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Question
The ground state energy of hydrogen atom is – 13∙6 eV. If an electron makes a transition from an energy level – 1∙51 eV to – 3∙4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.
Solution
We know,
`(hc)/λ=E_i −E_f`
`λ = (hc)/(E_i −E_f)`
`= (6.6 xx 10^(−34) xx 3xx10^8)/({−1.51−(−3.4)}xx1.6×10^(−19))`
`= (19.8 xx 10^(−26))/(3.024xx10^(−19))`
`= 6.5476 xx10^(−7)`
`= 6547 xx 10^(−10)`
`= 6547` Å
`E_n = E_1/n^2`
Where, E1 = Ground state energy
or, `n^2 = E_1/E_n`
`n =sqrt(E_1/E_n)`
`n_i = sqrt((−13.6)/(−1.51))= 3`
`n_f =sqrt((−13.6)/(−3.4)) = 2`
i.e. transition is from higher energy state n = 3 to the lower state n = 2. Therefore, it represents balmer series.
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