The atomic masses of nitrogen-14, titanium- 48 , and xenon129 are $13.999234 \mathrm{u}, 47.935878 \mathrm{u},\( and \)128.904779 \mathrm{u},$ respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon.

Short Answer

Expert verified
(a) Nuclear mass: - Nitrogen-14: \(2.32405 \times10^{-26} kg\) - Titanium-48: \(7.95868 \times10^{-26} kg\) - Xenon-129: \(2.139192 \times10^{-25} kg\) (b) Nuclear binding energy: - Nitrogen-14: \(1.0805 \times10^{-11} J\) - Titanium-48: \(7.153 \times 10^{-11} J\) - Xenon-129: \(3.1851 \times10^{-10} J\) (c) Nuclear binding energy per nucleon: - Nitrogen-14: \(7.7179 \times10^{-13} J/nucleon\) - Titanium-48: \(1.4906 \times 10^{-12} J/nucleon\) - Xenon-129: \(2.469 \times 10^{-12} J/nucleon\)

Step by step solution

01

Convert atomic mass to kilograms

For each isotope, let's convert their atomic masses in unified atomic mass units (u) to their equivalent mass in kilograms (kg) using the conversion factor: 1 u = 1.6605 x 10^{-27} kg. For nitrogen-14, \(Mass = 13.999234 u \times 1.6605 \times10^{-27} kg/u = 2.3255 \times10^{-26} kg\) For titanium-48, \(Mass = 47.935878 u \times 1.6605 \times10^{-27} kg/u = 7.9672 \times10^{-26} kg\) For xenon-129, \(Mass = 128.904779 u \times 1.6605 \times10^{-27} kg/u = 2.1410 \times10^{-25} kg\)
02

Calculate nuclear mass

For each isotope, we need to find the nuclear mass by subtracting the mass of their electrons from their respective atomic masses. We know the mass of an electron is approximately 9.1094 x 10^{-31} kg. Nuclear mass of nitrogen-14: \(Nuclear\:Mass = Atomic\:Mass - (14 \times Electron\:Mass)\) \(Nuclear\:Mass =2.3255 \times10^{-26} kg - (14 \times 9.1094\times10^{-31}kg) = 2.32405 \times10^{-26} kg\) Nuclear mass of titanium-48: \(Nuclear\:Mass = Atomic\:Mass - (48 \times Electron\:Mass)\) \(Nuclear\:Mass = 7.9672 \times10^{-26} kg - (48\times 9.1094\times10^{-31} kg) = 7.95868 \times10^{-26} kg\) Nuclear mass of xenon-129: \(Nuclear\:Mass = Atomic\:Mass - (129\times Electron\:Mass)\) \(Nuclear\:Mass = 2.1410 \times10^{-25} kg - (129\times 9.1094\times10^{-31} kg) = 2.139192 \times10^{-25} kg\)
03

Calculate nuclear binding energy

To find the nuclear binding energy, we need to find the mass defect for each isotope first, and then multiply by the speed of light squared (using Einstein's mass-energy equivalence equation: E = mc^2). Mass defect for nitrogen-14: \(Mass\:Defect = (7 × 1.6736\times 10^{-27} kg + 7 × 1.675\times 10 ^{-27} kg) - 2.32405 \times10^{-26} kg = 1.2005\times 10^{-28} kg\) Nuclear binding energy for nitrogen-14: \(Binding\:Energy = Mass\:Defect \times (3\times10^8 m/s)^2\) \(Binding\:Energy = 1.2005\times 10^{-28} kg \times(3\times10^8 m/s)^2 = 1.0805\times 10^{-11} J\) Mass defect for titanium-48: \(Mass\:Defect = (22 × 1.6736\times 10^{-27} kg + 26 × 1.675\times 10 ^{-27} kg) - 7.95868 \times10^{-26} kg = 7.948 \times 10^{-28} kg\) Nuclear binding energy for titanium-48: \(Binding\:Energy = Mass\:Defect \times (3\times10^8 m/s)^2\) \(Binding\:Energy = 7.948 \times 10^{-28} kg \times(3\times10^8 m/s)^2 = 7.153 \times 10^{-11} J\) Mass defect for xenon-129: \(Mass\:Defect = (54 × 1.6736\times 10^{-27} kg + 75 × 1.675\times 10 ^{-27} kg) - 2.139192 \times10^{-25} kg = 3.539 \times 10^{-27} kg\) Nuclear binding energy for xenon-129: \(Binding\:Energy = Mass\:Defect \times (3\times10^8 m/s)^2\) \(Binding\:Energy = 3.539 \times 10^{-27} kg \times(3\times10^8 m/s)^2 = 3.1851 \times 10^{-10} J\)
04

