Chapter 6: Q38P (page 290)

Calculate the wavelength, in centimeters, of the photon emitted under a hyperfine transition in the ground state (n=1) of deuterium. Deuterium is "heavy" hydrogen, with an extra neutron in the nucleus; the proton and neutron bind together to form a deuteron, with spin 1 and magnetic moment
$μ_{d l} = \frac{g_{d} e}{2 m_{d}} S_{d}$
he deuteron g-factor is 1.71.

Short Answer

Expert verified

The difference in energy is $λ_{d} ≃ 92 cm$

Step by step solution

Definehyperfine transition.

The interaction of electrons with the magnetic moments of the nucleus splits a spectral line of an atom or molecule into two or more closely spaced components.

Step 2: Draw the difference in energy of deuterium.

Given

$S_{e}^{2} = \frac{1}{2} (\frac{3}{2}) ℏ^{2} = \frac{3}{4} ℏ^{2}$

μ_{d} = \frac{g_{d} e}{2 m_{d}} S_{d}, E_{h f}^{1} = \frac{μ_{0} g_{d} e^{2}}{3 π m_{d} m_{e} a^{3}} <S_{d} \cdot S_{e}>

Spin $S_{d}^{2} = 1 (2) ℏ^{2} = 2 ℏ^{2}$

$<S_{d} \cdot S_{e}> = \{\begin{cases} \frac{1}{2} (\frac{3}{4} ℏ^{2} - \frac{3}{4} ℏ^{2} - 2 ℏ^{2}) = - ℏ^{2} \\ \frac{1}{2} (\frac{15}{4} ℏ^{2} - \frac{3}{4} ℏ^{2} - 2 ℏ^{2}) = \frac{1}{2} ℏ^{2} \end{cases}$

Difference of two state

$\frac{1}{2} ℏ^{2} - (- ℏ^{2}) = \frac{3}{2} ℏ^{2}$

$= \frac{g_{d} e^{2} ℏ^{2}}{2 π ε_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} ℏ^{2}}{4 π ε_{0} c^{2} m_{d} m_{e} a^{3}}$

$= \frac{2 g_{d} e^{2} ℏ^{4}}{c^{2} m_{d} m^{2} a^{4}}$

$= \frac{3}{2} \frac{g_{d}}{g_{p}} \frac{m_{p}}{m_{d}} Δ E_{hydrogen}$

$λ = \frac{c}{ν} = \frac{c h}{Δ E}$

localid="1657135928063" $\Rightarrow λ_{d} = \frac{2 g_{d}}{3 g_{p}} \frac{m_{d}}{m_{p}} λ_{h}, m_{d} = 2 m_{p}$

$\Rightarrow λ_{d} = \frac{4}{3} (\frac{5.59}{1.71}) \cdot 21 cm ≃ 92 cm$

Thus, the difference in energy is $λ_{d} ≃ 92 cm$

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Short Answer

Step by step solution

Definehyperfine transition.

Step 2: Draw the difference in energy of deuterium.

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