A light beam of power 1.5 mW and 400 nm wavelength incident on a cathode. If quantum efficiency is 0.1 % then, find out obtained photo current and number of photoelectron per second.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the energy of one photon
The energy of a photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h = 6.626 \times 10^{-34} \, \text{J s}\) (Planck's constant)
- \(c = 3 \times 10^{8} \, \text{m/s}\) (speed of light)
- \(\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}\)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the number of photons per second
The power of the light beam is given as \(P = 1.5 \, \text{mW} = 1.5 \times 10^{-3} \, \text{W}\).

The number of photons (\(N\)) per second can be calculated using the formula:
\[ N = \frac{P}{E} \]
Substituting the values:
\[ N = \frac{1.5 \times 10^{-3}}{4.97 \times 10^{-19}} \approx 3.02 \times 10^{15} \, \text{photons/s} \]

### Step 3: Calculate the number of photoelectrons emitted per second
The quantum efficiency is given as 0.1%, which means only 0.1% of the incident photons will result in the emission of photoelectrons.

Calculating the number of photoelectrons emitted:
\[ \text{Number of photoelectrons} = N \times \text{Quantum Efficiency} \]
\[ \text{Number of photoelectrons} = 3.02 \times 10^{15} \times 0.001 = 3.02 \times 10^{12} \, \text{electrons/s} \]

### Step 4: Calculate the photo current
The photo current (\(I\)) can be calculated using the formula:
\[ I = n \times e \]
Where:
- \(n = 3.02 \times 10^{12} \, \text{electrons/s}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\) (charge of an electron)

Substituting the values:
\[ I = (3.02 \times 10^{12}) \times (1.6 \times 10^{-19}) \approx 4.83 \times 10^{-7} \, \text{A} = 0.483 \, \mu A \]

### Final Answers:
- Number of photoelectrons emitted per second: \(3.02 \times 10^{12} \, \text{electrons/s}\)
- Photo current: \(0.483 \, \mu A\)

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