Question

A gold wire that is 0.8 mm in diameter and 34 cm long carries a current of 790 mA. How many electrons per second pass a given cross-section of the wire?

a) 1.256 X 10^19 electrons/s
b) 2.502 X 10^19 electrons/s
c) 5.004 X 10^19 electrons/s
d) 1.001 X 10^20 electrons/s

Answer 1

To find the number of electrons per second flowing through a wire, the formula I/q is used, where I is the current and q is the charge of an electron. The calculation yields 4.9375×1018 electrons/s, which does not match any of the given options, suggesting a potential error in the question or the provided options.

Step-by-step explanation:

To calculate the number of electrons per second that pass a given cross-section of a gold wire carrying a current, we can use the formula I = nqAv, where I is the current, n is the charge carrier density (number of free electrons per volume), q is the elementary charge of an electron, A is the cross-sectional area of the wire, and v is the drift velocity of the electrons.

First, we need to establish the charge of an electron, which is approximately q = 1.60×10−19 C. Next, we need to calculate the cross-sectional area A of the wire using the diameter provided. For a circular cross-section, the area is given by A = πr2 where r is the radius of the wire.

Once we have the current I and the charge of an electron q, we can rearrange the formula to solve for the number of electrons n passing a given point per second as n = I/q.

Given the current I = 790 mA or 790×10−3 A, and q = 1.60×10−19 C, we can perform the calculation:

n = I/q = (790×10−3 A)/(1.60×10−19 C)

When you complete this calculation, the result is approximately 4.9375×1018 electrons/s, which is not an option provided in the question, implying there might be a mistake in the question or in the options.