Let `n_r` and `n_b` be respectively the number of photons emitted by a red bulb and a blue blub of equal power in a given time.

Options

A. `n_r = n_b`
B. `n_rltn_b`
C. `n_rgtn_b`
D. The information is insufficient to get a relation between `(n_r) and (n_b)`.

Answer

To solve the problem, we need to establish the relationship between the number of photons emitted by a red bulb (`n_r`) and a blue bulb (`n_b`) given that both bulbs emit equal power in a given time.

### Step-by-Step Solution:

1. **Understanding Power and Energy**:
The power emitted by a light source is the rate at which it emits energy. If both bulbs emit equal power, we can express this as:
\[ P_{\text{red}} = P_{\text{blue}} \]

2. **Relating Power to Photons**:
The power can also be expressed in terms of the number of photons emitted and the energy of each photon. The total energy emitted in one second can be given by:
\[ P = n \cdot E \]
where \( n \) is the number of photons emitted and \( E \) is the energy of one photon.

3. **Energy of a Photon**:
The energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.

4. **Setting Up the Equation**:
For the red bulb:
\[ P_{\text{red}} = n_r \cdot \frac{hc}{\lambda_r} \]
For the blue bulb:
\[ P_{\text{blue}} = n_b \cdot \frac{hc}{\lambda_b} \]

5. **Equating Powers**:
Since both bulbs emit equal power:
\[ n_r \cdot \frac{hc}{\lambda_r} = n_b \cdot \frac{hc}{\lambda_b} \]

6. **Simplifying the Equation**:
We can cancel \( hc \) from both sides (since it is a common factor):
\[ n_r \cdot \frac{1}{\lambda_r} = n_b \cdot \frac{1}{\lambda_b} \]

7. **Rearranging the Equation**:
Rearranging gives us:
\[ n_r = n_b \cdot \frac{\lambda_r}{\lambda_b} \]

8. **Analyzing Wavelengths**:
We know that the wavelength of red light (\( \lambda_r \)) is greater than the wavelength of blue light (\( \lambda_b \)):
\[ \lambda_r > \lambda_b \]

9. **Conclusion**:
Since \( \frac{\lambda_r}{\lambda_b} > 1 \), it follows that:
\[ n_r > n_b \]

Thus, the number of photons emitted by the red bulb is greater than that emitted by the blue bulb.

### Final Answer:
\[ n_r > n_b \]