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Anika [276]
3 years ago
5

A monochromatic red laser beam emitting 1 mW at a wavelength of 638 nm is incident on a silicon solar cell. Find the following:

a. The number of photons per second incident on the cell b. The maximum possible efficiency of conversion of this laser c. beam to electricity
Physics
1 answer:
tia_tia [17]3 years ago
8 0

Answer:

Explanation:

First we calculate the energy of the photon

E=(Planck constant × speed of light in vacuum)÷ wave length

E=\frac{6.626*10^{34}*2.998*10^{8}  }{638*10^{-9} } = 3.114*10^{49}

Next we find the total energy per second

total energy= 1*10^{-3}W *\frac{1JS^{-1} }{1W}  = 1*10^{-3} JS^{-1}

Next we calculate the number the photon per second

= total energy ÷ energy of 1 photon

= \frac{1*10^{-3} JS^{-1}  }{3.114*10^{49} } =  3.21*10^{-53}  \ photons/sec

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Taking the first derivative:

\frac{d}{da} \kappa = \frac{d}{da}  (\frac{a}{a^2 + b^2})

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da}  (\frac{1}{a^2 + b^2} )

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1)  (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da}  (a^2+b^2)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a  (\frac{1}{(a^2 + b^2)^2} ) (2* a)

\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2}  -  2 a^2  (\frac{1}{(a^2 + b^2)^2} )

\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 -  2 a^2)

\frac{d}{da} \kappa = \frac{b^2 -  a^2}{(a^2 + b^2)^2}

This clearly will be zero when

a^2 = b^2

as both are greater (or equal) than zero, this implies

a=b

The second derivative is

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