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  • 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 depend to a certain extent on the nuclear mass. If nuclear motion is taken

 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 \mu, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here \mu = meM/ (me + M) where M is the nuclear mass and me 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 × 10–27 kg, the mass of 2H nucleus is 3.3374 × 10–27 kg, the mass of electron = 9.109 × 10-31 kg.)

 

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According to Bohr’s theory for a hydrogen-like atom of atomic number Z, the total energy of the electron in the nth state is

 E_{n}=-\left \{ \frac{\mu}{2h^{2}}\left (\frac{e^{2}}{4\pi \epsilon _{0}} \right )^{2} \right \}\frac{Z^{2}}{n^{2}}=\frac{\mu Z^{2}e^{4}}{8 \epsilon_{0}^{2}h^{2} }\left ( \frac{1}{n^{2}} \right )

Where signs are usual, and the u that occurs in Bohr’s formula is the reduced mass of the electron and proton.

For hydrogen atom:

\mu_{H}=\frac{m_{e}M_{H}}{\left ( m_{e}+M_{H} \right )}

For Deuterium

:\mu_{D}=\frac{m_{e}M_{D}}{\left ( m_{e}+M_{D} \right )}

Let \mu_{H} be the reduced mass of hydrogen and \mu_{D} that of deuterium. Then the frequency of the first Lyman line in hydrogen is

hf_{H}=\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}}\left ( 1-\frac{1}{4} \right )=\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}}\times \frac{3}{4}

Thus, the wavelength of the transition is

\lambda_{H}=\frac{3}{4}\frac{\mu_{H}e^{4}}{8\epsilon _{0}^{2}h^{2}c} 

The wavelength of the transition for the same line in deuterium is

\lambda_{D}=\frac{3}{4}\frac{\mu_{D}e^{4}}{8\epsilon _{0}^{2}h^{2}c}

\therefore Difference in wavelength \Delta \lambda=\lambda_{D}-\lambda_{H}

Hence, the percentage difference is

100 \times \frac{\Delta \lambda}{\lambda_{H}}=\frac{\lambda_{D}-\lambda_{H}}{\lambda_{H}}\times100=\frac{\mu_{D-\mu_{H}}}{\mu_{H}}\times 100\\\\ =\frac{\frac{m_{e}M_{D}}{\left ( m_{e}+M_{D} \right )}-\frac{m_{e}M_{H}}{\left ( m_{e}+M_{H} \right )}}{\frac{m_{e}M_{H}}{\left ( m_{e}+M_{H} \right )}}\times 100\\\\ =\left [ \left ( \frac{m_{e}+M_{H}}{m_{e}+M_{D}} \right )\frac{M_{D}}{M_{H}}-1 \right ]\times 100

Since me<<MH<<MD

\frac{\Delta \lambda}{\lambda_{H}}\times=\left [ \frac{M_{H}}{M_{D}}\times \frac{M_{D}}{M_{H}}\left ( \frac{1+\frac{m_{e}}{M_{H}}}{1+\frac{m_{e}}{M_{D}}} \right )-1 \right ]\times 100\\\\ =\left [ \left (1+\frac{m_{e}}{M_{H}} \right )\left (1+\frac{m_{e}}{M_{D}} \right )^{-1}-1 \right ]\times 100\\\\ =\left [ \left (1+\frac{m_{e}}{M_{H}} \right )\left (1+\frac{m_{e}}{M_{D}} \right )-1 \right ]\times 100\\\\ (By \; binomial\;theorem\; (1+x)^{n}=1+nx \; is \; |x|<1)\\\\ \frac{\Delta \lambda }{\lambda _{H}}\times 100=\left [ 1+\frac{m_{e}}{M_{H}}-\frac{m_{e}}{M_{D}}-\frac{(m_{e})^{2}}{M_{H}M_{D}}-1 \right ]\times 100\\\\ Neglecting \;\frac{(m_{e})^{2}}{M_{H}M_{D}}\;as \; it \; is\; very\; small

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