Question

In the lowest energy state of the hydrogen atom, the most probable distance of the single electron from the central proton (the nucleus) is$\mathit{r}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{\text{m}}$. (a) Compute the magnitude of the proton’s electric field at that distance. The component ${\mathbf{\mu }}_{s,z}$of the proton’s spin magnetic dipole moment measured on a z axis is $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{26}}\mathbf{}\mathbf{}\raisebox{1ex}{$\mathbf{\text{J}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{T}}$}\right.$. (b) Compute the magnitude of the proton’s magnetic field at the distance$\mathit{r}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathbf{}\mathbf{\text{m}}$on the z axis. (Hint: Use Eq. 29-27.) (c) What is the ratio of the spin magnetic dipole moment of the electron to that of the proton?

Short answer

(a) Magnitude of the proton’s electric field will be $5.3\times {10}^{11}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.$.

(b) Magnitude of the proton’s magnetic field will be $2.0\times {10}^{-2}\text{T}$.

(c) Ratio of the spin magnetic dipole moment of the electron to that of a proton is $6.6\times {10}^2$.

Step-by-step solution

Listing the given quantities:

Distance between electron and proton, $r=5.2\times {10}^{-11}\text{m}$

The component of the proton’s magnetic dipole moment along z-axis,${\mu }_{s,z}=1.4\times {10}^{-26}\raisebox{1ex}{$\text{\hspace{0.17em}J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$

Understanding the concepts of magnetic and electric field:

A moving charge produces both, an electric field as well as a magnetic field. The electric field is a region around a charge, in which any other charge experiences a force. Similarly, a magnetic field is a space around a magnet where any other magnet experiences a force.

The electric field is given due to a charge q at a separation r is given as-

$\mathrm{E}=\frac{}{4{\mathrm{\pi \epsilon }}_0}\times \left(\frac{\mathrm{q}}{{\mathrm{r}}^2}\right)$

The magnetic field is given as,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{\mu }}_{\mathrm{s},\mathrm{z}}}{2{\mathrm{\pi r}}^3}$

The Bohr Magneton,

${\mathrm{\mu }}_{\mathrm{b}}=\frac{\mathrm{eh}}{4{\mathrm{\pi m}}_{\mathrm{e}}}$

(a) Calculations of the magnitude of the proton’s electric field:

The electric field using Coulomb’s law is,

$\begin{array}{rcl}E& =& \frac{1}{4\pi {\epsilon }_0}\times \left(\frac{q}{r^2}\right)\\ & =& 9\times {10}^9\raisebox{1ex}{${\text{Nm}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{${\text{C}}^{\text{2}}$}\right.\times \left(\frac{1.6\times {10}^{-19}\text{\hspace{0.17em}C}}{{(5.2\times {10}^{-11}\text{\hspace{0.17em}m})}^2}\right)\\ & =& 5.3\times {10}^{11}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\end{array}$

Hence, the magnitude of the proton’s electric field will be $\begin{array}{rcl}& & 5.3\times {10}^{11}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\end{array}$.

(b) Calculations of the magnitude of the proton’s magnetic field:

You can write the magnetic field as.

$\begin{array}{rcl}B& =& \frac{{\mu }_0{\mu }_{s,z}}{2\pi r^3}\\ & =& \frac{4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{\hspace{0.17em}H}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.}\times 1.4\times {10}^{-26}{\displaystyle \raisebox{1ex}{$\text{\hspace{0.17em}J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}}{2\pi \times {(5.2\times {10}^{-11}\text{\hspace{0.17em}m})}^3}\\ & =& 2.0\times {10}^{-2}\text{T}\end{array}$

Hence, the magnitude of the proton’s magnetic field will be $\begin{array}{rcl}& & 2.0\times {10}^{-2}\text{T}\end{array}$.

(c) Calculations of the ratio of the spin magnetic dipole moment of the electron to that of proton:

You know that

${\mu }_b=\frac{eh}{4\pi m_e}$

And you have

${\mu }_{s,z}=1.4\times {10}^{-26}\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.$

Thus, you can take the ratio as $\frac{{\mu }_b}{{\mu }_{s,z}}$. You get

$\begin{array}{rcl}\frac{{\mu }_b}{{\mu }_{s,z}}& =& \frac{{\displaystyle \left(\frac{eh}{4\pi m_e}\right)}}{1.4\times {10}^{-26}{\displaystyle \raisebox{1ex}{$\text{\hspace{0.17em}J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}}\\ & =& \frac{{\displaystyle 9.27\times {10}^{-24}\raisebox{1ex}{${\displaystyle \text{\hspace{0.17em}J}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{T}}$}\right.}}{{\displaystyle 1.4}{\displaystyle \times }{\displaystyle {{\displaystyle 10}}^{-26}}{\displaystyle }\raisebox{1ex}{$\text{\hspace{0.17em}J}$}\!\left/ \!\raisebox{-1ex}{$\text{T}$}\right.}\\ & =& 6.6\times {10}^2\end{array}$

Hence, the ratio of the spin magnetic dipole moment of the electron to that of proton is $6.6\times {10}^2$.