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agile muller counter collects 10 to the power 8 electrons for discharge when the counting rate is 100 400 pounds per minute. what will be the average current in the circuit?

User Helium
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Final answer:

The average current in an Agile Muller counter circuit, which collects 10^8 electrons for each discharge at a rate of 100,400 counts per minute, is approximately 2.68 x 10^-5 milliamperes.

Step-by-step explanation:

The question asks about calculating the average current in an Agile Muller counter circuit. To find the current, we can use the formula I = Q / t, where I is the current in amperes (A), Q is the charge in coulombs (C), and t is the time in seconds (s) through which the charge flows.

Firstly, we need to convert the number of electrons to coulombs. We know that 1 electron has a charge of approximately 1.602 x 10^-19 coulombs. The counter collects 10^8 electrons for a discharge, so the total charge Q is:

Q = 10^8 electrons × 1.602 x 10^-19 C/electron

Q = 1.602 x 10^-11 C

The counting rate given is 100,400 counts per minute. To find the average current, we also need the time in seconds.

t = 1 minute / 100,400 counts

t = 60 seconds / 100,400

t ≈ 5.97 x 10^-4 seconds per count

Now, we can calculate the average current I using the charge Q and the time t.

I = Q / t

I = 1.602 x 10^-11 C / 5.97 x 10^-4 s

I ≈ 2.68 x 10^-8 A

To express this current in a more common unit, we can convert amperes to milliamperes (mA) by multiplying by 1000.

I ≈ 2.68 x 10^-5 mA

Therefore, the average current in the Agile Muller counter circuit is approximately 2.68 x 10^-5 mA.

User Novitzky
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