Question

Question #6c0db

Answer 1

We can switch the min current, `I_(min) = 0.32 mu A`, as 1 Amp is 1 Coulombs per second, into an equivalent number of electrons per second:

`n_e = (0.32xx10^(-6))/(e^(-)) "s^(-1)`

That's how many electrons that needs to pass each point along the circuit in one second in order to create `I_(min)`.

If every photon gives its energy to an electron, then:

`n_gamma = (0.32xx10^(-6))/(e^(-)) "s^(-1)`

So the incoming power is:

`P = n_gamma cdot (hc)/lambda approx 7 xx 10^(-7) \ W`