}$, the electron is initially present in an orbital that has no radial nodes but in which the sign of the wave function changes once across a plane. On absorbing a photon with an energy which is 75% of the ground state energy of hydrogen, the electron moves to a higher orbital. Find the sum of principal quantum numbers of the initial and final orbitals of the electron.

Answer 1

Initially, electron is in orbital with no radial nodes,
So $n-l-1=0$; and wavefunction changes sign once across a plane i.e., there is one angular node so $l=1$
$\Rightarrow n=2$
Hence initially electrons is in $2p$.
Now this electron absorbs $\frac{75}{100}\times 13.6\text{eV}$ or $\frac{3}{4}\times 13.6\text{eV}$ energy to move to higher orbit (shell)
In $H{e}^{+}$,
$E_1=-4\times 13.6\text{eV}$
$E_2=\frac{-4\times 13.6}{4}\text{eV}=-13.6\text{eV}$
$E_3=-\frac{4\times 13.6}{9}\text{eV}$
$E_4=-\frac{4\times 13.6}{16}\text{eV}=-\frac{13.6}{4}\text{eV}$
Hence we can see that $E_4-E_2=\frac{3}{4}\times 13.6\text{eV}$
Hence electron goes to the $4^{\text{th}}$ shell.
So, $n_{\text{initial}}+n_{\text{final}}=2+4=6$