Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV.

Solution

Given,
Wavelength of the monochromatic radiation,

λ

= 975 Å
Ionization energy for hydrogen atom = 13.6 eV

The energy of monochromatic radiation of wavelength 975 Å is,

E = \frac{hc}{λ} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{975 \times 10^{- 10} \times 1.6 \times 10^{- 9}} eV

= 12.75 eV

(∵ 1 eV = 1.6 \times 10^{- 19} J)

∴

12.75 = 13.6 (\frac{1}{1^{2}} - \frac{1}{n^{2}})

\frac{1}{n^{2}} = 1 - \frac{12.75}{13.6} = \frac{0.85}{13.6} = \frac{1}{16}

∴

n = 4

Thus, number of lines possible in the resultant spectrum = 6.

The longest wavelength will be emitted for transition from 4th orbit to 3rd orbit with an energy.

E_{n_{2}} - E_{n_{1}} = E_{0} Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

Atomic no. Z = 1

E_{4 \to 3} = 13.6 (\frac{1}{3^{2}} - \frac{1}{4^{2}}) = 13.6 (\frac{1}{9} - \frac{1}{16})

E_{4 \to 3} = 13.6 \times \frac{7}{144} eV = 13.6 \times \frac{7}{144} \times 1.6 \times 10^{- 19} J

The longest wavelength,

λ = \frac{hc}{E_{4 \to 3}}

= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8} \times 144}{13.6 \times 7 \times 1.6 \times 10^{- 19}} m

λ = 1.88 \times 10^{- 6} m = 18800 Å

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