# NCERT Physics Class 12 Exemplar Ch 12 Atoms Part 3

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 1st 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

1st 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

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