Question 42.10: CASE STUDY The Zeeman Effect in the Sun’s Hydrogen Spectrum ......
CASE STUDY The Zeeman Effect in the Sun’s Hydrogen Spectrum
In 1908, George Hale was the first person to exploit the Zeeman effect to measure the magnetic field in sunspots. Although Hale observed heavy elements such as iron, we’ll consider how we might observe the Zeeman effect in the Sun’s hydrogen spectrum.
A. Assume that the magnetic field strength in a typical sunspot is 0.1 T, and that only allowed transitions occur under the conditions in the Sun. Estimate the magnitude of the difference in wavelength Δλ between either of the two Hγ lines that result from the normal Zeeman effect and the original (B = 0) central line. What is \Delta \lambda / \lambda_0, where \lambda_0=434.047 nm is the wavelength of original line?
B. Hale reported that to see the Zeeman effect with his diffraction grating, he had to use the thirdorder part of the spectrum. If we wish to see the splitting of Hγ, with a resolving power that is three times the minimum required in the third order, how many rulings will our grating need?
C. Hale also reported that his grating had 567 rulings per millimeter. Assuming the same for our grating, what is the angular separation between the two lines?
A.
INTERPRET and ANTICIPATE
The difference in energy between the central line and either of the other two lines comes from Equation 42.30. This energy difference corresponds to a wavelength difference, which we can derive using calculus.
SOLVE
The allowed transitions have \Delta m=0, \pm 1. Start with Equation 42.30 to write an expression for the energy difference between the central line and the higher energy line generated by Δm = 1.
Write the energy of a photon in terms of its wavelength by using f = c/λ (Eq. 34.20) and E = hf (Eq. 40.8).
E=h f=\frac{h c}{\lambda} \quad \quad (2)Differentiate Equation (2) with respect to wavelength λ.
\frac{d E}{d \lambda}=-\frac{h c}{\lambda^2} \quad \quad (3)Eliminate dE from Equations (1) and (3) to arrive at an expression for dλ.
d \lambda=-\frac{\mu_{ Bohr } B}{h c} \lambda^2Use this expression to find the magnitude of the finite difference Δλ and \Delta \lambda / \lambda_0.
CHECK AND THINK
The Zeeman effect is very subtle. Next we’ll consider the quality of the diffraction grating required.
B.
INTERPRET and ANTICIPATE
This part of the problem requires a review of Section 36-5. The minimum resolving power needed is given by R=\lambda_{ av } / \Delta \lambda (Eq. 36.20), where \lambda_{av} is the average wavelength of two lines that are just barely separated, and Δλ is the difference in their wavelength. For these lines, \lambda_{ av } \approx \lambda_0, so R=\lambda_0 / \Delta \lambda.
SOLVE
The minimum resolving power needed is the inverse of the expression we found in part A.
The resolving power is related to the number of rulings according to Equation 36.21. (In this equation, m refers to the order number, not the quantum number.)
R = Nm \quad \quad (36.21)We’d like R to be three times the minimum required resolving power, and like Hale we plan to use the third-order part of the spectrum.
\begin{aligned}& R=N m=3 R_{\min } \\& N=\frac{3 R_{\min }}{m}=\frac{3}{3}\left(\frac{1}{2 \times 10^{-6}}\right) \\ & N=5 \times 10^5 \text { rulings }\end{aligned}C.
INTERPRET and ANTICIPATE
This part of the problem also requires information from Section 36-5. The dispersion is given by Equation 36.19,
which we can solve for Δθ. First, find the angular position of the third-order spectrum, using d sin θ = mλ(Eq. 36.15).
SOLVE
Find the distance between rulings, using the given 567 rulings per millimeter.
Both lines have very nearly the same wavelength, so their angular position is roughly the same.
\begin{aligned}& d \sin \theta=m \lambda \\& \theta=\sin ^{-1}\left(\frac{m \lambda}{d}\right)=\sin ^{-1}\left(\frac{3(434.047 nm )}{1.76 \times 10^3 nm }\right) \\& \theta=47.7^{\circ}\end{aligned}Now use Equation 36.19 for the dispersion. Substitute m = 3 (third-order part of spectrum) and information found previously.
\begin{aligned}& D=\frac{\Delta \theta}{\Delta \lambda}=\frac{m}{d \cos \theta} \\& \Delta \theta=\frac{m \Delta \lambda}{d \cos \theta}=\frac{3\left(8.8 \times 10^{-4} nm \right)}{\left(1.76 \times 10^3 nm \right)\left(\cos 47.7^{\circ}\right)} \\ & \Delta \theta=2.2 \times 10^{-6} rad\end{aligned}CHECK AND THINK
To see if this angular separation is sufficient to resolve the two lines, compare it to either line’s half-width. The angular separation between the two lines is about three times greater than their half-width. Lines are said to be barely resolved if the angular separation equals the half-width. In this case, the lines are better than barely resolved.
Although hydrogen is the most abundant element in the Universe and is very often observed in astronomy, there are reasons to observe other elements. Hale observed the Zeeman effect in the spectra of heavier elements in order to estimate the magnetic field strength at various depths within sunspots. He found that the magnetic fields in sunspots are about three to four orders of magnitude greater than the Earth’s magnetic field. The more we learn about atoms in the laboratory, the more we can learn about the Universe.