The ground state energy of hydrogen atom is `-13.6eV`. If an electron makes a transition form an energy level `-0.85 eV` to `-3.4 eV`, calculate the wavelength of spectral line emitted. To which series of hydrogen spectrum does this wavelength belongs?

Explanation

`E_(n)=-(13.6)/(n^(2))ev.` Here ground state energy for `n=1, E_(1)=-13.6 eV.`
Now electron transits from `E_(q)=-0.85 eV" to "E_(q)=-3.4 eV`
`-0.85=(-13.6)/(n_(p)^(2))`
`n_(p)^(2)=(13.6)/(0.85)=16`
`"Thus, "n_(r)=4`
`"Again, "-3.4=(13.6)/(n_(q)^(2))rArrn_(q)^(2)=(13.6)/(3.4)=4rArrn_(q)=2`
Thus, electron makes transition from n=4 to n=2. Hence it is Balmer series. Now `R=1.0974xx10^(7)m^(-1)`
`(1)/(lamda)=R((1)/(2^(2))-(1)/(n^(2)))rArr(1)/(lamda)=1.0974xx10^(7)`
`((1)/(2^(2))-(1)/(4^(2)))=(1.09xx10^(7)xx12)/(4xx16)rArr(1)/(lamda)=0.2057xx10^(7)`
`lamda=4.861xx10^(-7)`
`lamda=4861 Å`