Ncert Exercises & Solutions

The first four spectral lines in the Lyman series of a H-atom are $λ = 1218 \overset{o}{A}, 1028 \overset{o}{A}, 974.3 \overset{o}{A} and 951.4 \overset{o}{A}$ . If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Hint: The wavelength depends on the reduced mass.

Step 1: Find the relation between the wavelength and the reduced mass.

The total energy of the electron in the stationary state of the hydrogen atom is given by;

E_{n} = - \frac{{me}^{4}}{8 n^{2} ε_{0}^{2} h^{2}}

where signs are as usual and 'm' that occurs in the Bohr formula is the reduced mass of electron and proton in the hydrogen atom.

By Bohr' s model, {hν}_{if} = E_{n_{i}} - E_{n_{f}}

On simplifying, ν_{if} = \frac{{me}^{4}}{8 ε_{0}^{2} h^{3}} (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})

Since, λ \propto \frac{1}{μ}

Thus, λ_{if} \propto \frac{1}{μ} . . . . . . . . . . . . . . (i)

where μ is the reduced mass . (here, μ is used in place of m)

Step 2 : Fidn the reduced masses of hydrogen and deuterium .

Reduced mass for H = μ_{H} = \frac{m_{e}}{1 + \frac{m_{e}}{M}} = m_{e} (1 - \frac{m_{e}}{M})

Reduced mass for D = μ_{D} = \frac{m_{e}}{(1 - \frac{m_{e}}{2 M})} = m_{e} {(1 - \frac{m_{e}}{2 M})}^{- 1} = m_{e} (1 + \frac{m_{e}}{2 M})

Step 3 : Find the wavelength of deuterium .

If λ_{H} is the wavelength for hydrogen and λ_{D} is the wavelength for deuterium,

\frac{λ_{D}}{λ_{H}} = \frac{μ_{H}}{μ_{D}} ≃ {(1 + \frac{m_{e}}{2 M})}^{- 1} ≃ (1 - \frac{1}{2 \times 1840}) [from eq . (i)]

On substituting the values, we have λ_{D} = λ_{H} \times (0.99973)

Thus, lines are 1217.7 Å, 1027.7 Å, 974.04 Å, 951.143 Å .

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