Question

The $\mathbf{21}\mathbf{cm}$ Line: One of the most important windows to the mysteries of the cosmos is the $\mathbf{21}\mathbf{cm}$ line. With it astronomers map hydrogen throughout the universe. An important trait is that it involves a highly forbidden transition that is, accordingly, quite long-lived. But it is also an excellent example of the coupling of angular momentum. Hydrogen's ground state has no spin-orbit interaction—for$\mathit{l}\mathbf{=}\mathbf{0}$there is no orbit. However, the proton and electron magnetic moments do interact. Consider the following simple model.

(a) The proton seesitself surrounded by a spherically symmetric cloud of 1s electron, which has an intrinsic magnetic dipole moment/spin that of course, has a direction. For the purpose of investigating its effect the proton, treat this dispersed magnetic moment as behaving effectively like a single loop of current whose radius is${\mathbf{a}}_{\mathbf{0}}$then find the magnetic field at the middle of the loop in terms of e,$\mathit{\hslash }$,${\mathit{m}}_{\mathbf{e}}$ , ${\mathit{\mu }}_{\mathbf{0}}$and${\mathit{a}}_{\mathbf{0}}$.

(b) The proton sits right in the middle of the electron's magnetic moment. Like the electron the proton is a spin$\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$particle, with only two possible orientations in a magnetic field. Noting however, that its spin and magnetic moment are parallel rather than opposite, would the interaction energy be lower with the proton's spin aligned or anti-aligned with that of the electron?

(c) For the proton${\mathit{g}}_{\mathbf{\rho }}$.is 5.6. Obtain a rough value for the energy difference between the two orientations.

(d) What would be the wavelength of a photon that carries away this energy difference?

Short answer

(a)

The magnetic field that the proton would feel from the orbit of the electron is $\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{e}\sout{\mathrm{h}}}{4{\mathrm{\pi m}}_{\mathrm{e}}{\left({\mathrm{a}}_{\mathrm{o}}\right)}^3}$.

(b) When the magnetic moment/spin of the proton is aligned with the field of the electron and hence antialigned with the electron's spin; the electron would be in a low energy state.

(c) The energy difference between the two state is $2.22\times {10}^{-6}\mathrm{eV}$.

(d) The wavelength of the light that is emitted in the transition is $28.6$cm.

Step-by-step solution

Use Formula of magnetic field

The expression to determine the magnetic field Bat the center of the loop of radius and current I is given by,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{r}}$

Here,${\mathbf{\mu }}_{\mathbf{0}}$is the permeability of free space, I is the current and r is the radius of the loop.

The expression for the base current I is given by,

$\mathit{I}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{t}}$

The expression for the distance travelled d by the object of velocity v in time t is given by,

$\mathbf{d}\mathbf{=}\mathbf{vt}$

The expression for the quantization of the angular momentum equation is given by,

data-custom-editor="chemistry" $\mathbf{mvr}\mathbf{=}\mathbf{n}\sout{\mathbf{h}}$

Here, m is the mass of the object, v is the velocity of the object, r is the radius of the orbit anddata-custom-editor="chemistry" $\sout{\mathbf{h}}$ is the plank's constant.

Calculate the magnetic field at the center of the loop

The relation of magnetic field is expressed as,

$\begin{array}{rcl}\mathrm{B}& =& \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{r}}\\ & & \end{array}$

The magnetic moment $\mu$is expressed as

$\mu =IA=I\pi r^2$

So,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{2{\mathrm{\pi r}}^3}$ …(5)

The magnitude of electron moment $\mu$is expressed as,

$\mathrm{\mu }=\frac{\sqrt{3}\mathrm{e}\mathrm{\hslash }}{2{\mathrm{m}}_{\mathrm{e}}}$

Substitute the value of $\mu$in the equation (5).

