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 =  `(hc)/(lamda_1)`

Here,

c = Speed of light

h = Planck's constant

`therefore E_1 = (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

`E_2 = (hc)/( lamda)`

∴ `E_2 = 1242/550 = 2.228 = 2.26  eV`

The given range of wavelengths lies in the visible range.

`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.

`E_2 - E_3 = 13.6(1/4 - 1/9)`

=`(12. 6 xx 5)/30=1.9  eV`

`E_2- E_4 = 13.6 (1/4 - 1/16)`

= 2.55 eV

`E_2-E_2 = 13.6 (1/4 - 1/25)`

= `(10.5xx21)/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.

`lamda = (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.