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Chapter 23: Problem 28

Fully stripped (all electrons removed) sulfur \(\left({ }^{32} \mathrm{~S}\right)\) ions are accelerated in an accelerator from rest using a total voltage of \(1.00 \cdot 10^{9} \mathrm{~V}\). \({ }^{32} \mathrm{~S}\) has 16 protons and 16 neutrons. The accelerator produces a beam consisting of \(6.61 \cdot 10^{12}\) ions per second. This beam of ions is completely stopped in a beam dump. What is the total power the beam dump has to absorb?

Short Answer

Expert verified
Answer: The total power the beam dump must absorb is approximately 1.69 x 10^4 W or 16.9 kW.

Step by step solution

01

Calculate the charge of the sulfur ion

Since sulfur has 16 protons and is fully stripped (meaning no electrons are present), the charge of one sulfur ion is the charge of 16 protons. The charge of one proton is \(1.60 \times 10^{-19} \mathrm{C}\). So the charge of the sulfur ion can be calculated as: \(q = 16 \times 1.60 \times 10^{-19} \mathrm{C} = 2.56 \times 10^{-18} \mathrm{C}\)
02

Calculate the kinetic energy of one ion

The potential energy gained by one ion from the given voltage is equal to the kinetic energy of the ion. We can determine the kinetic energy using the equation: \(KE = qV\), where \(q\) is the charge of the ion and \(V\) is the voltage. Therefore, \(KE = (2.56 \times 10^{-18} \mathrm{C})(1.00 \times 10^{9} \mathrm{V}) = 2.56 \times 10^{-9} \mathrm{J}\)
03

Calculate the kinetic energy of all ions per second

We know that the accelerator produces \(6.61 \times 10^{12}\) ions per second. So, the total kinetic energy of all ions per second will be: \(KE_{total} = (6.61 \times 10^{12} \text{ ions/s})(2.56 \times 10^{-9} \mathrm{J/ion}) = 1.69 \times 10^{4} \mathrm{J/s}\)
04

Find the total power that the beam dump has to absorb

Since power is defined as the amount of energy transferred per unit time, the total power that the beam dump has to absorb can be inferred from the total kinetic energy of all ions per second. In this case, the power is equal to: \(P = 1.69 \times 10^{4} \mathrm{J/s}\) So, the total power the beam dump must absorb is approximately \(1.69 \times 10^{4} \mathrm{W}\) or \(16.9 \mathrm{kW}\).

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