Calculate the difference between the ionisation potentials of atomic hydrogen and atomic deuterium.

Explanation

Wave number `barv` of lines in the spectrum of an atom is given by
`bar v = ( mu e ^(4) Z ^(2))/( 8 epsi _(0) ^(2) ch ^(3)) [ (1)/( n _(1) ^(2) ) - ( 1)/( n _(2) ^(2)) ]`
where `mu = ( M m )/( M + m )`
The ionisation energy is the energy required to liberate an electron from a given orbit to an infinite from nucleus. It corresponds to energy transition from `n _(1) =1` to `n _(2) = oo.`
`therefore bar vc = v = ( mu e ^(4) Z ^(2))/( 8epsi _(0) ^(2) h ^(3))`
For hydrogen atom `(Z =1),`
` v _(H) = (e ^(4))/( 8 epsi _(0) ^(2) h ^(3)) [ ( M _(H) m )/(M _(H) + m ) ][ (1)/(1 ^(2)) - (1)/( oo^(2))]`
For deuterium atom `(Z =1),`
`v _(D) = (e ^(4))/( 8epsi _(0) ^(2) h ^(3)) [ ( M _(D) m )/( M _(D) + m ) ] [ (1)/( 1 ^(2)) - (1)/( oo^(2))]`
Hence difference between the ionisation energies of the two atoms,
`h (v _(D) -v _(H)) = (e ^(4))/(8epsi _(0) ^(2) h ^(2)) [ ( M _(D) m )/( M _(D) + m )- ( m _(H) m )/(M _(H) + m) ]`
`= (e ^(4) )/( 8epsi _(0) ^(2) h ^(2))[ (3680 m xx n )/( 3680 + m ) - (1840 xx m )/(1840 + m)]`
`= (e ^(4) m )/( 8epsi _(0) ^(2) h ^(2)) [ ( 3680)/(3681) - (1840)/(1841)]`
`= Rch [ (3680)/(3681) - (1840)/(1841)]`
`= 5 . 885 xx 10 ^(-22) J = ( 5.885 xx 10 ^(-22))/( 1. 6 xx 10 ^(-19))eV`
`=3.68 xx 10 ^(-3) eV`