Ncert Exercises & Solutions

Calculate the energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 in a hydrogen atom.

In hydrogen spectrum, the spectral lines are expressed in term of wave number

\bar{v}

obey the following formula

Wave number,

\bar{v} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) (where, R_{H} = Rydberg constant 109677 {cm}^{- 1})

\bar{v} = 109677 {cm}^{- 1} (\frac{1}{2^{2}} - \frac{1}{3^{2}})

\bar{v} = 108677 \times \frac{5}{36} = 15232.9 {cm}^{- 1}

\bar{v} = \frac{1}{λ}

or, λ = \frac{1}{v} = \frac{1}{15232.9} = 6.564 \times 10^{- 5} cm

Wavelength, λ = 6.564 \times 10^{- 7} m

Energy, E = \frac{hc}{λ}

= \frac{6.626 \times 10^{- 34} Js \times 3.0 \times 10^{8} {ms}^{- 1}}{6.564 \times 10^{- 7} m}

= 3.028 \times 10^{- 19} J

Frequency, v = \frac{c}{λ} = \frac{3.0 \times 10^{8} {ms}^{- 1}}{6.564 \times 10^{- 7} m}

= - 0.457 \times 10^{15} s^{- 1} = 4.57 \times 10^{14} s^{- 1}