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

487 nm wavelength will be absorbed .

Explanation:

The energies associated with 450 nm radiation  =  1242  / 450  =2.76 eV

Energy associated with 550 nm radiation  =  1242  / 550  =2.228=2.26eV.

The light comes under visible range,

Thus, n1 = 2 , n1 = 3,4,5,6 .....

E2 - E3 = 13.6 (1/4 - 1/9)

E2 - E3 = 12.6 x 5/ 30 = 1.9 eV

E2 - E4 = 13.6 (1/4 - 1/16) = 2.55 eV

E2 - E5 = 13.6 (1/4 - 1/25) = 2.85 eV

Only E2 - E4 comes in the range of energy provided. So the wave length corresponding to that energy will be absorbed.

λ = 1242 / 2.55 = 487 nm

Hence 487 nm wavelength will be absorbed .

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Answer 2

In the transmitted beam, the wavelength will have the least intensity of 487 nm.

Explanation:

• In the beam, the light component has the minimum wavelength, λ1 = 450 nm. Energy (E1) related to wavelength (λ1) is distributed between 550 nm and 450 nm and is shown as E1 = hc/ λ1.
• Here, h is the constant of Planck and c is the light’s speed. Therefore, E1 = 1242/450, which is equal to 2.76 eV.

       Using the same formula, $\mathrm{E} 2=\frac{1242}{550}=2.26 \mathrm{eV}$

• The wavelengths’ range in the visible range is given as n1 = 2, n2 = 3, 4, 5 and so on.
• Then, E2 – E3 becomes 1.9 eV, E2 – E4 becomes 2.55 eV and E2 – E2 becomes 2.856 eV.
• There will be the absorption of the light’s wavelength that has 2.55 eV.

        So, $\lambda=\frac{1242}{2.55}=487.5 \mathrm{nm}$

• So, in the transmitted beam, the wavelength 487 nm will have less intensity.