Question

How many photoelectrons are ejected per second in the experiment represented by the graph?

Answer 1

The maximum current produced in the experiment is, \(I=10 \mu \mathrm{A}\)

The rate of change flowing through a conductor is equal to the current.

So,

$$ I=\frac{\Delta q}{\Delta t} $$

solve for \(q\)

$$ \begin{aligned} &\Delta g=I \Delta t \\ &\Delta q=(10 \mu \mathrm{A})(1.0 \mathrm{~s}) \\ &\Delta q=10 \times 10^{-6} \mathrm{~A} \end{aligned} $$

Let \(N\) be the number of electron passing through conductor.

$$ \begin{aligned} \Delta q &=N e \\ N &=\frac{\Delta q}{e} \\ N &=\frac{10 \times 10^{-6} \mathrm{~A}}{1.60 \times 10^{-19} \mathrm{C}} \\ N &=6.25 \times 10^{13} \end{aligned} $$