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Chapter 19: Problem 45

A proton of mass \(m\) and charge \(+\mathrm{e}\) is moving in a circular orbit in a magnetic field with energy \(1 \mathrm{MeV}\). What should be the energy of \(\alpha\)-particle (mass \(=4 \mathrm{~m}\) and charge \(=+2 \mathrm{e}\) ), so that it can revolve in the path of same radius (A) \(1 \mathrm{MeV}\) (B) \(4 \mathrm{MeV}\) (C) \(2 \mathrm{MeV}\) (D) \(0.5 \mathrm{MeV}\)

Short Answer

Expert verified
The short answer to the question is: The energy of the 伪-particle should be (A) \(1 \mathrm{MeV}\), so that it can revolve in the path of the same radius as the proton.

Step by step solution

01

Equate centripetal force and magnetic force for the proton

First, we need to equate the centripetal force to the magnetic force in the case of the proton. The centripetal force is given by \(F_c = \frac{mv_p^2}{r}\), where \(m\) is the mass of the proton, \(v_p\) is its velocity, and \(r\) is the radius of its circular path. The magnetic force acting on a charged particle moving in a magnetic field can be expressed as \(F_B = qv_pB\), where \(q\) is the charge of the proton, which is equal to \(+e\), and \(B\) is the magnetic field strength. Now, let's equate these two forces: \[\frac{mv_p^2}{r} = ev_pB\]
02

Simplify the equation for the proton

We can now simplify the above equation for the proton as follows: \[mv_p = erB\] \[v_p = \frac{erB}{m}\]
03

Equate centripetal force and magnetic force for the 伪-particle

We will now do the same for the 伪-particle, which has a mass of \(4m\) and a charge equal to \(2e\). The centripetal force for the 伪-particle is \(F_c = \frac{4mv_{\alpha}^2}{r}\), and the magnetic force is \(F_B = 2ev_{\alpha}B\). Equate these forces: \[\frac{4mv_{\alpha}^2}{r} = 2ev_{\alpha}B\]
04

Simplify the equation for the 伪-particle

Now, you can simplify the above equation for the 伪-particle as follows: \[4mv_{\alpha} = 2erB\] \[v_{\alpha} = \frac{erB}{2m}\]
05

Find the ratio of velocities

We can now find the ratio of the velocities of the proton and the 伪-particle, as the radius of their circular paths is the same: \[\frac{v_p}{v_{\alpha}} = \frac{\frac{erB}{m}}{\frac{erB}{2m}} = \frac{2}{1}\] The velocity of the proton is twice the velocity of the 伪-particle: \[v_p = 2v_{\alpha}\]
06

Find the ratio of kinetic energies

The kinetic energy of each particle can be calculated using the formula: \[KE = \frac{1}{2}mv^2\] Hence, find the ratio of their kinetic energies: \[\frac{KE_p}{KE_{\alpha}} = \frac{\frac{1}{2}m(2v_{\alpha})^2}{\frac{1}{2}4m(v_{\alpha})^2}\]
07

Simplify the ratio of kinetic energies

Now, you can simplify the above expression for the ratio of kinetic energies, as follows: \[\frac{KE_p}{KE_{\alpha}} = \frac{4m(v_{\alpha})^2}{4m(v_{\alpha})^2} = 1\] The kinetic energies of the proton and the 伪-particle are equal: \[KE_p = KE_{\alpha}\]
08

Choose the correct option

Now, we are given that the energy of the proton is \(1 MeV\). Since their kinetic energies are equal, the energy of the 伪-particle must also be \(1 MeV\). Thus, the correct option is: (A) \(1 \mathrm{MeV}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
In circular motion, centripetal force plays a crucial role. It is the force that keeps an object moving in a circular path. Without this force, the object would travel in a straight line due to inertia. For an object with mass, like a proton, moving in a circle of radius \(r\), the centripetal force \(F_c\) is given by \(F_c = \frac{mv^2}{r}\). Here, \(m\) represents the mass of the object, and \(v\) is its velocity.

When an object moves within a magnetic field, the centripetal force needed for circular motion is often provided by the magnetic force. The balance between these forces ensures the object maintains its circular trajectory.
  • If \(F_c > F_B\), the object will spiral outwards.
  • If \(F_c < F_B\), the object will spiral inwards.
The equation \(\frac{mv^2}{r} = qvB\) shows how the centripetal force arises from the magnetic interaction, where \(q\) is the charge of the particle and \(B\) is the magnetic field strength.
Magnetic Force
Magnetic force is a key player in the behavior of charged particles in a magnetic field. This force acts perpendicular to both the velocity of the particle and the magnetic field. This perpendicular force is what causes the particle to move in a circular path rather than a straight line.

