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, λ₁ = 450 nm Energy associated (E₁) with wavelength (λ₁ ) is given by

E₁ =  ${\displaystyle \frac{hc}{{\lambda }_1}}$

Here,

c = Speed of light

h = Planck's constant

${\displaystyle \therefore E_1=\frac{1242}{450}}$

= 2.76 eV

Maximum wavelength of the light component present in the beam,  = 550 nm

Energy associated (E₂) with wavelength  is given by

${\displaystyle E_2=\frac{hc}{\lambda }}$

∴ ${\displaystyle E_2=\frac{1242}{550}=2.228=2.26 eV}$

The given range of wavelengths lies in the visible range.

${\displaystyle \therefore n_1=2,n_2=3,4,5....}$

Let E'₂ , E'₃ , E'₄ and E'₅ be the energies of  the 2nd, 3rd, 4th and 5th states, res pectively.

${\displaystyle E_2-E_3=13.6(\frac{1}{4}-\frac{1}{9})}$

=${\displaystyle \frac{12.6\times 5}{30}=1.9 eV}$

${\displaystyle E_2-E_4=13.6(\frac{1}{4}-\frac{1}{16})}$

= 2.55 eV

${\displaystyle E_2-E_2=13.6(\frac{1}{4}-\frac{1}{25})}$

= ${\displaystyle \frac{10.5\times 21}{100}=2.856 eV}$

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.

${\displaystyle \lambda =\frac{1242}{2.55}=487.5 nm}$

= 487 nm

The wavelength 487 nm will be absorbed by hydrogen gas. So, wavelength ​487 nm will have less intensity in the transmitted beam.