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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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Most popular questions from this chapter

A point charge of \(+2.0 \mu C\) is located at \((2.5 \mathrm{~m}, 3.2 \mathrm{~m})\) A second point charge of \(-3.1 \mu \mathrm{C}\) is located at \((-2.1 \mathrm{~m}, 1.0 \mathrm{~m})\). a) What is the electric potential at the origin? b) Along a line passing through both point charges, at what point(s) is (are) the electric potential(s) equal to zero?

Two protons at rest and separated by \(1.00 \mathrm{~mm}\) are released simultaneously. What is the speed of either at the instant when the two are \(10.0 \mathrm{~mm}\) apart?

The electric potential in a volume of space is given by \(V(x, y, z)=x^{2}+x y^{2}+y z\). Determine the electric field in this region at the coordinate (3,4,5) .

Show that an electron in a one-dimensional electri. cal potential \(V(x)=A x^{2},\) where the constant \(A\) is a positive real number, will execute simple harmonic motion about the origin. What is the period of that motion?

The electric potential energy of a continuous charge distribution can be found in a way similar to that used for systems of point charges in Section \(23.6,\) by breaking the distribution up into suitable pieces. Find the electric potential energy of an arbitrary spherically symmetrical charge distribution, \(\rho(r) .\) Do not assume that \(\rho(r)\) represents a point charge, that it is constant, that it is piecewise-constant, or that it does or does not end at any finite radius, \(r\). Your expression must cover all possibilities. Your expression may include an integral or integrals that cannot be evaluated without knowing the specific form of \(\rho(r) .\) (Hint: A spherical pearl is built up of thin layers of nacre added one by one.)
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