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

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Solution

The energies associated with 450nm radition

= $\frac{1242}{450} = 2.76 e V$

Energy associated with 550 nm radirion

= $\frac{1242}{550} = 2.228 = 2.26 e V$

The light comes under visible range, thus $n_{1} = 2, n_{1} = 3, 4, 5 . . . . . . . . . .$

$E_{2} - E_{3} = 13.6 (\frac{1}{4} - \frac{1}{9})$

= $\frac{12.6 \times 5}{30} = 1.9 e V$

$E_{2} - E_{4} = 13.6 (\frac{1}{4} - \frac{1}{16})$

= 2.55 eV

$E_{2} - E_{5} = 13.6 (\frac{1}{4} - \frac{1}{25})$

= $\frac{10.5 \times 21}{100} = 2.856 e V$

Only $E_{2} - E_{4}$ comes in ythe range of energy provided. So the wave length corresponding to that energy will he absorbed.

$λ = \frac{1242}{2.55} = 487.05 n m$

= 487 nm

487 nm wavelength will be absorbed

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