You are a technician testing the operation of a cyclotron. An alpha particle in the device moves in a circular path in a magnetic field $\overrightarrow{\mathbf{B}}$that is directed perpendicular to the path of the alpha particle. You measure the number of revolutions per second (the frequency f) of the alpha particle as a function of the magnetic field strength . Figure shows your results and the best straight-line fit to your data. (a) Use the graph in Figure to calculate the charge-to-mass ratio of the alpha particle, which has charge +2e .On the basis of your data, what is the mass of an alpha particle?(b) With B = 0.300T, what are the cyclotron frequencies f of a proton and of an electron? How do these f values compare to the frequency of an alpha particle? (c) With B = 0.300 T , what speed and kinetic energy does an alpha particle have if the radius of its path is 12.0cm ?

Step by step solution

01

Definition of kinetic energy

The term kinetic energy may be defined as the energy possess by the body by virtue of its motion.

02

Determine the mass of muon, speed of muon, frequency of electron and proton and kinetic energy of alpha particle.

The mass of the muon can be calculated as

$\begin{array}{r}\mathrm{m}=\frac{\mathrm{e}}{\mathrm{\pi }\left(\text{slope}\right)}\\ \text{slope}=\frac{23\times {10}^5\mathrm{Hz}-8.0\times {10}^5\mathrm{Hz}}{0.30\mathrm{T}-0.10\mathrm{T}}\\ \text{slope}=7.5\times {10}^6\mathrm{Hz}/\mathrm{T}\\ \mathrm{Now}\\ \mathrm{m}=\frac{1.60\times {10}^{-19}\mathrm{C}}{\mathrm{\pi }\left(7.5\times {10}^6\mathrm{Hz}/\mathrm{T}\right)}\\ \mathrm{m}=6.80\times {10}^{-27}\mathrm{kg}\end{array}$

Hence,the mass of muon is $\mathrm{m}=6.80\times {10}^{-27}\mathrm{kg}$,

The frequency of electron is calculated as

$\begin{array}{r}{\mathrm{f}}_{\mathrm{e}}=\frac{\left(1.602\times {10}^{-19}\mathrm{C}\right)(0.300\mathrm{T})}{2\mathrm{\pi }\left(9.11\times {10}^{-31}\mathrm{kg}\right)}\\ {\mathrm{f}}_{\mathrm{e}}=8.40\times {10}^6\mathrm{Hz}\\ {\mathrm{f}}_{\mathrm{p}}=\frac{\left(1.602\times {10}^{-19}\mathrm{C}\right)(0.300\mathrm{T})}{2\mathrm{\pi }\left(1.67\times {10}^{-27}\mathrm{kg}\right)}\\ {\mathrm{f}}_{\mathrm{p}}=4.58\times {10}^6\mathrm{Hz}\end{array}$

Hence, the frequency of electron and proton is $f_e=8.40\times {10}^9\mathrm{Hz},{\mathrm{f}}_p=4.58\times {10}^6\mathrm{Hz}$.

The frequency of alpha particle is half of frequency of proton and the frequency of alpha particle is $\frac{1}{3672}$frequency of electron.

$\begin{array}{r}{\mathrm{f}}_{\alpha }=\frac{1}{2}{\mathrm{f}}_p\\ {\mathrm{f}}_{\alpha }=\frac{1}{2}\left(4.58\times {10}^6\mathrm{Hz}\right)\\ {\mathrm{f}}_{\alpha }=2.3\times {10}^6\mathrm{Hz}\end{array}$

And

$\begin{array}{r}{\mathrm{f}}_{\mathrm{\alpha }}=\frac{1}{3672}{\mathrm{f}}_{\mathrm{e}}\\ {\mathrm{f}}_{\mathrm{\alpha }}=\frac{1}{3672}\left(8.40\times {10}^9\mathrm{Hz}\right)\\ {\mathrm{f}}_{\mathrm{\alpha }}=2.3\times {10}^6\mathrm{Hz}\\ \mathrm{The}\mathrm{kinetic}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{alpha}\mathrm{particle}\mathrm{is}\\ \mathrm{K}=\frac{1}{2}\frac{(\mathrm{RqB}{)}^2}{\mathrm{m}}\\ \mid \mathrm{K}=\frac{1}{2}\frac{(0.12\mathrm{m})\left(2\times 1.602\times {10}^{-19}\mathrm{C}\right)(0.30\mathrm{T}{)}^2}{6.80\times {10}^{-27}\mathrm{kg}}\\ \mathrm{K}=9.8\times {10}^{-18}\mathrm{J}\\ \mathrm{K}=6.12\times {10}^4\mathrm{eV}\end{array}$

Hence, the kinetic energy is$6.12\times {10}^4\mathrm{eV}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!