The magnetic force \(F_B\) on a particle with charge \(q\) is defined by the equation \(F_B = qvB\), where \(v\) is the velocity of the particle and \(B\) is the magnetic field strength. For example, a proton with a charge \(+e\) and an \(\alpha\)-particle with a charge \(+2e\) will experience different magnetic forces even if they move in the same magnetic field.
  • A stronger magnetic field increases the magnetic force.
  • A greater velocity also results in a stronger magnetic force.
This force does not do work on the particle, meaning it doesn't change the particle's speed, just its direction. Hence, the magnetic force is responsible for maintaining the circular motion of charged particles like protons and \(\alpha\)-particles.
Kinetic Energy
Kinetic Energy (KE) is the energy an object possesses due to its motion. Calculating the kinetic energy of a particle allows us to understand how energy is distributed in dynamic systems. The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\), where \(m\) is the object's mass, and \(v\) is its velocity.

When comparing a proton and an \(\alpha\)-particle moving in the same magnetic field and with the same circular path radius, we can derive their kinetic energies from the equative velocities adjusted for mass and charge differences.
  • For a proton with a mass \(m\) and velocity \(v_p\), \(KE_p = \frac{1}{2}mv_p^2\).
  • For an \(\alpha\)-particle with a mass \(4m\) and velocity \(v_{\alpha}\), \(KE_{\alpha} = \frac{1}{2}4m(v_{\alpha})^2\).
In our specific exercise, the kinetic energies equate because the radial path requires equivalent force balance and velocity adjustments, allowing both particles to orbit within the same magnetic conditions. Thus, an \(\alpha\)-particle can have the same kinetic energy as a proton when properly balanced within the magnetic field, in this case, \(1 \text{MeV}\).

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Most popular questions from this chapter

Hydrogen \(\left({ }_{1} \mathrm{H}^{1}\right)\), Deuterium \(\left({ }_{1} \mathrm{H}^{2}\right)\), singly ionised Helium \(\left({ }_{2} \mathrm{He}^{4}\right)^{+}\)and doubly ionised lithium \(\left({ }_{3} \mathrm{Li}^{6}\right)^{++}\) all have one electron around the nucleus. Consider an electron transition from \(n=2\) to \(n=1\). If the wavelengths of emitted radiation are \(\lambda_{1}, \lambda_{2}, \lambda_{3}\) and \(\lambda_{4}\) respectively then approximately which one of the following is correct? (A) \(4 \lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\) (B) \(\lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\) (C) \(\lambda_{1}=\lambda_{2}=4 \lambda_{3}=9 \lambda_{4}\) (D) \(\lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=4 \lambda_{4}\)

The 'rad' is the correct unit used to report the measurement of (A) the ability of a beam of gamma - ray photons to produce ions in a target. (B) the energy delivered by radiation to a target. (C) the biological effect of radiation. (D) the rate of decay of a radioactive source.

When photons of energy \(5 \mathrm{eV}\) strike the surface of a metal \(A\), the ejected photoelectrons have maximum kinetic energy \(K_{A} \mathrm{eV}\) and de Broglie wavelength \(\lambda_{A}\). The maximum kinetic energy of photoelectrons liberated from another metal \(B\) by photons of energy \(5.30 \mathrm{eV}\) is \(K_{B}=\left(K_{A}-1.5\right) . \mathrm{eV}\). If the de Broglie wavelength of these photoelectrons is \(\lambda_{B}=2 \lambda_{A}\), then find \(K_{A}\) and \(K_{B}\).

The binding energy per nucleon for the parent nucleus is \(E_{1}\) and that for the daughter nuclei is \(E_{2}\). Then (A) \(E_{1}=2 E_{2}\) (B) \(E_{2}=2 E_{1}\) (C) \(E_{1}>E_{2}\) (D) \(E_{2}>E_{1}\)

What is the value of Wein's displacement constant? (A) \(e^{A}\) (B) \(1 / e^{A}\) (C) \(\ln A\) (D) \(1 / \operatorname{In} A\)
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