Find the number of electrons emitted per second by a 24W source of monochromatic light of wavelength ` 6600 Å` , assuming 3 % efficiency for photoelectric effect (take `h= 6.6 xx 10^(-34)Js)`

Options

A. `48xx 10^(19) `
B. ` 48 xx 10^(17)`
C. `8 xx 10^(19)`
D. `24 xx 10^(17)`

Answer

To solve the problem of finding the number of electrons emitted per second by a 24W source of monochromatic light of wavelength 6600 Å, assuming 3% efficiency for the photoelectric effect, we can follow these steps:

### Step 1: Convert Wavelength to Meters
The wavelength given is 6600 Å. We need to convert this to meters:
\[ \lambda = 6600 \, \text{Å} = 6600 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the Energy of a Single Photon
The energy of a single 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)
- \( \lambda = 6600 \times 10^{-10} \, \text{m} \)

Substituting the values:
\[ E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{6600 \times 10^{-10}} \]
Calculating this gives:
\[ E = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} = 3 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the Total Power Available for the Photoelectric Effect
The total power of the source is given as 24 W. Since the efficiency of the photoelectric effect is 3%, the effective power used for the photoelectric effect is:
\[ P_{\text{effective}} = 0.03 \times 24 \, \text{W} = 0.72 \, \text{W} \]

### Step 4: Calculate the Number of Photons Emitted per Second
The number of photons emitted per second can be calculated using the formula:
\[ n = \frac{P_{\text{effective}}}{E} \]
Substituting the values:
\[ n = \frac{0.72}{3 \times 10^{-19}} = 2.4 \times 10^{18} \, \text{photons/s} \]

### Step 5: Conclusion
The number of electrons emitted per second is equal to the number of photons emitted per second, which is:
\[ \text{Number of electrons emitted per second} = 2.4 \times 10^{18} \]

### Final Answer
The number of electrons emitted per second by the source is approximately \( 2.4 \times 10^{18} \).

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