Question

A silicon solar cell behaves like a battery with a 0.46 V terminal voltage. Suppose that 1.0 W of light of wavelength 620 nm falls on a solar cell and that 50%% of the photons give their energy to charge carriers, creating a current. What is the solar cell's efficiency that is, what percentage of the energy incident on the cell is converted to electric energy?

Answer 1

We have that the percentage of the energy incident on the cell that is converted to electric energy is

$n=11\%$

From the question we are told that:

Voltage $V=0.46V$

Power of light $P=1.0W$

Wavelength $w=620nm$

50 \% of the photons give their energy to charge carriers,

Generally, the equation for number of Protons is mathematically given by

$N_p=\frac{P}{E}$

$N_p=\frac{P \lambda}{hc}$

$N_p=\frac{1}{(6.62*10^(-34)}*\frac{3*10^8}{(570*10^{-9}))}$

$N_p=2.87*10^{18}$

Generally, the equation for Number of electron is mathematically given by

$N_e=50 \% *n_3$

$N_e=0.5*2.87*10^{18}$

$N_e=1.43*10^{18}$

Therefore

Total current

$I= e*N_e$

Where

e=electron Charge

Therefore

$I=1.43*10^{18}*1.6*10^-{19}$

$I=0.230A$

Generally, the equation for Power is mathematically given by

$P=VI$

$P=0.46*0.230$

$P=0.1058W$

Therefore

Efficiency

$n=\frac{0.1058}{1}$

$n=0.1058$

$n=11\%$

In conclusion

The percentage of the energy incident on the cell that is converted to electric energy is

$n=11\%$

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