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Chapter 18: Problem 4

In an ion accelerator, \(3.0 \times 10^{13}\) helium-4 nuclei (charge \(+2 e\) ) per second strike a target. What is the beam current?

Short Answer

Expert verified
Answer: The beam current is \(9.6 \times 10^{-6} \text{A}\).

Step by step solution

01

Understand the given information and the problem

We are given: - Number of helium-4 nuclei striking the target per second: \(N = 3.0 \times 10^{13}\) nuclei/second - Charge of each helium-4 nucleus: \(q = +2e\) Our task is to find the beam current.
02

Use the formula for current

The formula for current (I) is given as follows, $$ I = n \cdot q $$ where "n" is the rate at which charges are flowing (number of helium-4 nuclei striking the target per second) and "q" is the charge of each helium-4 nucleus. In our case, we have: \(n = 3.0 \times 10^{13}\) nuclei/second \(q = +2e = +2 (1.6 \times 10^{-19} \text{C})\)
03

Calculate the current

Now, we will plug in the values of n and q into the formula to calculate the current: $$ I = n \cdot q = (3.0 \times 10^{13}) \times (+2 \times 1.6 \times 10^{-19}) $$ Solve for I: $$ I = 3.0 \times 10^{13} \times 3.2 \times 10^{-19} $$ Multiply the numbers: $$ I = 9.6 \times 10^{-6} $$ Therefore, the beam current is \(9.6 \times 10^{-6} \text{A}\).

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Most popular questions from this chapter

A string of 25 decorative lights has bulbs rated at \(9.0 \mathrm{W}\) and the bulbs are connected in parallel. The string is connected to a \(120-\mathrm{V}\) power supply. (a) What is the resistance of each of these lights? (b) What is the current through each bulb? (c) What is the total current coming from the power supply? (d) The string of bulbs has a fuse that will blow if the current is greater than 2.0 A. How many of the bulbs can you replace with 10.4-W bulbs without blowing the fuse?
A gold wire of 0.50 mm diameter has \(5.90 \times 10^{28} \mathrm{con}\) duction electrons/m". If the drift speed is \(6.5 \mu \mathrm{m} / \mathrm{s}\), what is the current in the wire?
A galvanometer has a coil resistance of \(50.0 \Omega\). It is to be made into an ammeter with a full-scale deflection equal to \(10.0 \mathrm{A} .\) If the galvanometer deflects full scale for a current of \(0.250 \mathrm{mA},\) what size shunt resistor should be used?
A voltmeter has a switch that enables voltages to be measured with a maximum of \(25.0 \mathrm{V}\) or \(10.0 \mathrm{V}\). For a range of voltages to $25.0 \mathrm{V},\( the switch connects a resistor of magnitude \)9850 \Omega$ in series with the galvanometer; for a range of voltages to \(10.0 \mathrm{V}\), the switch connects a resistor of magnitude \(3850 \Omega\) in series with the galvanometer. Find the coil resistance of the galvanometer and the galvanometer current that causes a full-scale deflection. [Hint: There are two unknowns, so you will need to solve two equations simultaneously.]
A current of \(2.50 \mathrm{A}\) is carried by a copper wire of radius $1.00 \mathrm{mm} .\( If the density of the conduction electrons is \)8.47 \times 10^{28} \mathrm{m}^{-3},$ what is the drift speed of the conduction electrons?
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