Question

A point source emitting photons of 2 eV energy and 1 W of power is kept at a distance of 1m from a small piece of a photoelectric material of area 10^-4 m² . If the efficiency of generation of photoelectrons is 10%, then the number of photoelectrons generated are 𝑓×10¹² per second. The value of 𝑓 is __________ (rounded off to two decimal places).
Given: 1eV = 1.6 × 10⁻¹⁹ J

Answer 1

Correct Answer: 2.3 - 2.7

We need to find the number of photoelectrons generated per second given the power emitted by the photon source, the energy of each photon, the area of the photoelectric material, and the efficiency of photoelectron generation.

 

1. Calculate the energy per second (power) received by the photoelectric material:

The power emitted by the point source is 1 W. The power \( P \) is related to the number of photons \( N_{\text{photons}} \) emitted per second by:

\[ P = N_{\text{photons}} \times E_{\text{photon}} \] where \( E_{\text{photon}} = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-19} \, \text{J} \) is the energy of each photon.

Now, we can solve for \( N_{\text{photons}} \):

\[ N_{\text{photons}} = \frac{P}{E_{\text{photon}}} = \frac{1 \, \text{W}}{3.2 \times 10^{-19} \, \text{J}} = 3.125 \times 10^{18} \, \text{photons/second} \]

1. Calculate the number of photons incident on the photoelectric material:

The total number of photons \( N_{\text{incident}} \) incident on the photoelectric material is proportional to the area \( A \) of the material and the distance \( r \) from the source. The intensity \( I \) at a distance \( r \) from the source is given by:

\[ I = \frac{P}{4 \pi r^2} \] where \( r = 1 \, \text{m} \) is the distance from the source. Therefore, the intensity at the photoelectric material is:

\[ I = \frac{1 \, \text{W}}{4 \pi (1 \, \text{m})^2} = \frac{1}{4 \pi} \, \text{W/m}^2 \]

The number of photons incident on the material per second is:

\[ N_{\text{incident}} = I \times A = \frac{1}{4 \pi} \times 10^{-4} = 7.96 \times 10^{-6} \, \text{photons/second} \]

1. Apply the efficiency factor for photoelectron generation:

The efficiency of generation of photoelectrons is 10%, so the number of photoelectrons generated per second \( N_{\text{photoelectrons}} \) is:

\[ N_{\text{photoelectrons}} = 0.10 \times N_{\text{incident}} = 0.10 \times 7.96 \times 10^{-6} = 2.3 \times 10^{12} \, \text{photoelectrons/second} \]

Final Answer:

The number of photoelectrons generated is approximately \( f \times 10^{12} \), where \( f \approx 2.3 \).

The value of \( f \) is approximately \( \boxed{2.3} \).