The electron, in a hydrogen atom, initially in a state of quantum number n₁, makes a transition to a state whose excitation energy, with respect to the ground state, is 10.2eV. If the wavelength, associated with the photon emitted in this transition, is 487 .5 nm, find the value of the quantum number, n₁ of the electron in its initial state.

Energy of an electron in the n^th orbit of H-atom,

$\begin{array}{rcl}{E}_n& =& -\frac{13.6}{n^2}\mathrm{eV}\\ {\mathrm{E}}_1& =& -13.6\mathrm{eV},{\mathrm{E}}_2=-3.4\mathrm{eV},{\mathrm{E}}_3=-1.51\mathrm{eV},\dots \end{array}$

$\text{ Clearly, }{E}_2-E_1=-3.4-(-13.6)=10.2\mathrm{eV}$

Thus the energy state n=2 has an excitation energy of 10.2 eV with respect to the ground state. Hence the electron is making a transition from state n = n₁ to the state n=2, where n₁ > 2.

$\text{ Now, }{E}_{n_1}-E_2=\frac{hc}{\lambda }=\frac{12400}{4875}\mathrm{eV}=2.55\mathrm{eV}$

$E_{n_1}=E_2+2.55\mathrm{eV}=-3.4+2.55=-0.85\mathrm{eV}$

$\begin{array}{rcl}\text{ Also, }{E}_{n_1}& =& -\frac{13.6}{n_1^2}\mathrm{eV}\\ \therefore -\frac{13.6}{n_1^2}& =& -0.85\\ \text{⇒ }{n}_1& =& \sqrt{\frac{13.6}{0.85}}=\sqrt{16}=4\end{array}$