1.

For the case of atomic hydrogen find: (a)the wavelength of the first three lines of the Balmer series, (b)the minimum resolving the first 20 lines of the Balmer series.

Answer» The energies are
`E_(H)((1)/(4)-(1)/(9))=(5)/(36)E_(H),E_(H)((1)/(4)-(1)/(16))=(3)/(16)E_(H)`
`E_(H)((1)/(4)-(1)/(25))=(21)/(100)E_(H)`
They correspond to wavelengths
`654.2nm, 484.6nm` and `433nm`
The `n^(th)` line the Balmer series has the energy
`E_(H)((1)/(4)-(1)/((n+2)^(2)))`
For `n= 19`, we get the wavelength `366.7450nm`
For `n= 20` we get the wavelength `366.4470nm`
To resolve these lines we require a resolving power of
`R~~(lambda)/(delta lambda)=(366.6)/(0.298)= 1.23xx10^(3)`


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