Question

A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?

Answer 1

Given:
Minimum wavelength of the light component present in the beam, ${\lambda }_1$ = 450 nm
Energy associated $\left(E_1\right)$ with wavelength $\left({\lambda }_1\right)$ is given by
E₁ = $\frac{hc}{{\lambda }_1}$
Here,
c = Speed of light
h = Planck's constant
$\therefore E_1=\frac{1242}{450}$
= 2.76 eV

Maximum wavelength of the light component present in the beam, ${\lambda }_2$ = 550 nm

Energy associated $\left(E_2\right)$ with wavelength $\left({\lambda }_2\right)$ is given by
E₂ = $\frac{hc}{{\lambda }_2}$
$\therefore E_2=\frac{1242}{550}=2.228=2.26\mathrm{eV}$
The given range of wavelengths lies in the visible range.
∴ n₁ = 2, n₂ = 3, 4, 5 ...
Let E'₂ , E'₃ , E'₄ and E'₅ be the energies of the 2nd, 3rd, 4th and 5th states, respectively.
$$\begin{gathered} E{\text{'}}_2-E{\text{'}}_3=13.6\left(\frac{1}{4}-\frac{1}{9}\right) \\ =\frac{12.6\times 5}{30}=1.9\mathrm{eV} \\ E{\text{'}}_2-E{\text{'}}_4=13.6\left(\frac{1}{4}-\frac{1}{16}\right) \\ =2.55\mathrm{eV} \\ E{\text{'}}_2-E{\text{'}}_5=13.6\left(\frac{1}{4}-\frac{1}{25}\right) \\ =\frac{10.5\times 21}{100}=2.856\mathrm{eV} \end{gathered}$$
Only, E'₂ − E'₄ comes in the range of the energy provided. So the wavelength of light having 2.55 eV will be absorbed.
$$\begin{gathered} \mathrm{\lambda }=\frac{1242}{2.55}=487.05\mathrm{nm} \\ =487\mathrm{nm} \end{gathered}$$
The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength ​487 nm will have less intensity in the transmitted beam.