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 can use Wien's Displacement Law. This law relates the temperature of a black body to the wavelength at which it emits maximum radiation.

### Step-by-Step Solution:

1. **Understand Wien's Displacement Law**:
According to Wien's Displacement Law, the wavelength (\( \lambda_{max} \)) at which the radiation is maximum is inversely proportional to the temperature (T) of the black body. The formula is given by:
\[ \lambda_{max} = \frac{b}{T} \]
where \( b \) is Wien's constant.

2. **Identify the Given Values**:
From the problem, we have:
- Temperature \( T = 3000 \, K \)
- Wien's constant \( b = 0.288 \, cm \cdot K = 0.288 \times 10^{-2} \, m \cdot K = 2.88 \times 10^{-3} \, m \cdot K \)

3. **Substitute the Values into the Formula**:
Now, we can substitute the values into the formula:
\[ \lambda_{max} = \frac{2.88 \times 10^{-3} \, m \cdot K}{3000 \, K} \]

4. **Calculate the Wavelength**:
Performing the calculation:
\[ \lambda_{max} = \frac{2.88 \times 10^{-3}}{3000} = 0.96 \times 10^{-6} \, m \]

5. **Convert the Wavelength to Angstroms**:
Since the answer is required in angstroms, we convert meters to angstroms. We know that:
\[ 1 \, \text{angstrom} = 10^{-10} \, m \]
Therefore:
\[ \lambda_{max} = 0.96 \times 10^{-6} \, m = 0.96 \times 10^{4} \, \text{angstrom} = 9600 \, \text{angstrom} \]

6. **Final Answer**:
The wavelength at which the electric bulb radiates maximum energy is:
\[ \lambda_{max} = 9600 \, \text{angstrom} \]