Question

A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.

Answer 1

(a)
The binding energy of hydrogen is given by
$E=\frac{13.6}{n^2}\mathrm{eV}$
For binding energy of 0.85 eV,
$$\begin{gathered} {n_2}^2=\frac{13.6}{0.85}=16 \\ {n}_2=4 \end{gathered}$$
For binding energy of 10.2 eV,
$$\begin{gathered} {n_1}^2=\frac{13.6}{10.2} \\ {n}_1=1.15 \\ \Rightarrow n_1=2 \end{gathered}$$
The quantum number of the upper and the lower energy state are 4 and 2, respectively.

(b) Wavelength of the emitted radiation $\left(\lambda \right)$ is given by

$\frac{1}{\lambda }=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$

Here,
R = Rydberg constant
n₁ and n₂ are quantum numbers.

$$\begin{gathered} \therefore \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{1}{4}-\frac{1}{16}\right) \\ \Rightarrow \mathrm{\lambda }=\frac{16}{1.097\times 3\times {10}^7} \\ =4.8617\times {10}^{-7} \\ =487\mathrm{nm} \end{gathered}$$