The Sun has a radius of 6.955 × 10⁸ km and a rotational period of 25.38 days. What is the difference in the wavelength of the peak of the observed spectrum between light received from the approaching side of the Sun and from the receding side of the Sun? Observations of the middle of the Sun place the spectral peak at 480 nm.

Question

Quizplus · Accepted Answer

Using the formula for the Doppler shift, we have 
$\Delta \lambda / \lambda = v_r/c$, 
where $\Delta \lambda$ is the difference in wavelength, $\lambda$ is the wavelength of the spectral peak at the middle of the Sun (480 nm), $v_r$ is the radial velocity of the approaching or receding side of the Sun, and $c$ is the speed of light.

The radial velocity of a point on the equator of the Sun is given by $v_r = \omega r$, where $r$ is the radius of the Sun (6.955 $	imes$ 10$^{5}$ km) and $\omega$ is the angular velocity in radians/s. We can find $\omega$ using $v = \omega r$, where $v$ is the tangential velocity of a point on the equator.

$v = 2 \pi r / T$, where $T$ is the rotational period of the Sun (25.38 days = 2.193 $	imes$ 10$^{6}$ s). 
Thus, $\omega = 2 \pi / T = 2.866 	imes 10^{-6}$ rad/s. 
$v_r$ at the equator is then $v_r = 1.96$ km/s.

The wavelength difference is then 
$\Delta \lambda = \lambda (v_r/c) = (480 	imes 10^{-9} \ 	ext{m})(1.96 	imes 10^3 \ 	ext{m/s} / 3 	imes 10^8 \ 	ext{m/s}) = 3.2 	imes 10^{-9}$ m, or 3.2 nm. 
Therefore, the answer is B.

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