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 `mu`, 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 `= 9.109 xx 10^(-31) kg`.)

Explanation

The total energy of the electron in the nth states of the hydrogen like atom of atomic number Z is given by
`E_(n)^(.) = - (uZ^(2)e^(4))/(8epsi_(0)^(2)h^(2))((1)/(n^(2)))`
where signs are as usual and the `mu` that occurs in the Bohr formula is the reduced mass of electron and proton.
Let `mu_(H)` be the reduced mass of hydrogen and `mu_(D)` that of Deutrium. Then, the frequency of the `1st` Lyman line in hydrogen is `hv_(H) = (mu_(H)e^(4))/(8epsi_(0)^(2)h^(2)) (1-(1)/(4)) = (mu_(H)e^(4))/(8epsi_(0)^(2)h^(2)) xx 3/4`
Thus, the wavelength of the transition is ` lambda_(H) = (3)/(4) (mu_(H)e^(4))/(4 8 epsi_(0)^(2)h^(3)c)`. The wavelength for
the same line in Deutrium is `lambda_(D) = (3)/(4)(mu_(D)e^(4))/(8 epsi_(0)^(2)h^(3)c)`
`:. Deltalambda = lambda_(D) - lambda_(H)`
Hence, the percentage difference is
`100 xx (Deltal)/(lambda_(H)) = (lambda_(D) - lambda_(H))/(lambda_(H)) xx 100 = (mu_(D) - mu_(H))/(mu_(H)) xx 100`
`= ((m_(e)M_(D))/((m_(e)+M_(D)))- (m_(e)M_(H))/((m_(e)-M_(H))))/((m_(e)M_(H))/((m_(e)+M_(H))))xx 100`
`=[((m_(e)+M_(H))/(m_(e)+M_(D)))(M_(D))/(M_(H))-1] xx 100`
Since, `m_(e) lt lt M_(H) lt lt M _(D)`
`(Deltalambda)/(lambda_(H)) xx 100=[(M_(H))/(M_(D))xx(M_(D))/(M_(H))((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`
[By binomial theorem, `(1+x)^(n) = 1+nx`is `|x| lt 1`]
`~~m_(e)[(1)/(M_(H))-(1)/(M_(D))]xx100`
`= 9.1 xx 10^(-31)[(1)/(1.6725 xx 10^(-27))- (1)/(3.3374 xx 10^(-27))] xx 100`
`= 9.1 xx 10^(-4) [0.5979 - 0.2996] xx 100`
`= 2.714 xx 10^(-2)%`