Question

Excited sodium atoms emit two closely spaced spectrum lines called the sodium doublet(Fig. 40-27) with wavelengths 588.995 nm and 589.592 nm. (a) What is the difference in energy between the two upper energy levels (n = 3, I = 1)? (b) This energy difference occurs because the electron’s spin magnetic moment can be oriented either parallel or anti-parallel to the internal magnetic field associated with the electron’s orbital motion. Use your result in (a) to find the magnitude of this internal magnetic field.

Short answer

a) The energy difference between the two upper energy levels $\left(\mathrm{n}=3,\mathrm{I}=1\right)$is 2.11 meV.

b) The magnitude of this internal magnetic field is 18 T.

Step-by-step solution

The given data:

Wavelengths of the two closely spaced lines are ${\mathrm{\lambda }}_1=588.995\mathrm{nm}\mathrm{and}{\mathrm{\lambda }}_2=589.592\mathrm{nm}$.

The given state, n = 3, I = 1

Understanding the concept of magnetic resonance and Plank’s relation:

Magnetic resonance, absorption or radiation by electrons or atomic nuclei in response to the use of other magnetic fields.

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

For the given closely spaced lines, the value of the energy difference between the two lines can be given using Planck's relation and the given data. Again, for the energy difference, there exists a magnetic field due to the resonance between them; this is calculated using the energy equation from the Stern-Gerlach experiment.

Formulas:

The energy of the photon due to Planck’s relation,

$∆\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (1)

Here, consider the known data below.

The Plank’s constant is,

$$\begin{gathered} \mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\left(6.242\times {10}^{18}\times 6.63\times {10}^{-34}\right)\mathrm{eV}.\mathrm{s} \\ =41.384\times {10}^{-16}\mathrm{eV}.\mathrm{s} \\ \mathrm{The}\mathrm{speed}\mathrm{of}\mathrm{light}\mathrm{is}, \\ \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =\left(3\times {10}^8\times {10}^9\right)\mathrm{nm}/\mathrm{s} \\ =3\times {10}^{17}\mathrm{nm}/\mathrm{s} \end{gathered}$$

The energy difference between two closely spaced lines from the Stern-Gerlach experiment is,

$∆\mathrm{E}=2{\mathrm{\mu }}_{\mathrm{B}}\mathrm{B}$ ….. (2)

Here, the Bohr magneton is,

${\mathrm{\mu }}_{\mathrm{B}}=5.788\times {10}^{-5}\mathrm{eV}/\mathrm{T}$

(a) Calculation of the energy difference between two upper levels:

Using the given data in equation (1), the energy difference between the two upper levels of the lines can be calculated is as follows.

$$\begin{gathered} ∆\mathrm{E}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \\ =\mathrm{hc}\left[\frac{1}{{\mathrm{\lambda }}_1}-\frac{1}{{\mathrm{\lambda }}_2}\right] \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} ∆\mathrm{E}=\left(41.384\times {10}^{-16}\mathrm{eV}.\mathrm{s}\right)\left(3\times {10}^{17}\mathrm{nm}/\mathrm{s}\right)\left[\frac{1}{588.995\mathrm{nm}}-\frac{1}{589.592\mathrm{nm}}\right] \\ =1240\mathrm{eV}.\mathrm{nm}\left[\frac{0.0016978}{\mathrm{nm}}-\frac{0.0016961}{\mathrm{nm}}\right] \\ =1240\mathrm{eV}.\mathrm{nm}\left[\frac{1.7\times {10}^{-6}}{\mathrm{nm}}\right] \\ =2.11\mathrm{meV} \end{gathered}$$

Hence, the value of the energy difference is 2.11 meV.

(b) Calculation of the magnitude of the internal magnetic field

Using the above value in equation (2), the magnitude of the given internal magnetic field can be calculated as follows:

$$\begin{gathered} \mathrm{B}=\frac{∆\mathrm{E}}{2{\mathrm{\mu }}_{\mathrm{B}}} \\ =\frac{2.11\times {10}^{-3}\mathrm{eV}}{2\left(5.788\times {10}^{-5}\mathrm{eV}/\mathrm{T}\right)} \\ =18\mathrm{T} \end{gathered}$$

Hence, the value of the magnetic field is 18 T.