The formula for the relativistic Doppler shift is

$\begin{align*}\lambda = \lambda_0 \sqrt{\frac{c - v}{c+ v}}\end{align*}$
where the sign of $v$ depends on whether the observer and the source are approaching or receding. In this problem, there are two sources, and one of them is receding and one of them is approaching, so the difference in wavelength is
$\begin{align}\Delta \lambda = \lambda_0 \left(\sqrt{\frac{c + v}{c - v}} - \sqrt{\frac{c - v}{c+ v}} \right) \label{eqn60:1}...$
where now $v$ is the speed (not the velocity) of a particle on the Sun's equator due to the Sun's rotation. Now,
$\begin{align*}\sqrt{\frac{c + v}{c - v}} = \frac{v+ c}{\sqrt{c^2 - v^2}}\end{align*}$
and
$\begin{align*}\sqrt{\frac{c - v}{c + v}} = \frac{v- c}{\sqrt{c^2 - v^2}}\end{align*}$
so, we can rewrite equation (1) as
$\setcounter{equation}{1}\begin{align}\Delta \lambda = \lambda_0 \left(\frac{v+ c}{\sqrt{c^2 - v^2}} - \frac{v- c}{\sqrt{c^2 -...$
Now, we guess that the speed $v$ is tiny compared to the speed of light. This is intuitive, because one normally does not think of relativistic speeds when one thinks of the Sun macroscopically. Thus, we set
$\begin{align*}\sqrt{c^2 - v^2} \approx c\end{align*}$
Then equation (2) becomes
$\begin{align*}\Delta \lambda = \frac{\lambda_0 \cdot 2 v}{c}\end{align*}$
We are given $\Delta \lambda = 1.8 \cdot 10^{-12}$ , $\lambda_0 = 122 \cdot 10^{-9}$ , so we can easily solve for $v$ . We find that

$\begin{align*}v = 2213 \mbox{ m/s} = \boxed{2.213} \mbox{ km/s}\end{align*}$
This is indeed a nonrelativistic speed, so the approximation that $\sqrt{c^2 - v^2} \approx c$ was a good one. Therefore, answer (B) is correct.

Solution to 1996 Problem 60