Find the number of photons emitted per second by a 25 W source of monochromatic light of wavelength `6600 Å`. What is the photoelectric current assuming 3% efficiency for photoelectric effect. Given `h = 6.6 xx 10^(-34) Js`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of a single photon.
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h = 6.6 \times 10^{-34} \, \text{Js}\) (Planck's constant),
- \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light),
- \(\lambda = 6600 \, \text{Å} = 6600 \times 10^{-10} \, \text{m}\) (wavelength).

Substituting the values:
\[ E = \frac{(6.6 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{6600 \times 10^{-10} \, \text{m}} \]

### Step 2: Calculate the number of photons emitted per second.
The power of the source is given as 25 W. The number of photons emitted per second can be calculated using the formula:
\[ N = \frac{P}{E} \]
where:
- \(P = 25 \, \text{W}\) (power of the source),
- \(E\) is the energy of a single photon calculated in Step 1.

### Step 3: Calculate the photoelectric current.
Given that the efficiency of the photoelectric effect is 3%, the number of photons that contribute to the photoelectric effect is:
\[ N_{photoelectric} = 0.03 \times N \]
The photoelectric current \(I\) can be calculated using the formula:
\[ I = N_{photoelectric} \times e \]
where \(e = 1.6 \times 10^{-19} \, \text{C}\) (charge of an electron).

### Step 4: Substitute and calculate the values.
Now we will perform the calculations step by step.

### Calculation:

1. **Energy of a single photon**:
\[ E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{6600 \times 10^{-10}} = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} \approx 3.00 \times 10^{-19} \, \text{J} \]

2. **Number of photons emitted per second**:
\[ N = \frac{P}{E} = \frac{25}{3.00 \times 10^{-19}} \approx 8.33 \times 10^{19} \, \text{photons/s} \]

3. **Number of photons contributing to photoelectric effect**:
\[ N_{photoelectric} = 0.03 \times N = 0.03 \times 8.33 \times 10^{19} \approx 2.50 \times 10^{18} \, \text{photons/s} \]

4. **Photoelectric current**:
\[ I = N_{photoelectric} \times e = (2.50 \times 10^{18}) \times (1.6 \times 10^{-19}) \approx 0.40 \, \text{A} \]

### Final Answer:
The number of photons emitted per second is approximately \(8.33 \times 10^{19}\) photons/s, and the photoelectric current is approximately \(0.40 \, \text{A}\).

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