Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 29: Problem 66

Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?

Short Answer

Expert verified
Answer: The approximate total energy of the emitted neutrino is 2.85 MeV.

Step by step solution

01

Identify the daughter nucleus

After electron capture, a proton in the parent nucleus transforms into a neutron, thus decreasing the atomic number (Z) by 1 while keeping the mass number (A) constant. So, the daughter nucleus will have an atomic number of 10 and a mass number of 22, which is \({}_{10}^{22}\mathrm{Ne}\).
02

Calculate the mass difference between parent and daughter nuclei

We can find the mass difference by using the atomic mass values for the parent (\({m}_{\mathrm{Na}}\)) and daughter (\({m}_{\mathrm{Ne}}\)) nuclei. Using a reference database, we find the atomic mass values as follows: \({m}_{\mathrm{Na}} = 21.9944364\) amu \({m}_{\mathrm{Ne}} = 21.9913849\) amu The mass difference (\(Δm\)) is obtained by subtracting the daughter nucleus mass from the parent nucleus mass: \(Δm = {m}_{\mathrm{Na}} - {m}_{\mathrm{Ne}} = 21.9944364 - 21.9913849 = 0.0030515\) amu
03

Convert the mass difference to energy

Now, we'll use the energy-mass equivalence principle (E=mc²) to convert the mass difference into energy. First, convert the mass difference from atomic mass units (amu) to kilograms (kg), using the conversion factor 1 amu = 1.66054 × 10^{-27} kg: \(Δm_\mathrm{kg} = 0.0030515 \times 1.66054 × 10^{-27} = 5.06895 × 10^{-30}\) kg Next, apply the energy-mass equivalence principle with the speed of light (c) value as 3 × 10^8 m/s: \(E = Δm_\mathrm{kg} \times c^2 = 5.06895 × 10^{-30} \times (3 × 10^8)^2 = 4.5620 × 10^{-13}\) Joules
04

Express the energy in a suitable unit and approximate the value

Nuclear reactions usually deal with energy in the unit of electronvolts (eV). So, we'll convert the calculated energy into MeV (Mega-electronvolts), using the conversion factor 1 Joule = 6.242 × 10^{12} eV: \(E_\mathrm{MeV} = 4.5620 × 10^{-13} \times 6.242 × 10^{12} \times 10^{-6} = 2.847\) MeV Approximately, the total energy of the emitted neutrino in the electron capture decay of \({}_{11}^{22}\mathrm{Na}\) is 2.85 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

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.
The first nuclear reaction ever observed (in 1919 by Ernest Rutherford) was \(\alpha+_{7}^{14} \mathrm{N} \right arrow \mathrm{p}+\mathrm{X} .\) (a) Show that the reaction product "X" must be \(\quad\) 8O. (b) For this reaction to take place, the \(\alpha\) particle must come in contact with the nitrogen nucleus. Calculate the distance \(d\) between their centers when they just make contact. (c) If the \(\alpha\) particle and the nitrogen nucleus are initially far apart, what is the minimum value of their kinetic energy necessary to bring the two into contact? Express your answer in terms of the elementary charge \(e\), the contact distance \(d\), and whatever else you need. (d) Is the total kinetic energy of the reaction products more or less than the initial kinetic energy in part (c)? Why? Calculate this kinetic energy difference.
Show that the spontaneous \(\alpha\) decay of \(^{19} \mathrm{O}\) is not possible.
An \(\alpha\) particle with a kinetic energy of \(1.0 \mathrm{MeV}\) is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the \(\alpha\) particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the \(\alpha\) particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an \(\alpha\) particle that will make contact with the gold nucleus?
Some types of cancer can be effectively treated by bombarding the cancer cells with high energy protons. Suppose \(1.16 \times 10^{17}\) protons, each with an energy of \(950 \mathrm{keV},\) are incident on a tumor of mass $3.82 \mathrm{mg} .\( If the quality factor for these protons is \)3.0,$ what is the biologically equivalent dose?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Kinematics Physics

Read Explanation

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

Read Explanation

Scientific Method Physics

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