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Deutarium was discovered in 1931by Harold Urey by measuring the small change in wavelength for a particular transition in .^1H and .^2H. 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=m_eM//(m_e+M) where M is the nuclear mass and m_e is the electronnic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in .^1H and .^2H. (Mass of ^1H nucleus is 1.6725xx10^(-27)kg, mass of .^2H nucleus is 3.3374xx10^(-27)kg, Mass of electron =9.109xx10^(-31)kg). |
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Answer» Solution :If we take into account the nuclear MOTION, the stationary state energies will be GIVEN by `E_(n)=-(mu Z^2e^4)/(8in_(0)^(2)h^2) (1/(n^2))` Let `mu_(H)` be reduced MASS of hydrogen `(H^1)` and `mu_(D)` be reduced mass of deuterium `(H^2)`. The frequency of 1st of LYMAN series in hydrogen is given by `hv_(H)=(mu_(H)e^4)/(8pi in_0^2h^2)(1/(1^2)-1/(2^2))= (3//4 mu_(H)e^4) /(8in_0^2h^2)` The wave number of transition is `barlambda_(H)=(3//4 mu_(H)e^4) /(8in_0^2h^3 c)` Similarly, for deuterium, `barlambda_(D)=(3//4 mu_(D)e^4) /(8in_0^2h^3 c)` `Deltabar lambda=barlambda_(D)-barlambda_(H)` `:.` The percentage difference in wave number is`(Deltabar(lambda))/(bar(lambda)_(H)) xx 100 = ((bar(lambda)_(D)-bar(lambda)_(H)) xx100)/(bar(lambda)_(H)) = (mu_(D)-mu_(H))/(mu_(H)) xx 100` USING `mu_(H)=(m_(e)M_(H))/(m_(e)+M_(H))` and `mu_(D)=(m_(e)M_(D))/(m_(e)+M_(D))`, we g et `(bar(Delta lambda))/(barlambda_(H))xx100=((m_(e)M_(D))/(m_(e)+M_(D))-(m_(e)M_(H))/(m_(e)+M_(H)))/(m_(e)M_(H)//(m_(e)+M_(H)))xx100` As `m_(e)lt lt M_(H)lt ltM_(D)` `:. (barDeltalambda)/(barlambda_(H))=[(M_(D))/(M_(H))xx(M_(H))/(M_(D))((1+m_(e)//M_(H))/(1+m_(e)//M_(D)))-1]xx100=[(1+(m_(e))/(M_(H)))(1+(m_(e))/(M_(D)))^(-1)-1]xx100` `=[1+(m_(e))/(M_(H))-(m_(e))/(M_(D))-1]xx100=m_(e)[1/(M_(H))-1/(M_(D))]xx100` `(bar(Deltalambda))/(bar(lambda)_(H))xx100=9.1xx10^(-31)[1/(1.6725xx10^(-27))-1/(3.3374xx10^(-27))]xx100` `=9.1xx10^(-4)(0.5979-0.2996)xx100=2.714xx10^(-2)%` |
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