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Chapter 19: Problem 43

A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)

Short Answer

Expert verified
Answer: The expression for the charge-to-mass ratio (q/m) is given by: $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$

Step by step solution

01

Kinetic energy gained by the particle

The charged particle is accelerated from rest through a potential difference \(\Delta V\). We can use the formula for energy conservation to say that the electric potential energy is converted into kinetic energy $$q \Delta V = \frac{1}{2}mv^2$$ Where \(q\) is the charge of the particle, \(m\) is the mass of the particle, and \(v\) is the final particle velocity.
02

Use the velocity selector condition

The velocity selector is designed such that the electric force acting on the charged particle is balanced by the magnetic force. This will allow only particles with a specific velocity \(v\) to pass straight through the selector. We can write the balance of the forces as $$ F_{electric} = F_{magnetic} $$ where \(F_{electric} = qE\) (electric force) and \(F_{magnetic} = qvB\) (magnetic force), so $$ q E = q v B $$ Now we can find the particle's velocity \(v\) in terms of the charge \(q\), field magnitudes \(E\) and \(B\), as $$v = \frac{E}{B}$$
03

Substitute the velocity expression in the energy conservation equation

Now that we have found an expression for the particle's velocity, we can substitute it into the energy conservation equation $$q \Delta V = \frac{1}{2}m \left(\frac{E}{B}\right)^2$$
04

Solve for the charge-to-mass ratio \((q/m)\)

We can now solve this equation for the charge-to-mass ratio \((q/m)\) by dividing through by \(m\) and moving the potential difference term \(\Delta V\) to the other side. $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$ Hence, we have derived an expression for the charge-to-mass ratio \((q/m)\) in terms of the potential difference \(\Delta V\), and field magnitudes \(E\) and \(B\) as $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$

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Most popular questions from this chapter

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing sulfur (atomic mass 32 u), manganese \((55 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field. The sulfur and manganese lines are separated by \(3.20 \mathrm{cm},\) and the unknown element makes a line between them that is \(1.07 \mathrm{cm}\) from the sulfur line. (a) What is the mass of the unknown clement? (b) Identify the element.
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