Question

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

${\mu }_{dl}=\frac{g_{d}e}{2m_d}{S}_d$

he deuteron g-factor is 1.71.

Short answer

The difference in energy is ${\lambda }_d\simeq 92\text{cm}$

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

$S_e^2=\frac{1}{2}\left(\frac{3}{2}\right){\hslash }^2=\frac{3}{4}{\hslash }^2$

Spin $S_d^2=1\left(2\right){\hslash }^2=2{\hslash }^2$

Difference of two state

$\frac{1}{2}{\hslash }^2-\left(-{\hslash }^2\right)=\frac{3}{2}{\hslash }^2$

$=\frac{g_d{e}^2{\hslash }^2}{2\pi {\epsilon }_0{c}^2{m}_d{m}_e{a}^3}$

$=\frac{2g_d{e}^2{\hslash }^2}{4\pi {\epsilon }_0{c}^2{m}_d{m}_e{a}^3}$

$=\frac{2g_d{e}^2{\hslash }^4}{c^2{m}_d{m}^2{a}^4}$

$=\frac{3}{2}\frac{g_d}{g_p}\frac{m_p}{m_d}\Delta E_{\text{hydrogen}}$

$\lambda =\frac{c}{\nu }=\frac{ch}{\Delta E}$

localid="1657135928063" $$\begin{gathered} \Rightarrow {\lambda }_d=\frac{2g_d}{3g_p}\frac{m_d}{m_p}{\lambda }_h, \\ {m}_d=2m_p \end{gathered}$$

$\Rightarrow {\lambda }_d=\frac{4}{3}\left(\frac{5.59}{1.71}\right)·21\text{cm}\simeq 92\text{cm}$

Thus, the difference in energy is${\lambda }_d\simeq 92\text{cm}$