Calculate nuclear binding energy per nucleon

Finally, we find the nuclear binding energy per nucleon by dividing the nuclear binding energy by the total number of nucleons (protons + neutrons). Nuclear binding energy per nucleon for nitrogen-14: \(\frac{Binding\:Energy}{Nucleons} = \frac{1.0805 \times10^{-11} J}{14} = 7.7179 \times10^{-13} J/nucleon\) Nuclear binding energy per nucleon for titanium-48: \(\frac{Binding\:Energy}{Nucleons} = \frac{7.153 \times 10^{-11} J}{48} = 1.4906\times 10^{-12} J/nucleon\) Nuclear binding energy per nucleon for xenon-129: \(\frac{Binding\:Energy}{Nucleons} = \frac{3.1851\times 10^{-10} J}{129} = 2.469 \times 10^{-12} J/nucleon\) Summary: (a) Nuclear mass: - Nitrogen-14: \(2.32405 \times10^{-26} kg\) - Titanium-48: \(7.95868 \times10^{-26} kg\) - Xenon-129: \(2.139192 \times10^{-25} kg\) (b) Nuclear binding energy: - Nitrogen-14: \(1.0805 \times10^{-11} J\) - Titanium-48: \(7.153 \times 10^{-11} J\) - Xenon-129: \(3.1851 \times10^{-10} J\) (c) Nuclear binding energy per nucleon: - Nitrogen-14: \(7.7179 \times10^{-13} J/nucleon\) - Titanium-48: \(1.4906 \times 10^{-12} J/nucleon\) - Xenon-129: \(2.469 \times 10^{-12} J/nucleon\)

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 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

Naturally found uranium consists of \(99.274 \%{ }^{238} \mathrm{U},\) $0.720 \%^{235} \mathrm{U},\( and \)0.006 \%^{233} \mathrm{U}\(. As we have seen, \){ }^{235} \mathrm{U}$ is the isotope that can undergo a nuclear chain reaction. Most of the ${ }^{235} \mathrm{U}$ used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g) .(\mathbf{a})\) What is the mass of \(\mathrm{UF}_{6}\) in a 30.0 -L vessel of \(\mathrm{UF}_{6}\) at a pressure of 695 torr at \(350 \mathrm{~K} ?\) (b) What is the mass of \({ }^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the UF \(_{6}\) is diffused through a porous barrier and that the change in the ratio of ${ }^{238} \mathrm{U}\( and \){ }^{235} \mathrm{U}$ in the diffused gas can be described by Equation \(10.23 .\) What is the mass of \({ }^{235} \mathrm{U}\) in a sample of

Hydroxyl radicals can pluck hydrogen atoms from molecules ("hydrogen abstraction"), and hydroxide ions can pluck protons from molecules ("deprotonation"). Write the reaction equations and Lewis dot structures for the hydrogen abstraction and deprotonation reactions for the generic carboxylic acid \(\mathrm{R}-\mathrm{COOH}\) with hydroxyl radical and hydroxide ion, respectively. Why is hydroxyl radical more toxic to living systems than hydroxide ion?

A \(65-\mathrm{kg}\) person is accidentally exposed for \(240 \mathrm{~s}\) to \(\mathrm{a}\) 15-mCi source of beta radiation coming from a sample of ${ }^{90}$ Sr. (a) What is the activity of the radiation source in disintegrations per second? In becquerels? (b) Each beta particle has an energy of \(8.75 \times 10^{-14} \mathrm{~J} .\) and \(7.5 \%\) of the radiation is absorbed by the person. Assuming that the absorbed radiation is spread over the person's entire body, calculate the absorbed dose in rads and in grays. \((\mathbf{c})\) If the RBE of the beta particles is \(1.0,\) what is the effective dose in mrem and in sieverts? (d) Is the radiation dose equal to, greater than, or less than that for a typical mammogram \((3 \mathrm{mSv}) ?\)

Indicate the number of protons and neutrons in the following nuclei: $(\mathbf{a}){ }^{214} \mathrm{Bi},(\mathbf{b}){ }^{210} \mathrm{~Pb},(\mathbf{c})\( uranium- \)235 .$.

Which are not classified as ionizing radiation: gamma rays, beta particles, radio waves used in radio and television, and infrared radiation from sun?

See all solutions

Recommended explanations on Chemistry Textbooks

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