$\begin{array}{l}\mathrm{B}=\frac{{\mathrm{\mu }}_0}{2{\mathrm{\pi r}}^3}\left(\frac{\sqrt{3}\mathrm{e}\mathrm{\hslash }}{2{\mathrm{m}}_{\mathrm{e}}}\right)\\ \mathrm{B}=\frac{\sqrt{3}{\mathrm{\mu }}_0\mathrm{e}\mathrm{\hslash }}{4{\mathrm{m}}_{\mathrm{e}}{\mathrm{\pi r}}^3}\end{array}$

Approximate r is equal to $a_0$the then,

$\mathrm{B}=\frac{\sqrt{3}{\mathrm{\mu }}_0\mathrm{e}\mathrm{\hslash }}{4{\mathrm{m}}_{\mathrm{e}}{{\mathrm{\pi a}}_0}^3}$

Hence the magnetic field that the proton would feel from the orbit of the electron is

$\mathrm{B}=\frac{\sqrt{3}{\mathrm{\mu }}_0\mathrm{e}\mathrm{\hslash }}{4{\mathrm{m}}_{\mathrm{e}}{{\mathrm{\pi a}}_0}^3}$

(b): Explanation of property of electron and is a type of angular momentum of the electron

The property of electron and is a type of angular momentum of the electron. When the spin is clockwise it is called the spin-up and when the spin is anticlockwise it is known as the anti-spin.

The alignment of the proton spin and that of the electron spin produces a low level interaction energy. The magnetic dipole has the lowest energy configuration in the magnetic field when it is aligned with the field. The lower interaction energy is with the proton, the magnetic field of the proton being aligned with the magnetic field and is produced by the electron. The spin of the proton is parallel to the magnetic field vector and is also aligned with the field.

The spin vector of an electron is parallel to the magnetic moment of the vector and is opposite to it. This suggests that the spin of the electron and the proton are anti-aligned to provide a low level of interaction energy.

The picture of the lowest energy configuration of the spin $\left(S\right)$and the magnetic moment ${\mathbf{\mu }}_{\mathbf{S}}$is shown below.

When the magnetic moment/spin of the proton is aligned with the field of the electron and hence antialigned with the electron's spin; the electron would be in a low energy state.

(c): Use Formula of energy difference between the two states of the proton

The energy difference between the two states of the proton is aligned versus anti-aligned and the equation is given by,

$\mathbf{U}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{j}}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{2}}{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\pi m}}_{\mathbf{e}}{\mathbf{m}}_{\mathbf{p}}{\left(a_0\right)}^{\mathbf{3}}}$

Here, $m_p$is the mass of the proton and $m_j$is the spin of the state.

Calculate the energy difference

$\mathrm{U}=\frac{{\mathrm{m}}_{\mathrm{j}}{\mathrm{g\mu }}_0{\mathrm{e}}^2{\sout{\mathrm{h}}}^2}{8{\mathrm{\pi m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}{\left({\mathrm{a}}_0\right)}^3}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{U}& =& \frac{\left(\pm {\displaystyle \frac{1}{2}}\right)\left(5.6\right)\left(4\mathrm{\pi }\times {10}^7\mathrm{N}/{\mathrm{A}}^2\right){\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2{\left(1.055\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\mathrm{\pi }\left(9.1\times {10}^{-31}\mathrm{kg}\right)\left(1.6\times {10}^{-16}\mathrm{C}\right){\left(5.29\times {10}^{-11}\mathrm{m}\right)}^3}\\ & =& \left(\pm \frac{1}{2}\right)3.55\times {10}^{-25}\end{array}$

The expression for the energy difference is calculated as,

$\begin{array}{rcl}∆\mathrm{U}& =& 3.55\times {10}^{-25}\mathrm{J}\\ & =& 3.55\times {10}^{-25}\mathrm{J}\left(\frac{{10}^{19}\mathrm{eV}}{1.6\mathrm{J}}\right)\\ & =& 2.22\times {10}^{-6}\mathrm{eV}\end{array}$

The energy difference between aligned and antialigned states would be twice this,$4.4\times {10}^{-6}\text{eV}$

Hence the energy difference between the two state is localid="1660138235244" $4.4\times {10}^{-6}\text{eV}$.

(d): Use Formula for the wavelength of the photon

The expression for the wavelength of the photon

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}}$

Step 2: Calculate the wavelength of the photon

$\begin{array}{c}\lambda =\frac{(6.63\times {10}^{-34}\text{\hspace{0.17em}Js})(3\times {10}^8\text{\hspace{0.17em}m/s})}{7\times {10}^{-25}\text{\hspace{0.17em}J}}\\ =0.286\text{\hspace{0.17em}m}\left(\frac{100\text{\hspace{0.17em}cm}}{1\text{\hspace{0.17em}m}}\right)\\ =28.6\text{\hspace{0.17em}cm}\end{array}$

Hence the wavelength of the light that is emitted in the transition is$28.6\mathrm{cm}$