When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from `0.26mum` to `0.13mum`. The ratio of the emissive powers of the body at the respective temperatures is

Options

A. `(16)/(1)`
B. `(4)/(1)`
C. `(1)/(4)`
D. `(1)/(16)`

Explanation

To find the ratio of the emissive powers of a black body at two different temperatures corresponding to maximum energy wavelengths, we can follow these steps:

### Step 1: Identify the given data
We have two wavelengths corresponding to maximum energy:
- Initial wavelength, \( \lambda_1 = 0.26 \, \mu m \)
- Final wavelength, \( \lambda_2 = 0.13 \, \mu m \)

### Step 2: Use Wien's Displacement Law
According to Wien's displacement law, the product of the wavelength corresponding to maximum energy (\( \lambda \)) and the temperature (\( T \)) is a constant:
\[ \lambda \cdot T = \text{constant} \]
This means:
\[ \lambda_1 \cdot T_1 = \lambda_2 \cdot T_2 \]

### Step 3: Rearranging the equation to find the temperature ratio
From the equation above, we can express the ratio of the temperatures:
\[ \frac{T_1}{T_2} = \frac{\lambda_2}{\lambda_1} \]

### Step 4: Substitute the values of wavelengths
Substituting the values of \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{T_1}{T_2} = \frac{0.13 \, \mu m}{0.26 \, \mu m} = \frac{1}{2} \]

### Step 5: Write the formula for emissive power
The emissive power \( E \) of a black body is given by:
\[ E = \sigma \cdot A \cdot T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant and \( A \) is the area. For a black body, emissivity \( \epsilon = 1 \).

### Step 6: Find the ratio of emissive powers
The ratio of emissive powers at temperatures \( T_1 \) and \( T_2 \) is:
\[ \frac{E_1}{E_2} = \frac{E_1}{E_2} = \frac{\sigma \cdot A \cdot T_1^4}{\sigma \cdot A \cdot T_2^4} = \frac{T_1^4}{T_2^4} \]

### Step 7: Substitute the temperature ratio into the emissive power ratio
Using the temperature ratio \( \frac{T_1}{T_2} = \frac{1}{2} \):
\[ \frac{E_1}{E_2} = \left(\frac{T_1}{T_2}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

### Conclusion
The ratio of the emissive powers of the body at the respective temperatures is:
\[ \frac{E_1}{E_2} = \frac{1}{16} \]