Gallium Arsenide solar cell has band gap energy 1.5eV. a)What is the required frequency a photon should have in order to excite an electron to jump over to conduction band? b) If the solar irradance of 10 x 10^-4 W is incident on the solar cell, calculate the number of photon enters per second.

Answer 1

a) To find the frequency of a photon required to excite an electron to jump to the conduction band, we can use the equation:

E = hf

where E is the energy of the photon (1.5 eV in this case), h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the photon.

First, we need to convert the energy from eV to Joules:

1.5 eV = 1.5 x 1.6 x 10^-19 J = 2.4 x 10^-19 J

Now, we can rearrange the equation to solve for the frequency:

f = E/h
f = (2.4 x 10^-19 J) / (6.626 x 10^-34 J s)
f ≈ 3.62 x 10^14 Hz

Therefore, the frequency of the photon required to excite an electron to the conduction band is approximately 3.62 x 10^14 Hz.

b) To calculate the number of photons entering the solar cell per second, we can use the formula:

Power = Number of photons x Energy of each photon x Number of photons

Given:
Power = 10 x 10^-4 W
Energy of each photon = 2.4 x 10^-19 J (from part a)
h = 6.626 x 10^-34 J s

Number of photons = Power / (Energy of each photon x f)

Number of photons = (10 x 10^-4) / (2.4 x 10^-19 x 3.62 x 10^14)
Number of photons ≈ 7.01 x 10^14

Therefore, approximately 7.01 x 10^14 photons enter the solar cell per second.