With what speed should a galaxy move outward with respect to Earth, so that the sodium-D line at wav

Q: Why did the wavelength increase? What does that physically mean?

The wavelength increased because the galaxy is moving away from Earth. As it moves away, each successive wave crest is emitted from a slightly farther distance, causing the waves to stretch out. This shift towards longer wavelengths is called 'Redshift'.

Q: Do I need to convert the wavelengths from Angstroms to meters before calculating?

No, you don't have to! In the formula $v = c \frac{\Delta \lambda}{\lambda}$, you are taking a ratio of two wavelengths. As long as both $\Delta \lambda$ and $\lambda$ are in the same units (like Angstroms), the units will cancel out perfectly.

Q: Is the formula $v = c \frac{\Delta \lambda}{\lambda}$ always valid for any speed?

No, this specific formula is an approximation valid only when the velocity $v$ is much less than the speed of light $c$ ($v \ll c$). For galaxies moving at a significant fraction of the speed of light, we must use the relativistic Doppler effect formula.

Q: What if the observed wavelength was $5884 \text{ \AA}$ instead of $5896 \text{ \AA}$?

If the observed wavelength was shorter than the actual wavelength, it would mean the light waves are being compressed. This is called 'Blueshift', and it indicates that the galaxy is moving *towards* Earth.

Q: Why is it specifically called the 'sodium-D line'?

The sodium-D line refers to a very bright and distinct pair of emission lines in the spectrum of sodium atoms. Because it is so recognizable and its exact wavelength is well-known in laboratories, astronomers frequently use it as a standard reference to measure the redshift of distant stars and galaxies.

Question

With what speed should a galaxy move outward with respect to Earth, so that the sodium-D line at wavelength $5890 	ext{ \AA}$ is observed at $5896 	ext{ \AA}$ ?

Options:

(A) $306 	ext{ km/s}$

(B) $322 	ext{ km/s}$

(C) $296 	ext{ km/s}$

(D) $336 	ext{ km/s}$

Sigma Percentile · Accepted Answer

Have you ever looked up at the night sky and wondered how we know that the universe is expanding? The secret lies hidden in the light reaching us from distant galaxies. In this problem, we are going to explore a beautiful application of the **Doppler Effect** for light, specifically focusing on a phenomenon known as **Redshift**.

### Analyzing the Setup

Imagine a galaxy far, far away. It emits light, and specifically, we are looking at the sodium-D line, a very famous emission line in physics. The actual wavelength of this light, $\lambda$, is $5890 	ext{ \AA}$. However, when astronomers on Earth observe this light through their telescopes, they measure its wavelength, $\lambda'$, to be $5896 	ext{ \AA}$.

Notice something interesting? The observed wavelength is *greater* than the actual wavelength. The light waves appear to have been stretched out. This stretching of light waves to longer (redder) wavelengths happens because the source (the galaxy) is moving *away* from the observer (Earth). This is the classic signature of **Redshift**.

### The Master Equation

To find out exactly how fast this galaxy is running away from us, we need to calculate the change in wavelength, denoted by $\Delta \lambda$.

$$\Delta \lambda = \lambda' - \lambda$$

Substituting our values:

$$\Delta \lambda = 5896 	ext{ \AA} - 5890 	ext{ \AA} = 6 	ext{ \AA}$$

Now, we bring in the Doppler shift formula for light. When the speed of the source $v$ is much less than the speed of light $c$ (which is usually the case, even for fast galaxies), the fractional change in wavelength is directly proportional to the ratio of the galaxy's speed to the speed of light:

$$\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$$

Rearranging this to solve for the velocity $v$, we get our master equation:

$$v = c \left( \frac{\Delta \lambda}{\lambda} ight)$$

### Final Calculation

Let's plug in the numbers. We know the speed of light $c$ is approximately $3 	imes 10^8 	ext{ m/s}$.

$$v = (3 	imes 10^8 	ext{ m/s}) 	imes \left( \frac{6 	ext{ \AA}}{5890 	ext{ \AA}} ight)$$

Notice how elegant this is: because we have a ratio of wavelengths, the units of Angstroms ($	ext{\AA}$) perfectly cancel out. We don't even need to convert them to meters!

$$v = \frac{18 	imes 10^8}{5890} 	ext{ m/s}$$

$$v \approx 305.6 	imes 10^3 	ext{ m/s}$$

Since $10^3 	ext{ m}$ is exactly $1 	ext{ km}$, we can write this as:

**$$v \approx 306 	ext{ km/s}$$**

And there we have it! The galaxy is hurtling away from us at a staggering speed of $306 	ext{ km/s}$. This simple yet profound calculation is exactly how Edwin Hubble first discovered that our universe is expanding.

Animated Solution for Physics - Waves: With what speed should a galaxy move outward with respect to Earth, so that the sodium-D line at wavelength $5890 \overset{˚}{A}$ is observed at $5896 \overset{˚}{A}$ ?

Select Answer:

Visualized Solution

Visualizing the Setup

Doppler Effect (Redshift)

Change in Wavelength

Doppler Shift Formula

Substituting Values

Calculation

Final Answer

The Way Forward

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equation

Final Calculation

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