A power radiated by a black body is `P_0`. and the wavelength corresponding to the maximum energy is around `lambda`. On changing the temperature of the black body, It was observed that the power radiated is increased to `256/81P_0`. The shift in the wavelength corresponding to the maximum energy will be

Options

A. `+lambda_0//4`
B. `+lambda_0//2`
C. `-lambda_0//4`
D. `-lambda_0//2`

Answer

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

### Step 1: Understand the relationship between power and temperature
According to Stefan-Boltzmann Law, the power radiated by a black body is given by:
\[ P \propto T^4 \]
This means that if the power changes, the temperature must also change.

### Step 2: Set up the initial conditions
Let:
- Initial power \( P_1 = P_0 \)
- Initial temperature \( T_1 \)
- New power \( P_2 = \frac{256}{81} P_0 \)
- New temperature \( T_2 \)

Using the relationship from Step 1:
\[ \frac{P_1}{P_2} = \left(\frac{T_1}{T_2}\right)^4 \]

Substituting the values:
\[ \frac{P_0}{\frac{256}{81} P_0} = \left(\frac{T_1}{T_2}\right)^4 \]

### Step 3: Simplify the equation
Cancel \( P_0 \) from both sides:
\[ \frac{81}{256} = \left(\frac{T_1}{T_2}\right)^4 \]

### Step 4: Take the fourth root
Taking the fourth root of both sides gives us:
\[ \frac{T_1}{T_2} = \left(\frac{81}{256}\right)^{1/4} \]

Calculating the fourth root:
\[ \frac{T_1}{T_2} = \frac{3}{4} \]

### Step 5: Express \( T_2 \) in terms of \( T_1 \)
From the above equation:
\[ T_2 = \frac{4}{3} T_1 \]

### Step 6: Apply Wien's Displacement Law
According to Wien's Displacement Law:
\[ \lambda T = b \]
where \( b \) is a constant.

For the initial conditions:
\[ \lambda_1 T_1 = b \]
For the new conditions:
\[ \lambda_2 T_2 = b \]

### Step 7: Relate the wavelengths
From the equations for \( \lambda_1 \) and \( \lambda_2 \):
\[ \lambda_1 T_1 = \lambda_2 T_2 \]

Substituting \( T_2 \):
\[ \lambda_1 T_1 = \lambda_2 \left(\frac{4}{3} T_1\right) \]

### Step 8: Solve for \( \lambda_2 \)
Cancel \( T_1 \) from both sides:
\[ \lambda_1 = \frac{4}{3} \lambda_2 \]

Rearranging gives:
\[ \lambda_2 = \frac{3}{4} \lambda_1 \]

### Step 9: Find the change in wavelength
The change in wavelength \( \Delta \lambda \) is given by:
\[ \Delta \lambda = \lambda_2 - \lambda_1 \]
Substituting \( \lambda_2 \):
\[ \Delta \lambda = \frac{3}{4} \lambda_1 - \lambda_1 \]
\[ \Delta \lambda = \frac{3}{4} \lambda_1 - \frac{4}{4} \lambda_1 \]
\[ \Delta \lambda = -\frac{1}{4} \lambda_1 \]

### Final Answer
The shift in the wavelength corresponding to the maximum energy is:
\[ \Delta \lambda = -\frac{1}{4} \lambda_0 \]

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