Two bulbs 'A' and 'B' emit red light and yellow light at 8000 "Å" and 4000 "Å" respectively. The number of photons emitted by both the bulbs per second is the same. If the red bulb is labelled as 100 watts, find the wattage of the yellow bulb.

Answer

To solve the problem, we need to find the wattage of the yellow bulb (bulb B) given that both bulbs emit the same number of photons per second, and the red bulb (bulb A) is rated at 100 watts.

### Step-by-Step Solution:

1. **Understand the relationship between energy, power, and wavelength:**
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light in meters.

2. **Convert the wavelengths from angstroms to meters:**
- For bulb A (red light):
\[ \lambda_1 = 8000 \, \text{Å} = 8000 \times 10^{-10} \, \text{m} = 8 \times 10^{-7} \, \text{m} \]
- For bulb B (yellow light):
\[ \lambda_2 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4 \times 10^{-7} \, \text{m} \]

3. **Calculate the energy of a photon emitted by bulb A:**
\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{8 \times 10^{-7} \, \text{m}} = \frac{1.9878 \times 10^{-25} \, \text{J m}}{8 \times 10^{-7} \, \text{m}} = 2.485 \times 10^{-19} \, \text{J} \]

4. **Calculate the number of photons emitted per second by bulb A:**
Since the power (P) of bulb A is 100 watts, the number of photons emitted per second (\(N_1\)) is given by:
\[ N_1 = \frac{P_1}{E_1} = \frac{100 \, \text{W}}{2.485 \times 10^{-19} \, \text{J}} \approx 4.02 \times 10^{20} \, \text{photons/s} \]

5. **Since both bulbs emit the same number of photons per second, we can set up the equation for bulb B:**
\[ N_2 = N_1 \]
where \(N_2\) is the number of photons emitted per second by bulb B.

6. **Calculate the energy of a photon emitted by bulb B:**
\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{4 \times 10^{-7} \, \text{m}} = \frac{1.9878 \times 10^{-25} \, \text{J m}}{4 \times 10^{-7} \, \text{m}} = 4.9695 \times 10^{-19} \, \text{J} \]

7. **Calculate the power of bulb B:**
Using the relationship \(N_2 = \frac{P_2}{E_2}\), we can express the power of bulb B as:
\[ P_2 = N_2 \cdot E_2 = N_1 \cdot E_2 \]
Substituting \(N_1\) and \(E_2\):
\[ P_2 = (4.02 \times 10^{20} \, \text{photons/s}) \cdot (4.9695 \times 10^{-19} \, \text{J}) \approx 200 \, \text{W} \]

### Final Answer:
The wattage of the yellow bulb (bulb B) is **200 watts**.