An electric bulb with a tungsten filament has an area of 0.20 `cm^2` and is raised to a temperature of the bulb to 3000 K emissivity of the filament is 0.40 and Stefan's constant is `5.7 xx 10^(-5) erg s^(-1) cm^(-1) K^(-4)` . The electrical energy consumed by the bulb is `E_1` watt. If due to fall in voltage, the temperature of the filament falls to 2800 K, then the wattage of the bulb is `E_2` watt. Calculate the value of `E_1 - E_2` , close to nearest integer.

Explanation

To solve the problem, we will use Stefan's law to calculate the electrical energy consumed by the bulb at two different temperatures and then find the difference between these two values.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Area of the filament, \( A = 0.20 \, \text{cm}^2 = 0.20 \times 10^{-4} \, \text{m}^2 \) (but we can keep it in cm² for calculations)
- Temperature at first state, \( T_1 = 3000 \, \text{K} \)
- Temperature at second state, \( T_2 = 2800 \, \text{K} \)
- Emissivity, \( \epsilon = 0.40 \)
- Stefan's constant, \( \sigma = 5.7 \times 10^{-5} \, \text{erg} \, \text{s}^{-1} \, \text{cm}^{-2} \, \text{K}^{-4} \)

2. **Calculate \( E_1 \) (Energy at 3000 K)**:
Using Stefan's law:
\[ E_1 = \epsilon \sigma A T_1^4 \]
Substituting the values:
\[ E_1 = 0.40 \times (5.7 \times 10^{-5}) \times (0.20) \times (3000)^4 \]

First, calculate \( (3000)^4 \):
\[ (3000)^4 = 81 \times 10^{12} \, \text{K}^4 \]

Now substitute:
\[ E_1 = 0.40 \times (5.7 \times 10^{-5}) \times (0.20) \times (81 \times 10^{12}) \]
\[ E_1 = 0.40 \times 5.7 \times 0.20 \times 81 \times 10^{7} \, \text{erg/s} \]
\[ E_1 = 0.40 \times 5.7 \times 0.20 \times 81 \times 10^{7} \approx 36.936 \, \text{W} \]

3. **Calculate \( E_2 \) (Energy at 2800 K)**:
Similarly, for \( T_2 = 2800 \, \text{K} \):
\[ E_2 = \epsilon \sigma A T_2^4 \]
Substituting the values:
\[ E_2 = 0.40 \times (5.7 \times 10^{-5}) \times (0.20) \times (2800)^4 \]

First, calculate \( (2800)^4 \):
\[ (2800)^4 = 61.4656 \times 10^{12} \, \text{K}^4 \]

Now substitute:
\[ E_2 = 0.40 \times (5.7 \times 10^{-5}) \times (0.20) \times (61.4656 \times 10^{12}) \]
\[ E_2 = 0.40 \times 5.7 \times 0.20 \times 61.4656 \times 10^{7} \, \text{erg/s} \]
\[ E_2 \approx 28.028 \, \text{W} \]

4. **Calculate \( E_1 - E_2 \)**:
\[ E_1 - E_2 = 36.936 - 28.028 \approx 8.908 \, \text{W} \]

Rounding to the nearest integer:
\[ E_1 - E_2 \approx 9 \, \text{W} \]

### Final Answer:
The value of \( E_1 - E_2 \) is approximately **9 watts**.