Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength `970.6 Å`. How many lines are possible in the resulting emission spectrum ? Calculate the longest wavelength amongst them. You may assume that the ionisation energy for hydrogen atom is 13.6 eV. Given Planck's constant `= 6.6 xx 10^(-34)Js, " " c = 3 xx 10(8) ms^(-1)`

Explanation

Wavelength of radiatio , `lambda 990.7Å`
=99m
As we know `DeltaE=(1242eV)/(lambda(nm))=(1242)/(99)=12.45 eV`
`DeltaE=13.6 Z^(2)(
(1)/(1^(2))-(1)/(n^(2)))`
`( :. n=1,Z` 1 becaise H-atom is in ground state)
`12.45=13.6(1)^(2)((1)/(1^(2))-(1)/(n^(2)))`
`rArr " " n=m`
Number of wavelength in resulting spectrum,
`(n(n-1))/(2)=(3(3-1))/(2)=3`
and we also know
`lambda prop(1)/(DeltaE)`
So for `DeltaE` to be minimum, it will in the trasnsition `3rarr2`.
Using the relation, `DeltaE=13.6 Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
`=13.6(1)^(2)((1)/(2^(2))-(1)/(3^(2)))=13.6 xx(5)/(36)`
So, the wavelength `lambda=(1242)/(1.88)=660.63 nm`