Question

Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ²H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ = m_eM/(m_e + M) where M is the nuclear mass and m e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ¹H and ²H. (Mass of ¹H nucleus is 1.6725 × 10^–27 kg, Mass of ²H nucleus is 3.3374 × 10^–27 kg, Mass of electron = 9.109 × 10^–31 kg.)

Answer 1

The total energy of the electron in n^th stable orbit in H or like atom

`E_n = (μZ^2e^4)/(8ε_0^2h^2n^2)`

μ = reduced mass of electron, proton and neutron (mass defect)

`E_H = (μ_H(1)^2e^4)/(8ε_0^2h^2) [1/n_1^2 - 1/n_2^2]`

= `(μ_He^4)/(8ε_0^2h^2) [1/1 - 1/2^2]`

= `(μ_He^4)/(8ε_0^2h^2) [3/4]`

`E = hv = h/λ` or `λ_h = h/E_H`

∴ `hv_H = (μ_He^4)/(8ε_0^2h^2) 3/4` 

`v_H = (μ_He^4)/(8ε_0^2h^2) * 3/4`

The percentage difference in the wavelength = `((λ_D - λ_H))/λ_H xx  100`

Percent change in wavelength 

% change `ΔE = [λ_D/λ_H - 1] xx 100`  (∵ ΔE = E₁ – E₂) ......(I)

`hv = (μ_e^4)/(8ε_0^2h^2) [1/n_2^2 - 1/n_2^2]`

`v = (μ_e^4)/(8ε_0^2h^3) [1/n_1^2 - 1/n_2^2]`

`c/λ = (μ_e^4)/(8ε_0^2h^3) [1/n_1^2 - 1/n_2^2]`

`1/λ = (μ_e^4)/(8ε_0^2h^3) [1/n_1^2 - 1/n_2^2]`

As μ = mass defect, e, εε₀c, and h are constants for an atom.

∴ `λ ∞ 1/h`

So equation 1^st can be written as percentage change in the wavelength = `[μ_H/μ_D - 1] xx 100`

For Deutrium: `μ_D = (m_eM_D)/((m_e + M_D))`

Let μ_H be the reduced mass of hydrogen and μ_D that of Deuctrium. Then, the frequency of the 1^st Lyman line in hydrogen is `hf_H = (μ_He^4)/(8ε_0^2h^2) (1 - 1/4) = (μ_He^4)/(8ε_0^2h^2) xx 3/4`

Thus, the wavelength of the transition is `λ_H = 3/4 (μ_He^4)/(8ε_0^2h^3c)`. The wavelength of the transition for the same line in Deuterium is `λ_D = 3/4 (μ_De^4)/(48ε_0^2h^3c)`

∴ The difference in wavelength Δλ = λ_D – λ_H

Hence, the percentage difference is `100 xx (Δλ)/λ_H = (λ_D - λ_H)/λ_H xx 100 = (μ_D - μ_H)/μ_H xx 100`

= `((m_eM_D)/((m_e + M_D)) - (m_eM_H)/((m_e + M_H)))/((m_eM_H)/((m_e + M_H))) xx 100`

= `[((m_e + M_H)/(m_e + M_D)) M_D/M_H - 1] xx 100`

Since, m_e <<  M_H << M_D

`(Δλ)/λ_H xx 100 = [M_H/M_D xx M_D/M_H ((1 + m_e/M_H)/(1 + m_e/M_D)) - 1] xx 100`

= `[(1 + m_e/M_H)(1 + m_e/M_D)^-1 - 1] xx 100` 

= `[(1 + m_e/M_H)(1 - m_e/M_D) - 1] xx 100`   ......[By binomial theorem, (1 + x)ⁿ = 1 + nx is |x| > 1]

`(Δλ)/λ_H xx 100 = [1 + m_e/M_H - m_e/M_D - (m_e)^2/(M_HM_D) - 1] xx 100`

Neglecting `(m_e)^2/(M_HM_D)`, as it is very small.

∵ `μ = (m_eM)/((M + m_e))`  .....(Given)

∴ Percentage change in wavelength = `[((m_eM_H)/((M_H + m_e)))/((m_e + M_D)/((M_D + m_e))) - 1]100`

`(Δλ)/λ_H xx 100 = [M_H/M_D ((M_D + m_e))/((M_H + m_e)) - 1] xx 100`

= `[M_H/M_D (M_D (1 + m_e/M_D))/(M_H (1 + m_e/M_H)) - 1] xx 100`

= `[(1 + m_e/M_D)(1 + m_e/M_H)^-1 - 1] xx 100`

= `[(1 + m_e/M_D)(1 - m_e/M_H) - 1] xx 100`

M_e << M_D so neglecting the higher degree term

`(Δλ)/λ_H xx 100 = [1 - m_e/M_H + m_e/M_D - (m_em_e)/(M_D * M_H) - 1] xx 100`

= `m_e [1/M_D - 1/M_H] xx 100`

= `9.1 xx 10^-31  [1/(3.3374 xx 10^-27) - 1/(1.6725 xx 10^-27)] xx 100`

= `(9.1 xx 10^(-31+2))/10^-27 [(1.6725 - 3.3374)/(3.3374 xx 1.6725)]`

`(Δλ xx 100)/λ_H = (- 9.1 xx 10^(-29 + 27) xx 0.6649)/(3.3374 xx 1.6725) = (-6.05059 xx 10^-2)/5.5180`

`(Δλ xx 100)/λ_H = - 1.084 xx 10^-2%` Decrease in wavelength.

(–) sign shows that λ_D < λ_H.