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Current and Avogadro's Number 
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#1
Sep1808, 06:23 PM

P: 142

1. The problem statement, all variables and given/known data
The electron beam emerging from a certain highenergy electron accelerator has a circular cross section of radius 1.20 mm. (a) The beam current is 7.75 ľA. 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 m3 (c) Over what time interval does Avogadro's number of electrons emerge from the accelerator? s 2. Relevant equations [tex]J=I/A[/tex] [tex]I_{avg} = nqv_{d}A 3. 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. 


#2
Sep1808, 07:50 PM

HW Helper
P: 5,343

So what's the definition of an ampere? 


#3
Sep1808, 08:02 PM

P: 142

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.



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