Temperature of an ordinary bulb is `3000^@K`. At what wave length will it emit maximum energy –

Options

A. 10Å
B. 100Å
C. `10^3Å`
D. `10^4Å`

Answer

To solve the problem of finding the wavelength at which an ordinary bulb emitting at a temperature of 3000 K emits maximum energy, we will use Wien's Displacement Law.

### Step-by-Step Solution:

**Step 1: Understand Wien's Displacement Law**
Wien's Displacement Law states that the wavelength of maximum emission (λm) of a black body is inversely proportional to its absolute temperature (T). The formula is given by:
\[ \lambda_m = \frac{b}{T} \]
where \( b \) is Wien's displacement constant, approximately \( 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \).

**Step 2: Identify the Given Values**
- Temperature \( T = 3000 \, \text{K} \)
- Wien's constant \( b = 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \)

**Step 3: Substitute the Values into the Formula**
Now, we will substitute the values of \( b \) and \( T \) into the equation:
\[ \lambda_m = \frac{2.898 \times 10^{-3}}{3000} \]

**Step 4: Perform the Calculation**
Calculating \( \lambda_m \):
\[ \lambda_m = \frac{2.898 \times 10^{-3}}{3000} = \frac{2.898}{3} \times 10^{-6} \]
\[ \lambda_m = 0.966 \times 10^{-6} \, \text{m} \]

**Step 5: Convert to Angstroms**
Since \( 1 \, \text{m} = 10^{10} \, \text{Å} \), we can convert the wavelength from meters to angstroms:
\[ \lambda_m = 0.966 \times 10^{-6} \, \text{m} \times 10^{10} \, \text{Å/m} \]
\[ \lambda_m = 9660 \, \text{Å} \]

**Step 6: Round Off to the Nearest Significant Figure**
Rounding off, we can say:
\[ \lambda_m \approx 10,000 \, \text{Å} \]

### Conclusion
The wavelength at which the ordinary bulb emits maximum energy is approximately \( 10,000 \, \text{Å} \).