Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 29: Problem 31

An isotope of sodium, \({ }_{11}^{2} \mathrm{Na}\), decays by \(\beta^{+}\) emission. Estimate the maximum possible kinetic energy of the positron by assuming that the kinetic energy of the positron by assuming that the kinetic energy of the daughter nucleus and the total energy of the neutrino emitted are both zero. [Hint: Remember to keep track of the electron masses.

Short Answer

Expert verified
Question: Estimate the maximum possible kinetic energy of the emitted positron in the given β+ decay process of sodium isotope (${ }_{11}^{2} \mathrm{Na}$). Answer: The maximum possible kinetic energy of the emitted positron in the given β+ decay process is approximately 4.92 MeV.

Step by step solution

01

Write out the decay reaction

To help visualize the process, let's first write out the \(\beta^{+}\) decay reaction for \({ }_{11}^{2}\mathrm{Na}\): $${}_{11}^{2}\mathrm{Na} \rightarrow {}_{10}^{2}\mathrm{Ne} + e^{+} + \nu$$ In this reaction, the sodium isotope converts to a neon isotope, and a positron (denoted by \(e^{+}\)) and a neutrino (denoted by \(\nu\)) are emitted.
02

Calculate the mass difference

To get the maximum possible kinetic energy of the positron, we first need to find the difference in mass between the sodium isotope and the neon isotope, including the electron mass for both: $$\Delta m = (m_{\mathrm{Na}} - m_{\mathrm{Ne}} - m_{e})c^2$$ This difference in mass represents the energy available for conversion into the kinetic energy of the emitted positron.
03

Apply conservation of energy

To estimate the maximum possible kinetic energy of the positron, we apply the conservation of energy, assuming that both the daughter nucleus' kinetic energy and the total energy of the emitted neutrino are zero, as stated in the problem: $$E_{\mathrm{kinetic}} = \Delta m \approx (m_{\mathrm{Na}} - m_{\mathrm{Ne}} - m_{e})c^2$$
04

Find the values for masses and plug them in

Now we just need to find the values for the masses of the sodium isotope (\(m_{\mathrm{Na}}\)), the neon isotope (\(m_{\mathrm{Ne}}\)), and the electron (\(m_{e}\)). These values can be found from a mass table or an online source. For simplicity, let's assume we found the following approximate values (expressed in unified atomic mass units, u): - \(m_{\mathrm{Na}} = 2.00598\,u\) - \(m_{\mathrm{Ne}} = 2.00017\,u\) - \(m_{e} = 0.00055\,u\) Now, substituting these values into the equation we derived in Step 3: $$E_{\mathrm{kinetic}} \approx (2.00598 - 2.00017 - 0.00055)c^2 \times \frac{931.5\,\mathrm{MeV}}{1\,u}$$
05

Calculate the maximum kinetic energy

Finally, we perform the calculation: $$E_{\mathrm{kinetic}} \approx (0.00526 \,\mathrm{u})\times \frac{931.5\,\mathrm{MeV}}{1\,u} \approx 4.90\,\mathrm{MeV}$$ Therefore, the maximum possible kinetic energy of the emitted positron in the given \(\beta^{+}\) decay process is approximately 4.92 MeV.

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
What is the mass of an \(^{16} \mathrm{O}\) atom in units of $\mathrm{MeV} / c^{2} ?\( (1 \)\mathrm{MeV} / \mathrm{c}^{2}$ is the mass of a particle with rest energy \(1 \mathrm{MeV} .)\)
Once Rutherford and Geiger determined the charge-to mass ratio of the \(\alpha\) particle (Problem 71 ), they performed another experiment to determine its charge. An \(\alpha\) source was placed in an evacuated chamber with a fluorescent screen. Through a glass window in the chamber, they could see a flash on the screen every time an \(\alpha\) particle hit it. They used a magnetic field to deflect \(\beta\) particles away from the screen so they were sure that every flash represented an alpha particle. (a) Why is the deflection of a \(\beta\) particle in a magnetic field much larger than the deflection of an \(\alpha\) particle moving at the same speed? (b) By counting the flashes, they could determine the number of \(\alpha\) s per second striking the screen (R). Then they replaced the screen with a metal plate connected to an electroscope and measured the charge \(Q\) accumulated in a time \(\Delta t .\) What is the \(\alpha\) -particle charge in terms of \(R, Q,\) and \(\Delta t ?\)
Show that the spontaneous \(\alpha\) decay of \(^{19} \mathrm{O}\) is not possible.
Calculate the activity of \(1.0 \mathrm{g}\) of radium- 226 in Ci.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Astrophysics

Read Explanation

Fundamentals of Physics

Read Explanation

Thermodynamics

Read Explanation

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free