If the filament of a 100 W bulb has an area `0.25cm^2` and behaves as a perfect block body. Find the wavelength corresponding to the maximum in its energy distribution. Given that Stefan's constant is `sigma=5.67xx10^(-8) J//m^(2)s K^(4)`.

Options

A. 8751.23
B. 2898.14
C. 9971.9
D. 7055.5

Answer

To solve the problem, we need to find the wavelength corresponding to the maximum in the energy distribution of a 100 W bulb filament, which behaves as a perfect black body. We will use Stefan's law and Wien's displacement law to find the solution step by step.

### Step 1: Calculate the energy radiated per unit area (E)
The energy radiated per unit area (E) can be calculated using the formula:
\[ E = \frac{P}{A} \]
where \(P\) is the power of the bulb and \(A\) is the area of the filament.

Given:
- \(P = 100 \, W\)
- \(A = 0.25 \, cm^2 = 0.25 \times 10^{-4} \, m^2\)

Substituting the values:
\[ E = \frac{100}{0.25 \times 10^{-4}} = \frac{100}{0.000025} = 4 \times 10^6 \, J/(m^2 \cdot s) \]

### Step 2: Use Stefan-Boltzmann Law to find the temperature (T)
According to Stefan-Boltzmann law:
\[ E = \sigma T^4 \]
where \(\sigma = 5.67 \times 10^{-8} \, J/(m^2 \cdot s \cdot K^4)\).

Now, substituting the value of \(E\):
\[ 4 \times 10^6 = 5.67 \times 10^{-8} T^4 \]

Rearranging to find \(T^4\):
\[ T^4 = \frac{4 \times 10^6}{5.67 \times 10^{-8}} = 7.055 \times 10^{13} \]

Now, taking the fourth root to find \(T\):
\[ T = (7.055 \times 10^{13})^{1/4} \approx 2898.14 \, K \]

### Step 3: Use Wien's Displacement Law to find the wavelength (\(\lambda_{max}\))
Wien's displacement law states:
\[ \lambda_{max} = \frac{b}{T} \]
where \(b = 2.89 \times 10^{-3} \, m \cdot K\).

Substituting the value of \(T\):
\[ \lambda_{max} = \frac{2.89 \times 10^{-3}}{2898.14} \approx 9.9719 \times 10^{-7} \, m \]

Converting to angstroms (1 m = \(10^{10}\) angstroms):
\[ \lambda_{max} \approx 9.9719 \times 10^{-7} \, m \times 10^{10} \, \text{angstrom/m} \approx 9971.9 \, \text{angstrom} \]

### Final Answer
The wavelength corresponding to the maximum in its energy distribution is approximately:
\[ \lambda_{max} \approx 9971.9 \, \text{angstrom} \]