Question

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380 nm to 780 nm).

Answer 1

Given:
Possible transitions:
From n₁ = 1 to n₂ = 3
n₁ = 2 to n₂ = 4

(a) Here, n₁ = 1 and n₂ = 3
Energy, $E=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
$$\begin{gathered} E=13.6\left(\frac{1}{1}-\frac{1}{9}\right) \\ =13.6\times \frac{8}{9}.....\left(1\right) \\ \mathrm{Energy}\left(E\right)\mathrm{is}\mathrm{also}\mathrm{given}\mathrm{by} \\ E=\frac{hc}{\mathrm{\lambda }} \end{gathered}$$

Here, h = Planck constant
c = Speed of the light
$\lambda$ = Wavelength of the radiation
$$\begin{gathered} \therefore E=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{\lambda }.....\left(2\right) \\ \mathrm{Equating}\mathrm{equation}s\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have} \\ \Rightarrow \mathrm{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times 9}{13.6\times 8} \\ =0.027\times {10}^{-7}=103\mathrm{nm} \end{gathered}$$

(b) Visible radiation comes in Balmer series.
As 'n' changes by 2, we consider n = 2 to n = 4.

Energy, $E_1=13.6\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$
$$\begin{gathered} =13.6\times \left(\frac{1}{4}-\frac{1}{16}\right) \\ =2.55\mathrm{eV} \end{gathered}$$
If ${\lambda }_1$ is the wavelength of the radiation, when transition takes place between quantum number n = 2 to n = 4, then
255 = $\frac{1242}{{\lambda }_1}$
or λ_1​ = 487 nm