How many electrons emerge per second?

[Ithryndil]

Homework Statement

The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.20 mm.
(a) The beam current is 7.75 µA. Find the current density in the beam assuming it is uniform throughout.
correct check mark A/m2

(b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam.
correct check mark m-3

(c) Over what time interval does Avogadro's number of electrons emerge from the accelerator?
s

Homework Equations

$$J=I/A$$
$$I_{avg} = nqv_{d}A<br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> Part a and b are straight forward. <br /> <br /> For part a I have: 1.71 A/m^2<br /> <br /> For part b I have: 3.565 x 10^10 m^-3<br /> <br /> I am having issues with part c. I know I need to figure out how many electrons are leaving the wire per second and then from there it should be a straight division problem using the 6.022 x 10^23 for Avogradro's number.$$

 

[LowlyPion]

Homework Statement

The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.20 mm.
(a) The beam current is 7.75 µA. Find the current density in the beam assuming it is uniform throughout.
correct check mark A/m2

(b) The speed of the electrons is so close to the speed of light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam.
correct check mark m-3

(c) Over what time interval does Avogadro's number of electrons emerge from the accelerator?
s

Homework Equations

$$J=I/A$$
$$I_{avg} = nqv_{d}A$$

The Attempt at a Solution

Part a and b are straight forward.

For part a I have: 1.71 A/m^2

For part b I have: 3.565 x 10^10 m^-3

I am having issues with part c. I know I need to figure out how many electrons are leaving the wire per second and then from there it should be a straight division problem using the 6.022 x 10^23 for Avogradro's number.

Think of it as a bucket. How long to fill'er up.

So what's the definition of an ampere?

 

[Ithryndil]

Right...an ampere is a coulomb per second. So we take the current which is a coulomb per second and divide it by the elementary charge to get the number of electrons per second. Afterwards it's a simple division of avogadro's number by the aforementioned number...far easier than I anticipated. Sometimes your mind can just be clouded.