In an experiment on photoelectric effect of light wavelength 400 nm is incident on a metal plate at rate of 5W. The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every `10^6` photons is able to eject a photoelectron, find the photocurrent in the cirucuit.

Explanation

To find the photocurrent in the circuit due to the photoelectric effect, we can 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:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) (wavelength) = \( 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)

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

Calculating this gives:
\[ E \approx 4.97 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the number of photons incident per second
The number of photons \( N \) incident per second can be calculated using the formula:
\[ N = \frac{P}{E} \]
Where \( P \) (power) = \( 5 \, \text{W} \).

Substituting the values:
\[ N = \frac{5 \, \text{W}}{4.97 \times 10^{-19} \, \text{J}} \]

Calculating this gives:
\[ N \approx 1.01 \times 10^{19} \, \text{photons/s} \]

### Step 3: Calculate the number of photoelectrons emitted
Given that on average one out of every \( 10^6 \) photons ejects a photoelectron, the number of photoelectrons \( n_e \) emitted per second is:
\[ n_e = \frac{N}{10^6} \]

Substituting the value of \( N \):
\[ n_e = \frac{1.01 \times 10^{19}}{10^6} \approx 1.01 \times 10^{13} \, \text{photoelectrons/s} \]

### Step 4: Calculate the photocurrent
The photocurrent \( I \) can be calculated using the formula:
\[ I = n_e \cdot e \]
Where \( e \) (charge of an electron) = \( 1.6 \times 10^{-19} \, \text{C} \).

Substituting the values:
\[ I = (1.01 \times 10^{13} \, \text{photoelectrons/s})(1.6 \times 10^{-19} \, \text{C}) \]

Calculating this gives:
\[ I \approx 1.616 \times 10^{-6} \, \text{A} \approx 1.62 \, \mu\text{A} \]

### Final Answer
The photocurrent in the circuit is approximately \( 1.62 \, \mu\text{A} \).

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