The temperature of an ordinary electric bulb is round 3000 K. At what wavelength will it radiate maximum enegy? Given b=0.288cmK

Options

A. `8600Å`
B. `9600Å`
C. `4655Å`
D. `1000Å`

Answer

To find the wavelength at which an ordinary electric bulb radiates maximum energy, we will use Wien's Displacement Law. This law states that the wavelength at which the radiation is maximum (λ_max) is inversely proportional to the temperature (T) of the black body. The formula is given by:

\[ \lambda_{\text{max}} = \frac{b}{T} \]

where:
- \( \lambda_{\text{max}} \) is the wavelength at which the maximum energy is radiated,
- \( b \) is a constant (Wien's displacement constant),
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values:**
- Temperature \( T = 3000 \, K \)
- Wien's displacement constant \( b = 0.288 \, \text{cm K} \)

2. **Convert the constant \( b \) to meters:**
- Since \( 1 \, \text{cm} = 0.01 \, \text{m} \), we convert:
\[ b = 0.288 \, \text{cm K} = 0.288 \times 0.01 \, \text{m K} = 0.00288 \, \text{m K} \]

3. **Use Wien's Displacement Law to calculate \( \lambda_{\text{max}} \):**
\[ \lambda_{\text{max}} = \frac{b}{T} = \frac{0.00288 \, \text{m K}}{3000 \, K} \]

4. **Perform the calculation:**
\[ \lambda_{\text{max}} = \frac{0.00288}{3000} = 0.00000096 \, \text{m} = 9.6 \times 10^{-7} \, \text{m} \]

5. **Convert the wavelength to Angstroms:**
- Since \( 1 \, \text{m} = 10^{10} \, \text{Å} \):
\[ \lambda_{\text{max}} = 9.6 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} = 9600 \, \text{Å} \]

### Final Answer:
The wavelength at which the electric bulb radiates maximum energy is \( 9600 \, \text{Å} \).