The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and th

Q: Why do we assume stars behave like black bodies?

Stars are excellent approximations of black bodies because they are massive, opaque spheres of gas. Any light that enters a star is absorbed and scattered countless times before it can escape, meaning the light it eventually emits is almost entirely thermal radiation dependent only on its surface temperature, perfectly mimicking a theoretical black body.

Q: Why is the ratio inverted in the formula?

According to Wien's Displacement Law, $\lambda_m \propto \frac{1}{T}$. Because they are inversely proportional, when you set up a ratio of temperatures like $\frac{T_1}{T_2}$, it will equal the inverse ratio of their wavelengths, $\frac{\lambda_2}{\lambda_1}$.

Q: Do I need to convert nanometers to meters for this calculation?

In this specific problem, no. Since we are taking a ratio of two wavelengths, the units of nanometers ($\text{nm}$) will cancel out in the numerator and denominator, leaving a dimensionless ratio. You only need to convert to meters if you are calculating the exact temperature using Wien's constant $b$.

Q: What does a ratio of 0.69 mean physically?

A ratio of $0.69$ means that the Sun's surface temperature is only about $69\%$ of the North Star's surface temperature. It proves that the North Star is significantly hotter than our Sun.

Q: Why does a shorter wavelength mean a hotter star?

Higher temperatures mean the atoms and electrons in the star are vibrating and colliding with much more energy. This higher kinetic energy corresponds to higher frequency and shorter wavelength photons being emitted. Therefore, blue light (short wavelength) indicates a much hotter source than red light (long wavelength).

Question

The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and the north star is

Options:

(A) 1.46

(B) 0.69

(C) 1.21

(D) 0.83

Sigma Percentile · Accepted Answer

### The Cosmic Thermometer

Have you ever wondered how astronomers can confidently state the temperature of a star that is millions of light-years away? We cannot send a thermometer into the heart of the cosmos. Instead, we rely on the most profound messenger in the universe: **light**.

When you look up at the night sky, stars do not just appear as white dots; they have subtle hues. Some, like Betelgeuse, shine with a brilliant blue-white intensity. Others, like Antares, glow with a deep, ominous red. Our own Sun sits somewhere in the middle, bathing us in a warm yellow-white glow. These colors are not random; they are direct, undeniable signatures of the star's surface temperature.

### The Idealization: Black Body Radiation

In physics, we often use idealized models to understand complex realities. For stars, the perfect model is a **Black Body**. A black body is a theoretical object that absorbs all radiation that falls upon it and, conversely, is a perfect emitter of radiation.

When a black body is heated, it emits a continuous spectrum of light. However, it does not emit all wavelengths equally. There is always a specific wavelength where the intensity of the emitted light reaches an absolute peak. This peak wavelength is the key to unlocking the star's thermal secrets.

### Wien's Displacement Law: The Master Key

In 1893, the brilliant physicist Wilhelm Wien formulated a law that beautifully connects the color of a glowing object to its temperature. **Wien's Displacement Law** states that the wavelength corresponding to maximum emission ($\lambda_m$) is inversely proportional to the absolute temperature ($T$) of the black body.

Mathematically, this is expressed as:

$$\lambda_m T = b$$

Where $b$ is Wien's displacement constant (approximately $2.898 	imes 10^{-3} 	ext{ m}\cdot	ext{K}$).

The physical intuition here is crucial: **as an object gets hotter, the peak of its emission spectrum shifts to shorter wavelengths**. Think of a piece of iron in a blacksmith's forge. As it heats up, it first glows a dull red (long wavelength). As it gets hotter, it turns yellow, and eventually, if heated enough, it glows a blinding white or even blue (short wavelength).

### Analyzing the Setup: Sun vs. North Star

In our specific problem, we are given the peak emission wavelengths for two celestial bodies:

1.  **The Sun:** $\lambda_{	ext{sun}} = 510 	ext{ nm}$
2.  **The North Star (Polaris):** $\lambda_{	ext{star}} = 350 	ext{ nm}$

Notice that the North Star's peak wavelength ($350 	ext{ nm}$) is significantly shorter than the Sun's ($510 	ext{ nm}$). The $350 	ext{ nm}$ wavelength sits right at the edge of the visible spectrum, bordering on ultraviolet, which explains why the North Star appears bluish-white compared to our yellowish Sun.

### The Master Equation: Setting up the Ratio

We are tasked with finding the ratio of the surface temperature of the Sun ($T_{	ext{sun}}$) to the surface temperature of the North Star ($T_{	ext{star}}$).

Using Wien's Displacement Law, we can write two equations:

$$\lambda_{	ext{sun}} T_{	ext{sun}} = b$$
$$\lambda_{	ext{star}} T_{	ext{star}} = b$$

Since both expressions equal the same constant $b$, we can equate them:

$$\lambda_{	ext{sun}} T_{	ext{sun}} = \lambda_{	ext{star}} T_{	ext{star}}$$

Rearranging this to find the ratio of the temperatures, we get:

$$\frac{T_{	ext{sun}}}{T_{	ext{star}}} = \frac{\lambda_{	ext{star}}}{\lambda_{	ext{sun}}}$$

**This is the critical trap!** Because the relationship is inversely proportional, the ratio of the temperatures is the *inverse* of the ratio of their wavelengths. A common silly mistake is to write $\frac{T_{	ext{sun}}}{T_{	ext{star}}} = \frac{\lambda_{	ext{sun}}}{\lambda_{	ext{star}}}$, which would lead to a completely incorrect physical conclusion.

### Final Calculation: Crunching the Numbers

Now, we simply substitute the given values into our derived ratio equation:

$$\frac{T_{	ext{sun}}}{T_{	ext{star}}} = \frac{350 	ext{ nm}}{510 	ext{ nm}}$$

Notice how the units of nanometers ($	ext{nm}$) perfectly cancel out, leaving us with a dimensionless ratio.

$$\frac{T_{	ext{sun}}}{T_{	ext{star}}} = \frac{35}{51}$$

Performing the division:

$$\frac{T_{	ext{sun}}}{T_{	ext{star}}} \approx 0.68627...$$

Rounding to two decimal places to match our options, we get:

$$\frac{T_{	ext{sun}}}{T_{	ext{star}}} \approx 0.69$$

### Conclusion: The Physical Significance

Our calculated ratio of $0.69$ is less than $1$. This mathematically confirms our earlier physical intuition: because the Sun has a longer peak wavelength, it must have a lower surface temperature than the North Star. Specifically, the Sun's surface is only about $69\%$ as hot as the surface of the North Star.

This simple yet elegant application of Wien's Displacement Law is a cornerstone of astrophysics, allowing us to classify stars across the universe simply by looking at the colors of the light they send us across the dark expanse of space.

Select Answer:

Visualized Solution

$Visualizing the Stars$

$Wien’s Displacement Law$

$Setting up the Ratio$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Heat Transfer

The Cosmic Thermometer

The Idealization

Wien's Displacement Law

Analyzing the Setup

The Master Equation

Final Calculation

Conclusion

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