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 `(lambda_(A)//lambda_(B))` of their wavelengths `lambda_(A)` and `lambda_(B)` at which the peaks oc cur in their respective radiation curves is :

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two spherical stars, A and B. The radius of star A is 400 times that of star B, and the power emitted by star A is \(10^4\) times the power emitted by star B.

### Step 2: Use the Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the power emitted by a black body is given by:
\[ P = A \cdot \epsilon \cdot \sigma \cdot T^4 \]
where:
- \(P\) is the power,
- \(A\) is the surface area,
- \(\epsilon\) is the emissivity,
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(T\) is the temperature.

For a sphere, the surface area \(A\) is given by \(4\pi R^2\).

### Step 3: Write the power equations for both stars
Let the radius of star B be \(R_B\). Then the radius of star A is \(R_A = 400 R_B\).

The power emitted by star A (\(P_A\)) and star B (\(P_B\)) can be expressed as:
\[ P_A = 4\pi R_A^2 \cdot \epsilon_A \cdot \sigma \cdot T_A^4 \]
\[ P_B = 4\pi R_B^2 \cdot \epsilon_B \cdot \sigma \cdot T_B^4 \]

### Step 4: Set up the ratio of powers
Given that \(P_A = 10^4 P_B\), we can substitute the expressions for \(P_A\) and \(P_B\):
\[ 4\pi (400 R_B)^2 \cdot \epsilon_A \cdot \sigma \cdot T_A^4 = 10^4 \cdot (4\pi R_B^2 \cdot \epsilon_B \cdot \sigma \cdot T_B^4) \]

### Step 5: Simplify the equation
Cancel \(4\pi\) and \(\sigma\) from both sides:
\[ (400^2 R_B^2) \cdot \epsilon_A \cdot T_A^4 = 10^4 \cdot (R_B^2 \cdot \epsilon_B \cdot T_B^4) \]
This simplifies to:
\[ 160000 \cdot \epsilon_A \cdot T_A^4 = 10^4 \cdot \epsilon_B \cdot T_B^4 \]

### Step 6: Rearrange to find the temperature ratio
Rearranging gives:
\[ \frac{T_A^4}{T_B^4} = \frac{10^4 \cdot \epsilon_B}{160000 \cdot \epsilon_A} \]
This can be simplified to:
\[ \frac{T_A^4}{T_B^4} = \frac{10^4}{16 \cdot 10^4} \cdot \frac{\epsilon_B}{\epsilon_A} = \frac{1}{16} \cdot \frac{\epsilon_B}{\epsilon_A} \]

Assuming both stars have similar emissivity (\(\epsilon_A \approx \epsilon_B\)), we can ignore the emissivity factor:
\[ \frac{T_A^4}{T_B^4} = \frac{1}{16} \]
Taking the fourth root gives:
\[ \frac{T_A}{T_B} = \frac{1}{2} \]

### Step 7: Use Wien's Law to find the wavelength ratio
Wien's Law states that:
\[ \lambda_{max} \cdot T = b \]
where \(b\) is a constant. Therefore, we can write:
\[ \lambda_A \cdot T_A = \lambda_B \cdot T_B \]
Thus, the ratio of the wavelengths is:
\[ \frac{\lambda_A}{\lambda_B} = \frac{T_B}{T_A} \]
Substituting the temperature ratio:
\[ \frac{\lambda_A}{\lambda_B} = \frac{2}{1} = 2 \]

### Final Answer
The ratio of the wavelengths at which the peaks occur in their respective radiation curves is:
\[ \frac{\lambda_A}{\lambda_B} = 2 \]

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