A black body emits maximum radiation of wavelength is `lambda_1`= 2000 Å at a certain temperature `T_1`. On increasing the temperature, the total energy of radiation emitted is increased 16 times at temperature `T_2` .If `lambda_2` is the wavelength corresponding to which maximum radiation emitted at temperature `T_2` Calculate the value of `(lambda_1/lambda_2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between energy and temperature
According to Stefan-Boltzmann law, the energy emitted by a black body is proportional to the fourth power of its absolute temperature. Therefore, we can express the energies at temperatures \( T_1 \) and \( T_2 \) as:
\[ E_1 \propto T_1^4 \]
\[ E_2 \propto T_2^4 \]

### Step 2: Set up the equation based on the problem statement
The problem states that the total energy of radiation emitted at temperature \( T_2 \) is 16 times that at temperature \( T_1 \):
\[ E_2 = 16 E_1 \]
Using the proportionality from Step 1, we can write:
\[ T_2^4 = 16 T_1^4 \]

### Step 3: Solve for the ratio of temperatures
Taking the fourth root of both sides gives us:
\[ T_2 = 2 T_1 \]
This means that the temperature \( T_2 \) is twice that of \( T_1 \).

### Step 4: Apply Wien's Displacement Law
Wien's Displacement Law states that the product of the wavelength at which the emission is maximum (\( \lambda \)) and the absolute temperature (\( T \)) is a constant:
\[ \lambda_1 T_1 = \lambda_2 T_2 \]

### Step 5: Substitute the known values
From our earlier result, we know that \( T_2 = 2 T_1 \). Substituting this into the equation gives:
\[ \lambda_1 T_1 = \lambda_2 (2 T_1) \]

### Step 6: Simplify the equation
We can cancel \( T_1 \) from both sides (assuming \( T_1 \neq 0 \)):
\[ \lambda_1 = 2 \lambda_2 \]

### Step 7: Find the ratio \( \frac{\lambda_1}{\lambda_2} \)
Rearranging the equation gives:
\[ \frac{\lambda_1}{\lambda_2} = 2 \]

### Final Answer
Thus, the value of \( \frac{\lambda_1}{\lambda_2} \) is 2.

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