# Question 6c0db

May 4, 2017

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:

$P = {n}_{\gamma} \cdot \frac{h c}{\lambda} \approx 7 \times {10}^{- 7} \setminus W$