Power radiated by a black body is `P_0` and the wavelength corresponding to maximum energy is around `lamda_0`, On changing the temperature of the black body, it was observed that the power radiated becames `(256)/(81)P_0`. The shift in wavelength corresponding to the maximum energy will be

Options

A. `+lamda_0/4`
B. `+lamda_0/2`
C. `-lamda_0/4`
D. `-lamda_0/2`

Answer

To solve the problem step by step, we will use the principles of black body radiation, specifically Stefan-Boltzmann law and Wien's displacement law.

### Step 1: Understand the relationship between power and temperature
According to the Stefan-Boltzmann law, the power radiated by a black body is directly proportional to the fourth power of its absolute temperature (T). This can be expressed as:
\[ P \propto T^4 \]

### Step 2: Set up the equation for the change in power
Let \( P_0 \) be the initial power and \( T_0 \) be the initial temperature. After changing the temperature, the new power \( P' \) is given as:
\[ P' = \frac{256}{81} P_0 \]

Using the relationship from Step 1, we can write:
\[ \frac{P'}{P_0} = \left(\frac{T'}{T_0}\right)^4 \]
Substituting the values:
\[ \frac{256}{81} = \left(\frac{T'}{T_0}\right)^4 \]

### Step 3: Solve for the new temperature
Taking the fourth root of both sides gives:
\[ \frac{T'}{T_0} = \left(\frac{256}{81}\right)^{1/4} \]
Calculating the fourth root:
\[ \frac{T'}{T_0} = \frac{4}{3} \]
Thus, we find:
\[ T' = \frac{4}{3} T_0 \]

### Step 4: Apply Wien's Displacement Law
Wien's displacement law states that the wavelength corresponding to the maximum energy (\( \lambda \)) and temperature (T) is given by:
\[ \lambda T = b \]
where \( b \) is a constant.

For the initial state:
\[ \lambda_0 T_0 = b \]

For the new state:
\[ \lambda' T' = b \]

### Step 5: Relate the new wavelength to the initial wavelength
Substituting \( T' = \frac{4}{3} T_0 \) into the equation for the new state:
\[ \lambda' \left(\frac{4}{3} T_0\right) = b \]
From the initial state, we know:
\[ \lambda_0 T_0 = b \]

Setting the two equations equal gives:
\[ \lambda' \left(\frac{4}{3} T_0\right) = \lambda_0 T_0 \]

### Step 6: Solve for the new wavelength
Dividing both sides by \( T_0 \):
\[ \lambda' \frac{4}{3} = \lambda_0 \]
Thus, we can find \( \lambda' \):
\[ \lambda' = \frac{3}{4} \lambda_0 \]

### Step 7: Calculate the shift in wavelength
The shift in wavelength (\( \Delta \lambda \)) is given by:
\[ \Delta \lambda = \lambda' - \lambda_0 \]
Substituting the values:
\[ \Delta \lambda = \frac{3}{4} \lambda_0 - \lambda_0 \]
\[ \Delta \lambda = \frac{3}{4} \lambda_0 - \frac{4}{4} \lambda_0 \]
\[ \Delta \lambda = -\frac{1}{4} \lambda_0 \]

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