Ncert Exercises & Solutions

When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula

$\bar{v} = 109677 [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]$

What points of Bohr's model of an atom can be used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term.

The two important points of Bohr's model that can be used to derive the given formula are as follows

(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.

(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state to lower stationary state to higher stationary state respectively.

Derivation The energy of the electron in the n^th stationary state is given by the expression,

E_{n} = - R_{H} (\frac{1}{n^{2}}) n = 1, 2, 3 . . . . . . . . . . . (i)

Where, R_H is called Rydberg constant and its value is

2.18 \times 10^{- 18} J

. The energy of the lowest state, also called the ground state, is

E_{n} = - 2.18 \times 10^{- 18} (\frac{1}{1^{2}}) = - 2.18 \times 10^{- 18} J . . . . . . . . . . . . (ii)

The energy gap between the two orbits is given by the equation,

∆ E = E_{f} - E_{i}

On combining Eqs. (i) and (iii)

∆ E = (- \frac{R_{H}}{n_{f}^{2}}) - (- \frac{R_{H}}{n_{i}^{2}})

Where, n_i and n_f stand for initial orbit and final orbit.

∆ E = R_{H} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}] = 2.18 \times 10^{- 18} J [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]

Frequency, v associated with the absorption and emission of the photon can be calculated as follows

v = \frac{∆ E}{h} = \frac{R_{H}}{h} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]

\Rightarrow v = \frac{2.18 \times 10^{- 18} J}{6.626 \times 10^{- 34} Js} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]

v = 3.29 \times 10^{15} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}] Hz

\Rightarrow \bar{v} = \frac{v}{c} = \frac{329 \times 10^{15}}{3 \times 10^{8} {ms}^{- 1}} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]

\bar{v} = 1.09577 \times 10^{7} [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}] m^{- 1}

\bar{v} = 109677 [\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}] {cm}^{- 1}

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