Ncert Exercises & Solutions

Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ²H. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If the nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass.

Such a system is equivalent to a single particle with a reduced mass $μ$ , revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $μ = m_{e} M / (m_{e} + M)$ , where M is the nuclear mass and m_e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ¹H and ²H. (mass of ¹H nucleus is $1.6725 \times 10^{- 27}$ kg, mass of ²H nucleus is $3.3374 \times 10^{- 27}$ kg, Mass of electron $= 9.109 \times 10^{- 31} kg$ .)

Hint: The wavelength of the transition depends on the reduced mass.

Step 1: Find the wavelength of the transition for Hydrogen and Deuterium.

The total energy of the electron in the nth states of the hydrogen-like atom of atomic number Z is given by

E_{n} = - \frac{μ Z^{2} e^{2}}{8 ε_{0}^{2} h^{2}} (\frac{1}{n^{2}})

where signs are usual and the

μ

that occurs in the Bohr formula is the reduced mass of electron and proton.

Let

μ_{H}

be the reduced mass of hydrogen and

μ_{D}

that of Deuterium. Then, the frequency of the 1st Lyman line in hydrogen is:

{hν}_{H} = \frac{μ_{H} e^{4}}{8 ε_{0}^{2} h^{2}} (1 - \frac{1}{4}) = \frac{μ_{H} e^{4}}{8 ε_{0}^{2} h^{2}} \times \frac{3}{4}

Thus, the wavelength of the transition is

λ_{H} = \frac{3}{4} \frac{μ_{H} e^{4}}{8 ε_{0}^{2} h^{3} c} .

The wavelength of the transition of the same line in Deuterium is

λ_{D} = \frac{3}{4} \frac{μ_{D} e^{4}}{8 ε_{0}^{2} h^{3} c}

∴ ∆ λ = λ_{D} - λ_{H}

Step 2: Find the percentage difference.

Hence, the percentage difference is:

100 \times \frac{∆ λ}{λ_{H}} = \frac{λ_{D} - λ_{H}}{λ_{H}} \times 100 = \frac{μ_{D} - μ_{H}}{μ_{H}} \times 100

= \frac{\frac{m_{e} M_{D}}{(m_{e} + M_{D})} - \frac{m_{e} M_{H}}{(m_{e} - M)}}{\frac{m_{e} M_{H}}{(m_{e} + M_{H})}} \times 100

= [(\frac{m_{e} + M_{H}}{m_{e} + M_{D}}) \frac{M_{D}}{M_{H}} - 1] \times 100

Since, m_{e} < < M_{H} < < M_{D}

\frac{∆ λ}{λ_{H}} \times 100 = [\frac{M_{H}}{M_{D}} \times \frac{M_{D}}{M_{H}} (\frac{1 + \frac{m_{e}}{M_{H}}}{1 + \frac{m_{e}}{M_{D}}}) - 1] \times 100

= [(1 + \frac{m_{e}}{M_{H}}) {(1 + \frac{m_{e}}{M_{D}})}^{- 1} - 1] \times 100 ≃ [1 + \frac{m_{e}}{M_{H}} - \frac{m_{e}}{M_{D}} - 1] \times 100

[By binomial theorem, {(1 + x)}^{n} = 1 + nx is |x| < 1]

\approx m_{e} [\frac{1}{M_{H}} - \frac{1}{M_{D}}] \times 100

= 9.1 \times 10^{- 31} [\frac{1}{1.6725 \times 10^{- 27}} - \frac{1}{3.3374 \times 10^{- 27}}] \times 100

= 9.1 \times 10^{- 4} [0.5979 - 0.2996] \times 100

= 2.714 \times 10^{- 2} %

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