The wavelengths of the spectral lines in the | vedclass

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(C) The Rydberg constant $R$ for an atom is given by $R = \frac{\mu e^{4}}{8 \epsilon_{0}^{2} h^{3} c}$,where $\mu$ is the reduced mass of the electro

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The mass of the two nuclei is different.

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(C) The Rydberg constant $R$ for an atom is given by $R = \frac{\mu e^{4}}{8 \epsilon_{0}^{2} h^{3} c}$,where $\mu$ is the reduced mass of the electron-nucleus system. The reduced mass $\mu$ is defined as $\mu = \frac{m_{e} M}{m_{e} + M}$,where $m_{e}$ is the mass of the electron and $M$ is the mass of the nucleus. Since the mass of the deuterium nucleus $(M_{D} \approx 2M_{H})$ is different from the mass of the hydrogen nucleus $(M_{H})$,the reduced mass $\mu$ for deuterium is different from that of hydrogen. Consequently,the Rydberg constant $R$ and the resulting wavelengths of the spectral lines $(\frac{1}{\lambda} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}))$ differ slightly between the two.

The wavelengths of the spectral lines in the spectrum of deuterium $(_{1}H^{2})$ differ slightly from those in the spectrum of hydrogen $(_{1}H^{1})$ because ...

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