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Question
The ground state energy of hydrogen atom is −13.6 eV. If an electron make a transition from an energy level −0.85 eV to −1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
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Solution
Since energy of Hydrogen atom is given as
`E_n = (-13.6)/n^2 eV`
Ground state energy E1 = −13.6eV
Now electron makes a transition from energy level
np = −0.85 eV to nq = −1.51 eV
Now,
`E_(n_p) = (-13.6)/n_p^2 eV`
`n_p^2 = (-13.6)/-0.85 = 16`
`n_p = 4`
Again,`E_(n_p) = (-13.6)/n_q^2 eV`
`n_p^2 = (-13.6)/(-1.51) =9`
`n_q = 3`
Thus, we have transition from n = 4 to n = 3.
Since transition corresponds to the transition of an electron from some higher energy state to an orbit having n = 3. It is Paschen series of the hydrogen spectrum. Wavelength is given as
`1/lambda = R(1/3^2 - 1/x^2)`
Where R is Rydberg’s constant
`R = 1.097 xx 10^7 m^-1 1/lambda = 1.097 xx 10^7 [1/9 - 1/16]`
`= 1.097 xx 10^7 [7/(9 xx 16)]`
`lambda = (9 xx 16)/(1.097 xx 10^7 xx 7) = 18.752 xx 19^-7 m = 1875 nm`
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List-II (Wavelengths (nm)) |
||
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