For atoms of light and heavy hydrogen `(H and D)` fine the difference,
(a) between the binding energies of their electrons in the ground state.
(b) between the wavelength of first lines of the Lyman series.

Explanation

The difference between the binding energies is
`DeltaE_(b)=E_(b)(D)-E_(b)(H)`
`=Delta(m)/(1+(m)/(M))(e^(4))/(2 ħ^(2))+(m)/(1+(m)/(2M))(e^(4))/(2 ħ^(2))`
`=(me^(4))/(2 ħ^(2))((m)/(2M))`
Substitution gives `Delta E_(b)= 3.7 meV`
For the first line the Lyman series
`(2pi ħC)/(lambda)= ħR((1)/(4)-(1)/(4))=(3)/(4) ħR`
or `lambda=(8pi c)/(3R)=(8pi c)/(3R)=(8 pi ħc)/(3E_(b))`
Hence `DeltaE_(b)=E_(b)(D)-E_(b)(H)`
`=Delta(m)/(1+(m)/(M))(e^(4))/(2 ħ^(2))+(m)/(1+(m)/(2M))(e^(4))/(2 ħ^(2))`
`=(me^(4))/(2 ħ^(2))((m)/(2M))`
Substitution gives `Delta E_(b)= 3.7 meV`
For the first line the Lyman series
`(2pi ħC)/(lambda)= ħR((1)/(4)-(1)/(4))=(3)/(4) ħR`
or `lambda=(8pi c)/(3R)=(8pi c)/(3R)=(8 pi ħc)/(3E_(b))`
Hence
`lambda_(H)-lambda_(D)=(8 pi ħc)/(3)((1)/(E_(b)(H))-(1)/(E_(b)(D)))`
`=(8pi ħc)/(3((me^(4))/(2 ħ^(2)))).(m)/(2M)`
`=(m)/(2M)xxlambda_(1)`
(where `lambda_(1)` is the wavelength of the first line of Lyman series without considering nuclear motion).
Substitution gives (see.21 for `lambda_(1)`) using `lambda_(1)= 121nm`
`Delta lambda=33p m`