Two spherical stars A and B emit black body radiation. The radius of A is 400 times that of B and A

Q: Why do we assume stars are perfect black bodies?

While stars aren't perfect black bodies, their emission spectra closely approximate black body radiation, making it a very useful and accurate model for calculating their properties like temperature and power.

Q: What does the peak wavelength actually tell us?

The peak wavelength corresponds to the color of the star where it emits the most energy. Hotter stars peak at shorter wavelengths (blue), while cooler stars peak at longer wavelengths (red).

Q: Why is the power proportional to the fourth power of temperature?

This is a fundamental result of the Stefan-Boltzmann Law, derived from integrating Planck's law of black body radiation over all wavelengths and angles.

Q: Can we use this method for any two objects?

Yes, as long as both objects can be reasonably approximated as black bodies and you know the ratio of their radii and emitted powers.

Q: What if the stars had the same power but different radii?

If the powers were the same, the star with the larger radius would have to be cooler (to emit the same total power over a larger area), meaning its peak wavelength would be longer.

Question

Two spherical stars $A$ and $B$ emit black body radiation. The radius of $A$ is 400 times that of $B$ and $A$ emits $10^4$ times the power emitted from $B$. The ratio $\left(\frac{\lambda_A}{\lambda_B}\right)$ of their wavelengths $\lambda_A$ and $\lambda_B$ at which the peaks occur in their respective radiation curves is

Sigma Percentile · Accepted Answer

The universe is a vast, dark canvas, but it is illuminated by the brilliant light of stars. In physics, we often model these stars as perfect **black bodies**—objects that absorb and emit all frequencies of electromagnetic radiation perfectly.

In this thrilling problem, we are tasked with comparing two such stars, $A$ and $B$. Star $A$ is an absolute behemoth, with a radius 400 times larger than Star $B$. Not only is it larger, but it also emits a staggering $10^4$ times more power! Our mission is to find the ratio of their peak emission wavelengths, $\frac{\lambda_A}{\lambda_B}$.

### The Power of a Star: Stefan-Boltzmann Law

To begin our journey, we need a mathematical tool that connects the power a star emits to its physical properties. Enter the **Stefan-Boltzmann Law**. This fundamental law states that the total power $P$ radiated by a black body is proportional to its surface area $A$ and the fourth power of its absolute temperature $T$.

$$P = \sigma A T^4$$

Since our stars are spherical, their surface area is given by $A = 4\pi R^2$. Substituting this into our equation, we see that the power is proportional to the square of the radius and the fourth power of the temperature:

$$P \propto R^2 T^4$$

### The Color of a Star: Wien's Displacement Law

Now, we have a relationship for power, but the question asks about wavelengths. How do we bridge this gap? We use **Wien's Displacement Law**. This elegant law tells us that the wavelength at which a black body emits the most intense radiation (the peak wavelength, $\lambda_m$) is inversely proportional to its absolute temperature.

$$\lambda_m T = b \implies T \propto \frac{1}{\lambda_m}$$

This means that hotter stars peak at shorter wavelengths (appearing blue), while cooler stars peak at longer wavelengths (appearing red).

### Merging the Laws

We now have two powerful relationships. Let's merge them by substituting the temperature proportionality from Wien's Law into our power equation from the Stefan-Boltzmann Law.

$$P \propto R^2 \left(\frac{1}{\lambda_m}ight)^4$$

$$P \propto \frac{R^2}{\lambda_m^4}$$

This single equation beautifully links the total power emitted by a star to its radius and its peak wavelength!

### The Final Calculation

Since we want to find the ratio of the wavelengths, let's rearrange our master equation to solve for $\lambda_m$. By taking the fourth root of both sides, we get:

$$\lambda_m \propto \frac{R^{1/2}}{P^{1/4}}$$

Now, we can set up the ratio for Star $A$ and Star $B$:

$$\frac{\lambda_A}{\lambda_B} = \left(\frac{R_A}{R_B}ight)^{1/2} \left(\frac{P_B}{P_A}ight)^{1/4}$$

Notice that because power is in the denominator, the ratio is $P_B / P_A$, not $P_A / P_B$. This is a classic trap, so we must be careful! We are given that $R_A = 400 R_B$ and $P_A = 10^4 P_B$. Let's substitute these values:

$$\frac{\lambda_A}{\lambda_B} = (400)^{1/2} \left(\frac{1}{10^4}ight)^{1/4}$$

The square root of 400 is 20. The fourth root of $10^4$ is 10, so the fourth root of $1/10^4$ is $1/10$.

$$\frac{\lambda_A}{\lambda_B} = 20 	imes \frac{1}{10} = 2$$

And there we have it! The ratio of their peak wavelengths is exactly **2**. This means Star $A$ peaks at a wavelength twice as long as Star $B$, making it appear significantly redder despite being much larger and more powerful.

Enter Numerical Value:

Visualized Solution

$R_{A} = 400 R_{B}, P_{A} = 1 0^{4} P_{B}$

$P = σ A T^{4}$

$λ_{m} T = b$

$P \propto \frac{R ^{2}}{λ _{m}^{4}}$

$λ_{m} \propto \frac{R ^{1/2}}{P ^{1/4}}$

$\frac{λ _{A}}{λ _{B}} = (\frac{R _{A}}{R _{B}})^{1/2} (\frac{P _{B}}{P _{A}})^{1/4}$

$\frac{λ _{A}}{λ _{B}} = 2$

$Final Answer: 2$

The Sigma Insight: Heat Transfer

The Power of a Star

The Color of a Star

Merging the Laws

The Final Calculation

Similar Questions

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

Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

If the temperature of the sun were to increase from $T$ to $2 T$ and its radius from $R$ to $2 R$ , then the ratio of the radiant energy received on earth to what it was previously, will be

The graph, shown in the diagram, represents the variation of temperature ( $T$ ) of the bodies, $x$ and $y$ having same surface area, with time ( $t$ ) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies

Assuming the sun to be a spherical body of radius $R$ at a temperature of $T$ K, evaluate the total radiant power, incident on earth, at a distance $r$ from the sun. where, $r_{0}$ is the radius of the earth and $σ$ is Stefan's constant.

A black body is at a temperature of $2880 K$ . The energy of radiation emitted by this body with wavelength between $499 nm$ and $500 nm$ is $U_{1}$ , between $999 nm$ and $1000 nm$ is $U_{2}$ and between $1499 nm$ and $1500 nm$ is $U_{3}$ . The Wien constant, $b = 2.88 \times 1 0^{6} nm-K$ . Then,