To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,
I = q*N
Here q is Flow of electric charge in one second and N the number of electron flow per second.
A the same time the power is described as the applied voltage for the current.
P = VI
We know the charge of electron, [tex]q = 1.602 * 10^{-19}[/tex] Coulombs, then the current is
[tex]I = (1.602*10^{-19})(10^{15})[/tex]
[tex]I = 0.1602 mA[/tex]
And the power in the Beam is
[tex]P = VI[/tex]
[tex]P = (350*10^3)(0.1602)[/tex]
[tex]P = 0.05607 Watts[/tex]
