Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in $1 \mathrm{H}$ and $2 \mathrm{H}$. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If 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 $\mu=\mathrm{me} \mathrm{M} /(\mathrm{me}+\mathrm{M})$ where $\mathrm{M}$ is the nuclear mass and $\mathrm{me}$ is the electronic mass. Estimate the percentage difference in wavelength for the 1 st line of the Lyman series in $1 \mathrm{H}$ and $2 \mathrm{H}$. (Mass of $1 \mathrm{H}$ nucleus is $1.6725 \times 10^{-27} \mathrm{~kg}$, Mass of $2 \mathrm{H}$ nucleus is $3.3374 \times 10^{-27} \mathrm{~kg}$, Mass of electron $=9.109 \times 10^{-31}$ kg.)

Home » Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in and . This is because the wavelength of transition depends to a certain extent on the nuclear mass. If 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 , revolving around the nucleus at a distance equal to the electron-nucleus separation. Here where is the nuclear mass and is the electronic mass. Estimate the percentage difference in wavelength for the 1 st line of the Lyman series in and . (Mass of nucleus is , Mass of nucleus is , Mass of electron kg.)

Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in $1 \mathrm{H}$ and $2 \mathrm{H}$ . This is because the wavelength of transition depends to a certain extent on the nuclear mass. If 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 $\mu=\mathrm{me} \mathrm{M} /(\mathrm{me}+\mathrm{M})$ where $\mathrm{M}$ is the nuclear mass and $\mathrm{me}$ is the electronic mass. Estimate the percentage difference in wavelength for the 1 st line of the Lyman series in $1 \mathrm{H}$ and $2 \mathrm{H}$ . (Mass of $1 \mathrm{H}$ nucleus is $1.6725 \times 10^{-27} \mathrm{~kg}$ , Mass of $2 \mathrm{H}$ nucleus is $3.3374 \times 10^{-27} \mathrm{~kg}$ , Mass of electron $=9.109 \times 10^{-31}$ kg.)

Atoms | CBSE | Class 12 | NCERT Exemplar | Physics

Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in $1 \mathrm{H}$ and $2 \mathrm{H}$ . This is because the wavelength of transition depends to a certain extent on the nuclear mass. If 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 $\mu=\mathrm{me} \mathrm{M} /(\mathrm{me}+\mathrm{M})$ where $\mathrm{M}$ is the nuclear mass and $\mathrm{me}$ is the electronic mass. Estimate the percentage difference in wavelength for the 1 st line of the Lyman series in $1 \mathrm{H}$ and $2 \mathrm{H}$ . (Mass of $1 \mathrm{H}$ nucleus is $1.6725 \times 10^{-27} \mathrm{~kg}$ , Mass of $2 \mathrm{H}$ nucleus is $3.3374 \times 10^{-27} \mathrm{~kg}$ , Mass of electron $=9.109 \times 10^{-31}$ kg.)

The energy of an electron in the nth state is given by the expression,

$E_ n=-\mu Z^{2} \mathrm{e}^{4} / 8 \varepsilon_ 0^{2} h^{2}\left(1 / n^{2}\right)$

For hydrogen atom we have,

$\mu _{H}=\mathrm{m_eM_h} /(\mathrm{m_e}+\mathrm{M_h})$

For deuterium atom we have,

$\mu _{D}=\mathrm{m_eM_d} /(\mathrm{m_e}+\mathrm{M_d})$

The wavelength of the transition for hydrogen can be calculated as,
$\lambda_ H=3 / 4 \mu \mathrm{He}^{4} / 8 \varepsilon_ 0^{2} h^{2} c$

The wavelength of the transition for deuterium can be calculated as, $\lambda_ D=3 / 4 \mu D e^{4} / 8 \varepsilon_ 0^{2} h^{2} c$

The difference between the wavelength can be calculated as, $\lambda=\lambda_ H-\lambda_ D=2.714 \times 10^{-2} \%$

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