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

Explanation

The total energy of the electron in the stationary states of the hydrogen atom is given by
`E_(n)= - (me^(4))/(8n^(2)epsi_(0)^(2)h^(2))`
where signs are as usual and the m that occurs in the Bohr formula is the reduced mass of electron and proton in hydrogen atom.
By Bohr's model,
`hv_(if) = E_(n_(i)) - E_(n_(f))`
On simplifying,
`v_(if) = (me^(4))/(8epsi_(0)^(2)h^(3)) ((1)/(n_(f)^(2)) - (1)/(n_(i)^(2)))`
Since, `lambda prop (1)/(mu)`
Thus, `lambda_(if) prop (1)/(mu)`
where `mu` is the reduced mass. (here, `mu` is used in place of m)
Reduced mass for `H = mu_(H) = (m_(e))/(1+(m_(e))/(M)), m_(e) (1+(m_(e))/(M))`
Reduced mass for `D = mu_(0), m_(e) (1-(m_(e))/(2M))`
`= m_(e) (1-(m_(e))/(2M))(1+(m_(e))/(2M))`
If for hydrogen deuterium, the wavelength is `(lambda_(H))/(lambda_(D))`
`(lambda_(D))/(lambda_(H)) = (mu_(H))/(lambda_(D)) = (1+(m_(e))/(2M))^(-1) = (1-(1)/(2xx 1840))` [From Eq. (i)]
`lambda_(D) = lambda_(H) xx (0.99973)`
On substiuting the values, we have
Thus, lines are `1217.7Å, 1027.7Å, 974.04Å, 951.143Å`