# NCERT Physics Class 12 Exemplar Ch 12 Atoms Part 3 (For CBSE, ICSE, IAS, NET, NRA 2023)

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## Long Answer

**24** The first four spectral lines in the Lyman serics of a H-atom are , and . If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Ans:

Reduced mass for

Reduced mass for

If for Hydrogen/Deuterium the wavelength is

Thus lines are

**25** 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 depend 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 *M* 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 .)

Ans:

Taking into account the nuclear motion, the stationary state energies shall be,

Let be the reduced mass of Hydrogen and that of Deutrium. Then the frequency of the

1^{st} Lyman line in Hydrogen is Thus the wavelength of the transition is The wavelength of the transition for the same line in Deuterium is

Hence the percentage difference is

Since