/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 What is the average binding en... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 12

What is the average binding energy per nucleon for 40 18 Ar?

Short Answer

Expert verified
Answer: [Value obtained from the Step 4 calculation] MeV/nucleon

Step by step solution

01

Determine the number of protons and neutrons

For 40 18 Ar, the number of protons (atomic number) is 18, and the number of neutrons can be calculated by subtracting the atomic number from the mass number: 40 - 18 = 22 neutrons.
02

Calculate the mass difference

Now, we need to calculate the mass difference between the combined mass of individual protons and neutrons and the actual mass of the nucleus. The mass of a proton is roughly 1.007276466812 amu and the mass of a neutron is roughly 1.008664915 amu. The mass of Argon-40 nucleus is 39.9623831238 amu. We can use these values to find the mass difference: mass difference = (18 * 1.007276466812 amu + 22 * 1.008664915 amu) - 39.9623831238 amu
03

Calculate the total binding energy

To calculate the total binding energy, we'll use the mass difference and the speed of light (c = 2.99792458脳10^8 m/s). We'll also need to convert the mass difference from amu to kg. One amu is approximately equal to 1.660539040脳10^-27 kg. Thus: total binding energy = mass difference * (1.660539040脳10^-27 kg/amu) * (2.99792458脳10^8 m/s)^2
04

Calculate the average binding energy per nucleon

Now, we need to divide the total binding energy by the total number of nucleons (protons + neutrons) to find the average binding energy per nucleon: average binding energy per nucleon = total binding energy / (18 + 22) Calculate this value using the numbers obtained in the previous steps to get the average binding energy per nucleon for 40 18 Ar.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Show that \(c^{2}=931.494\) MeV/u. [Hint: Start with the conversion factors to SI units for MeV and atomic mass units.]
Radium- 226 decays as ${ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} .\( If the \){ }_{88}^{226}$ Ra nucleus is at rest before the decay and the \({ }_{86}^{222} \mathrm{Rn}\) nucleus is in its ground state, estimate the kinetic energy of the \(\alpha\) particle. (Assume that the \({ }_{86}^{222} \mathrm{Rn}\) nucleus takes away an insignificant fraction of the kinetic energy.)
Radon gas (Rn) is produced by the \(\alpha\) decay of radium 226 88 28 A How many neutrons and how many protons are present in the nucleus of the isotope of Rn produced by this decay? (b) In the air in an average size room in a student basement apartment in Ithaca, NY, there are about \(10^{7}\) Rn nuclei. The Rn nucleus itself is radioactive; it too decays by emitting an \(\alpha\) particle. The half-life of Rn is 3.8 days. How many \(\alpha\) particles per second are emitted by decaying Rn nuclei in the room?
The nucleus in a \({ }_{7}^{12} \mathrm{~N}\) atom captures one of the atom's electrons, changing the nucleus to \({ }_{6}^{12} \mathrm{C}\) and emitting a neutrino. What is the total energy of the emitted neutrino? [Hint:You can use the classical expression for the kinetic energy of the ${ }_{6}^{12} \mathrm{C}$ atom and the extremely relativistic expression for the kinetic energy of the neutrino.]
In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Wave Optics

Read Explanation

Atoms and Radioactivity

Read Explanation

Energy Physics

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

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.