Hydrogen atom is ground state is excited by mean of monochromatic radiation of wavelength `975Å` How many different lines are possible in the resulting spectrum ? Calculate the longest wavelength for hydrogen atom as ionization energy is`13.6 eV`

Explanation

The energy of an electron in nth orbit of hydrogen `(Z = 1)` is given by ` E_(n) = - (R h c)/(n^(2))`
`:. Lonization energy = E_(oo) - E_(1) = R h c = 13.6 e V`
`:. E_(n) (13.6)/(n^(2)) e V`
Therefore, energy of electron in ground state `(n = 1)`,
`E_(1) = - (13.6)/(1) = - 13.6 e V`
energy of electron in the first excited state `(n = 2)`.
`E_(2) = - (13.6)/(2^(2)) = - (13.6)/(4) = - 3.4 e V`
The energy of electron in the second excited state `(n = 3)`.
`E_(3) = - (13.6)/(3^(3)) = (13.6)/(9) = - 1.511 e V`

The energy of electron in the third excited state `(n = 4)`.
`E_(4) = - (13.6)/(4^(2)) = - (13.6)/(16) = - 0.85 e V`
Energy of incident photon of wavelength
`lambda = 975 Å = 975 xx 10^(-10) m`
`E = (hc)/(lambda) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(975 xx 10^(-10)) J`
`= (6.63 xx 10^(-34) xx 3 xx 10^(8))/(975 xx 10^(-10)) xx 1.6 xx 10^(-19) e V = 12.75 e V`
When incident photon of this energy is absorbed by hydrogen atom, let its ground state electron accupy `(n - 1)th` excited state or nth orbit. Then,
`E = - (Rhc)/(n^(2)) - ((- Rhc)/(1^(2))) = Rhc (1 - (1)/(n^(2)))`
i.e. , `12.75 e V = 13.6 e V (1 - (1)/(n^(2)))`
This gives `n = 4` .
That is the electron is excited to the third excited state . The emission spectrum will contain the transition `(4to3)` for which the energy different is minimum.
i.e. `E_(min) = E_(4) - E_(3) = - 0.85 - (- 1.511) e V`
`= 1.511 - 0.85 = 0.661 e V`
`= 0.661 xx 1.6 xx 10^(-19) J`
`:. lambda _(min) = (hc)/(E_(min)) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(975 xx 0.661 xx 1.6 xx 10^(-19)) = 18.807 xx 10^(-7) m`
`= 18807 Å`