To solve the problem of finding the number of electrons emitted per second by a 24 W source of monochromatic light of wavelength 6600 Å with 3% efficiency for the photoelectric effect, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Given Values**:
- Power of the light source, \( P = 24 \, \text{W} \)
- Wavelength of light, \( \lambda = 6600 \, \text{Å} = 6600 \times 10^{-10} \, \text{m} \)
- Efficiency for the photoelectric effect, \( \eta = 3\% = 0.03 \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \, \text{Js} \)
2. **Calculate the Energy of a Single Photon**:
The energy of a photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where \( c \) is the speed of light, \( c = 3 \times 10^8 \, \text{m/s} \).
Substituting the values:
\[
E = \frac{(6.6 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{6600 \times 10^{-10} \, \text{m}}
\]
3. **Simplify the Calculation**:
First, convert \( \lambda \) to meters:
\[
\lambda = 6600 \times 10^{-10} \, \text{m} = 6.6 \times 10^{-7} \, \text{m}
\]
Now, substituting this back into the energy equation:
\[
E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{6.6 \times 10^{-7}}
\]
The \( 6.6 \) in the numerator and denominator cancels out:
\[
E = \frac{(3 \times 10^8)}{(10^{-7})} = 3 \times 10^{-19} \, \text{J}
\]
4. **Calculate the Effective Power for the Photoelectric Effect**:
Since only 3% of the power is used for the photoelectric effect:
\[
P_{\text{effective}} = 0.03 \times P = 0.03 \times 24 \, \text{W} = 0.72 \, \text{W}
\]
5. **Calculate the Number of Electrons Emitted per Second**:
The number of electrons emitted per second, \( n \), can be calculated using:
\[
n = \frac{P_{\text{effective}}}{E}
\]
Substituting the values:
\[
n = \frac{0.72 \, \text{W}}{3 \times 10^{-19} \, \text{J}} = \frac{0.72}{3 \times 10^{-19}} = 2.4 \times 10^{18} \, \text{electrons/second}
\]
### Final Answer:
The number of electrons emitted per second is \( n = 2.4 \times 10^{18} \).
