The maximum energy in thermal radiation from a source occurs at the wavelength 4000Å. The effective temperature of the source

Options

A. 7325 K
B. 800 K
C. `10^(4)K `
D. `10^(6)K`

Answer

To find the effective temperature of a source emitting thermal radiation at a maximum wavelength of 4000 Å (angstroms), we can use Wien's Displacement Law. This law states that the wavelength at which the emission of radiation is maximized is inversely proportional to the absolute temperature of the black body.

### Step-by-Step Solution:

1. **Understand Wien's Displacement Law**:
\[ \lambda_{max} \cdot T = b \]
where:
- \(\lambda_{max}\) is the wavelength at which the maximum energy is emitted,
- \(T\) is the absolute temperature in Kelvin,
- \(b\) is Wien's displacement constant, approximately \(2.898 \times 10^{-3} \, \text{m} \cdot \text{K}\).

2. **Convert Wavelength from Angstroms to Meters**:
The given wavelength is \(4000 \, \text{Å}\).
\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \]
Therefore,
\[ \lambda_{max} = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4 \times 10^{-7} \, \text{m} \]

3. **Rearrange the Formula to Solve for Temperature**:
\[ T = \frac{b}{\lambda_{max}} \]

4. **Substitute the Values into the Formula**:
Using \(b = 2.898 \times 10^{-3} \, \text{m} \cdot \text{K}\) and \(\lambda_{max} = 4 \times 10^{-7} \, \text{m}\):
\[ T = \frac{2.898 \times 10^{-3}}{4 \times 10^{-7}} \]

5. **Calculate the Temperature**:
\[ T = \frac{2.898}{4} \times 10^{4} = 0.7245 \times 10^{4} \approx 7.245 \times 10^{3} \, \text{K} \approx 7245 \, \text{K} \]

6. **Final Answer**:
The effective temperature of the source is approximately \(7245 \, \text{K}\).