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Chapter 27: Problem 30

An electron with energy equal to \(4.00 \cdot 10^{2} \mathrm{eV}\) and an electron with energy equal to \(2.00 \cdot 10^{2} \mathrm{eV}\) are trapped in a uniform magnetic field and move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

Short Answer

Expert verified
Answer: The ratio of the radii of the orbits of the two electrons is 鈭�2.

Step by step solution

01

Formula for the radius of the circular orbit

The formula for the radius (r) of the circular orbit an electron with charge (q) and velocity (v) moving in a plane perpendicular to a magnetic field (B) is given by: r = mv/(qB), where m is the mass of the electron.
02

Formula for the energy of the electron

The energy (E) of an electron is given by the formula: E = (1/2)mv^2
03

Express the velocity in terms of energy

Solving the energy formula for velocity, we get: v = sqrt(2E/m)
04

Substitute the velocity expression back into the radius formula

Substituting the expression for velocity into the equation for the radius, we get: r = m(sqrt(2E/m))/(qB)
05

Simplify the expression for radius

Simplifying the expression, we get: r = sqrt(2mE)/(qB)
06

Ratio of the radii of the electrons' orbits

Let r鈧� and r鈧� be the radii of the orbits for the electrons with energies E鈧� and E鈧� respectively. The ratio of the radii is given by: r鈧�/r鈧� = sqrt(2mE鈧�)/(qB) * (qB)/sqrt(2mE鈧�)
07

Simplify the ratio expression

Canceling the common terms (qB) and (2m) in the numerator and denominator yields: r鈧�/r鈧� = sqrt(E鈧�/E鈧�)
08

Calculate the ratio

Given the energies E鈧� = 4.00*10虏 eV and E鈧� = 2.00*10虏 eV, we can now calculate the ratio of the radii: r鈧�/r鈧� = sqrt((4.00*10虏)/(2.00*10虏))
09

Simplify the result

Simplifying the result, we get: r鈧�/r鈧� = sqrt(2) So the ratio of the radii of the orbits of the two electrons is 鈭�2.

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Most popular questions from this chapter

An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction \(),\) the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

Which of the following has the largest cyclotron frequency? a) an electron with speed \(v\) in a magnetic field with magnitude \(B\) b) an electron with speed \(2 v\) in a magnetic field with magnitude \(B\) c) an electron with speed \(v / 2\) in a magnetic field with magnitude \(B\) d) an electron with speed \(2 v\) in a magnetic field with magnitude \(B / 2\) e) an electron with speed \(v / 2\) in a magnetic field with magnitude \(2 B\)

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.
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