Two spherical black bodies A and B,having radii `r_(A)` and `r_(B) =2r_(A)` emit radiation with peak intensities at wavelength 400 nm and 800 nm respectively. If their temperature are `T_(A)` and `T_(B)` respectively in Kelvin scale, their emissive powers are `E_(A)` and `E_(B)` then :

Options

A. `T_(A)/(T_(B))=2`
B. `T_(A)/(T_(B))=4`
C. `E_(A)/(E_(B))=8`
D. `E_(A)/(E_(B))=4`

Explanation

To solve the problem, we will use Wien's Displacement Law and Stefan-Boltzmann Law. Let's break down the solution step-by-step.

### Step 1: Apply Wien's Displacement Law
Wien's Displacement Law states that the product of the peak wavelength of emission (\( \lambda_{max} \)) and the absolute temperature (\( T \)) of a black body is a constant. Mathematically, it can be expressed as:

\[ \lambda_{max} \cdot T = b \]

where \( b \) is Wien's displacement constant.

For bodies A and B, we have:

\[ \lambda_{max,A} \cdot T_A = \lambda_{max,B} \cdot T_B \]

Given:
- \( \lambda_{max,A} = 400 \, \text{nm} \)
- \( \lambda_{max,B} = 800 \, \text{nm} \)

Substituting these values into the equation gives:

\[ 400 \cdot T_A = 800 \cdot T_B \]

### Step 2: Simplify the Equation
We can simplify the equation from Step 1:

\[ \frac{T_A}{T_B} = \frac{800}{400} = 2 \]

This means:

\[ T_A = 2T_B \]

### Step 3: Use Stefan-Boltzmann Law
The Stefan-Boltzmann Law states that the emissive power \( E \) of a black body is proportional to the fourth power of its absolute temperature and its surface area:

\[ E \propto A \cdot T^4 \]

The surface area \( A \) of a sphere is given by:

\[ A = 4\pi r^2 \]

For bodies A and B, the emissive powers can be expressed as:

\[ E_A \propto 4\pi r_A^2 \cdot T_A^4 \]
\[ E_B \propto 4\pi r_B^2 \cdot T_B^4 \]

### Step 4: Calculate the Ratio of Emissive Powers
To find the ratio of emissive powers \( \frac{E_A}{E_B} \):

\[ \frac{E_A}{E_B} = \frac{4\pi r_A^2 \cdot T_A^4}{4\pi r_B^2 \cdot T_B^4} \]

The \( 4\pi \) cancels out:

\[ \frac{E_A}{E_B} = \frac{r_A^2 \cdot T_A^4}{r_B^2 \cdot T_B^4} \]

Given \( r_B = 2r_A \):

\[ \frac{E_A}{E_B} = \frac{r_A^2 \cdot T_A^4}{(2r_A)^2 \cdot T_B^4} \]

This simplifies to:

\[ \frac{E_A}{E_B} = \frac{r_A^2 \cdot T_A^4}{4r_A^2 \cdot T_B^4} = \frac{T_A^4}{4T_B^4} \]

### Step 5: Substitute \( T_A \) in Terms of \( T_B \)
From Step 2, we know \( T_A = 2T_B \):

\[ \frac{E_A}{E_B} = \frac{(2T_B)^4}{4T_B^4} = \frac{16T_B^4}{4T_B^4} = 4 \]

### Final Result
Thus, the ratio of the emissive powers of bodies A and B is:

\[ \frac{E_A}{E_B} = 4 \]