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?

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Solution

The energies associated with 450 nm radiation $= \frac{1242}{450} = 2.76 e v$
Energy associated with 550 nm radiation $\frac{1242}{550} = 2.26 e v$

The light comes under the visible range
Thus,
$n_{1} = 2, n_{2} = 3, 4, 5.......$
$E_{2} - E_{3} = 13.6 (1 / 2^{2} - 1 / 3^{2}) = 1.9 e v$
$E_{2} - E_{4} = 13.6 (1 / 2^{2} - 1 / 4^{2}) = 2.55 e v$
$E_{2} - E_{5} = 13.6 (1 / 2^{2} - 1 / 5^{2}) = 2.856 e v$

Only $E_{2} - E_{4}$ comes in the range of energy provided. So the wavelength corresponding to that energy will be absorbed.

$λ = \frac{1242}{2.55} = 487.05 n m = 487 n m$
Therefore, 487 nm wavelength will be absorbed.

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