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Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ""^(1)H and ""^(2)H. 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 u, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here mu=m_(e)M//(m_(e)+M) where M is the nuclear mass and m_(e) is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ""^(1)H and ""^(2)H. (Mass of ""^(1)H nucleus is 1.6725 xx 10^(-27) kg, Mass of ""^(2)H nucleus is 3.3374 xx 10^(-27) kg Mass of electron =9.709xx10^(-31)kg) |
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Answer» Solution :For HYDROGEN and DEUTERIUM atoms, according to formula `lambda_(d)=(1-(m_(e))/(M))(1+(m_(e))/(2M))_lambda_(h)` `:. lambda_(d)=(1-(9.109xx10^(-31))/(1.6725xx10^(-27)))(1+(9.109xx10^(-31))/(3.3374xx10^(-27)))(1218)` `:. Lambda_(d)=(1-5.446xx10^(-4))(1+2.7294xx10^(-4))(1218)` `:. lambda_(d)=(1-0.005446)(1+0.0002729)(1218)` :. lambda_(d)=(0.9994554) (1.0002729)(1218)` :. lambda_(d)=1217.6688Å` A For first line of Lyman SERIES of deuterium atom, percentage decrease in its wavelength (RELATIVE to corresponding wavelength for hydrogen atom) `=(Delta lambda)/(lambda_(h))xx100%` `(lambda_(h)-lambda_(d))/(lambda_(h))xx100%` `=((1218-1217.6688)/(1218))xx100%` `=(0.3312)/(1218)xx100%` `2.719xx10^(-2)%` `=0.02719%` |
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