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Question
The first four spectral lines in the Lyman series of a H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
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Solution
Let µH and µD be the reduced masses of electrons for hydrogen and deuterium respectively.
We know that `1/λ = R[1/n_f^2 - 1/n_1^2]`
As ni and nf are fixed for by mass series for hydrogen and deuterium.
`λ ∝ 1/R` or `λ_D/λ_H = R_H/R_D` ......(i)
`R_R = (m_ee^4)/(8ε_0ch^3) = (µ_He^4)/(8ε_0ch^3)`
`R_D = (m_ee^4)/(8ε_0ch^3) = (µ_De^4)/(8ε_0ch^3)`
∴ `R_H/R_D = µ_H/µ_D` ......(ii)
From equations (i) and (ii)
`λ_D/λ_H = µ_H/µ_D` ......(iii)
Reduced mass for hydrogen,
`µ_H = m_e/(1 + m_e/M) ≃ m_e(1 - m_e/M)`
Reduced mass for deuterium,
`µ_D = (2M * m_e)/(2M(1 + m_e/(2M))) ≃ m_e(1 - m_e/M)`
Where M is the mass of proton
`µ_H/µ_D = (m_e(1 - m_e/(2M)))/(m_e(1 - m_e/(2M))) = (1 - m_e/M)(1 - m_e/(2M))^-1`
= `(1 - m_e/M)(1 + m_e/(2M))`
⇒ `µ_H/µ_D = (1 - m_e/(2M))`
or `µ_H/µ_D = (1 - 1/(2 xx 1840))` = 0.99973 .....(iv) (∵ M = 1840 me)
From (iii) and (iv)
`λ_D/λ_H` = 0.99973, λD = 0.99973 λH
Using λH = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å, we get
λD = 1217.7 Å, 1027.7 Å, 974.04 Å, 951.1 Å
Shift in wavelength (λH – λD) = 0.3 Å.
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