The first four spectral lines in the Lyman series of a H-atom are `2 = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å`. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Explanation

Formula of Rydberg.s constant is `R=(me^(4))/(8 epsi_(0)^(2)h^(2)c)`
While deriving above formula, nucleus was assumed theoretically to have infinite mass (as compared to mass of electrons) so that it can be considered stationary at the centre of atom. But we know, in practice, all the nuclei have some definite masses. Hence Rydberg.s constant does not remain same for the atoms of different elements. Practically we find that Rydberg.s constant varies slightly for atoms of different elements. Under these circumstances, Rydberg.s constant is defined as under.
`R=(mue^(4))/(8epsi_(0)^(2)h^(3)c) ...(1)`
where `mu`= reduced mass of an electron
`mu=(Mm_(e))/(M+m_(e))` where `m_(e)`= mass of electron
M=mass of nucleus of atom.
Note : Here, nucleus and electron become a binary system, in which both of them revolve around their common centre of mass. Such a system of two particles can be replaced by a Mme single particle of mass `(Mm_(e))/(M+m_(e))` at a distance from above common centre of mass. Such a mass is called "reduced mass" of electron and it is shown by symbol `mu`. (Its SI unit is also kg).
Now, in formula `(1)/(lambda)=R((1)/(m^(2))-(1)/(n^(2)))` taking m = 1 for Lyman series and for its four spectral lines taking n=2,3,4,5, since here for a given spectral line, term `((1)/(m^(2))-(1)/(n^(2)))` remains contant.
Thus `(1)/(lambda) prop R`
`or R prop (1)/(lambda)`
But `R prop mu` [ From equation (1)]
`:. mu prop (1)/(lambda)`
`:. (mu_(h))/(mu_(d))=(lambda_(d))/(lambda_(h))` (where h for hydrogen d for deuterium)
`:. lambda_(d)=((mu_(h))/(mu_(d)))lambda_(h)....(3)`
`rArr` Reduced mass of electron in H-atom (from equation (2))
`mu_(h)=(Mm_(e))/(M+m_(e))` (Where M= mas of nucleus in H-atom =Mass of one proton `=1.67xx10^(-27)kg)`
`:. mu_(h)=(Mm_(e))/(M(1+(m_(e))/(M)))=(m_(e))/(1+(m_(e))/(m))=m_(e)(1+(m_(e))/(M))^(-1)`
`:. mu_(h)=m_(e)(1-(M_(e))/(M))...(4)` (According to binomial theorem)
Similarly, reduced mass of electron in deuteriun atom
`mu_(d)=(2Mm_(e))/(2M+m_(e))` (where 2M= mass of nucleus of deuterium atom `m_(p)+m_(n)=2m_(p)=2xx1067xx10^(-27)kg)`
`:.mu_(d)=(2Mm_(e))/(2M(1+(m_(e))/(2M)))`
`:.mu_(d)=(m_(e))/(2M(1+(m_(e))/(2M)))....(5)`
From equation (3),(4),(5)
`lambda_(d)=(1-(m_(e))/(M))(1+(m_(e))/(2M))lambda_(h).....(6)`
Now, in above equation ,
`(1-(m_(e))/(M))(1+(m_(e))/(2M))=(1-(9.1xx10^(-31))/(1.67xx10^(-27)))xx(1+(9.1xx10^(-31))/(2xx1.67xx10^(-31)))`
`=(1-0.005449)(1+0.002724)`
`(0.9994551)(1.0002724)`
`=0.9997274...(7)`
From equation (6) and (7),
`lambda_(d)=0.9997274 lambda_(h) ....(8)`
When we substitute four values of `lambda_(h)` in equation (8), corresponding values of `lambda_(d)` are obtained as follows: