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.

Answer 1

Let µ_H and µ_D 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 n_i and n_f 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 m_e)

From (iii) and (iv)

`λ_D/λ_H` = 0.99973, λ_D = 0.99973 λ_H

Using λ_H = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å, we get

λ_D = 1217.7 Å, 1027.7 Å, 974.04 Å, 951.1 Å

Shift in wavelength (λ_H – λ_D) = 0.3 Å.