Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectivel

Q: Why do we use Wien's Displacement Law here?

The problem gives us the wavelength at which the emission intensity is maximum ($\lambda_m$). Wien's Displacement Law is the specific physical principle that relates this peak wavelength to the absolute temperature of the black body.

Q: Does the energy radiated depend on the volume or the surface area?

It depends strictly on the surface area. Radiation is a surface phenomenon, so we use $A = 4\pi R^2$ for a sphere, not the volume formula.

Q: What if the bodies were not perfect black bodies?

If they were not perfect black bodies, the problem would have provided their emissivities ($e$). The Stefan-Boltzmann Law would then be modified to $E = e \sigma A T^4$. Since the problem states to consider them as black bodies, we assume $e = 1$.

Q: Why did we take the ratio of temperatures as $\frac{\lambda_B}{\lambda_A}$ and not $\frac{\lambda_A}{\lambda_B}$?

Because Wien's Displacement Law states that $\lambda_m$ and $T$ are *inversely* proportional ($\lambda_m \propto \frac{1}{T}$). Therefore, $\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A}$.

Q: Is it necessary to convert the radius from cm to meters?

No, because we are taking a ratio ($\frac{R_A}{R_B}$). As long as both radii are in the same units, the conversion factors will cancel out perfectly.

Question

Two spherical bodies $A$ (radius $6	ext{ cm}$ ) and $B$ (radius $18	ext{ cm}$) are at temperatures $T_1$ and $T_2$, respectively. The maximum intensity in the emission spectrum of $A$ is at $500	ext{ nm}$ and in that of $B$ is at $1500	ext{ nm}$. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by $A$ to that of $B$?

Sigma Percentile · Accepted Answer

The problem asks us to find the ratio of the rate of total energy radiated by two spherical black bodies, $A$ and $B$. We are given their radii and the wavelengths at which their emission spectra peak. This is a classic problem that beautifully combines two fundamental laws of black body radiation: Wien's Displacement Law and the Stefan-Boltzmann Law.

### Analyzing the Setup

Imagine two glowing spheres. Sphere $A$ is smaller, with a radius $R_A = 6	ext{ cm}$, and it emits radiation that peaks at a shorter wavelength, $\lambda_A = 500	ext{ nm}$. Sphere $B$ is larger, with a radius $R_B = 18	ext{ cm}$, and its radiation peaks at a longer wavelength, $\lambda_B = 1500	ext{ nm}$.

Because sphere $A$ peaks at a shorter wavelength, our intuition should immediately tell us that it must be hotter than sphere $B$. But exactly how much hotter? That's where Wien's Displacement Law comes in.

### Finding the Temperature Ratio

Wien's Displacement Law states that the wavelength corresponding to maximum intensity ($\lambda_m$) is inversely proportional to the absolute temperature ($T$) of the black body. Mathematically, it is expressed as:

$$ \lambda_m T = b $$

where $b$ is Wien's constant. From this, we can see that $T \propto \frac{1}{\lambda_m}$. Let's use this proportionality to find the ratio of their temperatures:

$$ \frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A} $$

Substituting the given values:

$$ \frac{T_A}{T_B} = \frac{1500	ext{ nm}}{500	ext{ nm}} = 3 $$

This confirms our intuition: body $A$ is exactly three times hotter than body $B$.

### The Master Equation: Stefan-Boltzmann Law

Now that we have the temperature ratio, we need to find the ratio of the total energy radiated per second. The Stefan-Boltzmann Law tells us that the rate of energy radiated ($E$) by a black body is proportional to its surface area ($A$) and the fourth power of its absolute temperature ($T$):

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

Since both bodies are spherical, their surface area is given by $A = 4\pi R^2$. Substituting this into our equation gives:

$$ E = \sigma (4\pi R^2) T^4 $$

This means the energy radiated is proportional to the square of the radius and the fourth power of the temperature:

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

### Final Calculation

We can now set up the ratio for the energy radiated by body $A$ to that of body $B$:

$$ \frac{E_A}{E_B} = \left(\frac{R_A}{R_B}ight)^2 \left(\frac{T_A}{T_B}ight)^4 $$

Let's carefully substitute the values we know. The ratio of their radii is $\frac{R_A}{R_B} = \frac{6}{18} = \frac{1}{3}$, and the ratio of their temperatures is $\frac{T_A}{T_B} = 3$.

$$ \frac{E_A}{E_B} = \left(\frac{1}{3}ight)^2 (3)^4 $$

$$ \frac{E_A}{E_B} = \left(\frac{1}{9}ight) 	imes 81 $$

**$$ \frac{E_A}{E_B} = 9 $$**

Despite being three times smaller in radius (which reduces its surface area by a factor of 9), body $A$ is three times hotter. Because the radiated energy depends on the *fourth power* of temperature, this temperature difference overwhelmingly compensates for the smaller size, resulting in body $A$ radiating 9 times more energy than body $B$.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Wien’s Displacement Law$

$Temperature Ratio$

$Stefan-Boltzmann Law$

$Ratio of Radiated Energy$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Heat Transfer

Analyzing the Setup

Finding the Temperature Ratio

The Master Equation

Final Calculation

Similar Questions

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 body with area $A$ and temperature $T$ and emissivity $e = 0.6$ is kept inside a spherical black body. What will be the maximum energy radiated?

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

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

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

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,

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wien’s Displacement Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stefan-Boltzmann Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Radiated Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward