Which has the larger binding energy, fluorine- 19 or oxygen-17?

Short Answer

Expert verified
Answer: Fluorine-19 (F-19) has a larger binding energy than oxygen-17 (O-17).

Step by step solution

01

Find the mass defect for F-19 and O-17

To find the mass defect, we need to know the masses of F-19, O-17, protons, and neutrons. The atomic masses of F-19 and O-17 are approximately 18.9984u and 16.9991u, respectively. The masses of a proton and a neutron are approximately 1.0073u and 1.0087u, respectively (u stands for atomic mass units). For F-19, there are 9 protons and 10 neutrons. The total mass is: Total mass of F-19 constituents = (9 × 1.0073u) + (10 × 1.0087u) = 19.1534u The mass defect of F-19 is: Mass defect of F-19 = Total mass of F-19 constituents - mass of F-19 = 19.1534u - 18.9984u = 0.155u For O-17, there are 8 protons and 9 neutrons. The total mass is: Total mass of O-17 constituents = (8 × 1.0073u) + (9 × 1.0087u) = 17.1308u The mass defect of O-17 is: Mass defect of O-17 = Total mass of O-17 constituents - mass of O-17 = 17.1308u - 16.9991u = 0.1317u
02

Convert mass defect to energy

To convert the mass defect to energy, we use Einstein's equation: E = mc^2. In this case, we only need the energy conversion factor for atomic mass units (u): 1u = 931.5 MeV/c^2. For F-19, the binding energy is: Binding energy of F-19 = Mass defect of F-19 × energy conversion factor = 0.155u × 931.5 MeV/c^2 = 144.3825 MeV For O-17, the binding energy is: Binding energy of O-17 = Mass defect of O-17 × energy conversion factor = 0.1317u × 931.5 MeV/c^2 = 122.69055 MeV
03

Compare the binding energies of F-19 and O-17

Based on our calculations, the binding energies are as follows: - Binding energy of F-19: 144.3825 MeV - Binding energy of O-17: 122.69055 MeV Since the binding energy of F-19 is greater than the binding energy of O-17, fluorine-19 has a larger binding energy than oxygen-17.

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

To determine the \(K_{s p}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), a solid sample is used, in which some of the iodine is present as radioactive I-131. The count rate of the sample is \(5.0 \times 10^{11}\) counts per minute per mole of \(\mathrm{I}\). An excess amount of \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) is placed in some water, and the solid is allowed to come to equilibrium with its respective ions. A 150.0-mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{s p}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\).

The cleavage of ATP (adenosine triphosphate) to ADP (adenosine diphosphate) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) may be written as follows: It is interesting to determine which bond (the \(\mathrm{P}-\mathrm{O}\) bond marked a or the \(\mathrm{O}\) - \(\mathrm{P}\) bond marked \(\mathbf{b}\) ) is cleaved by hydrolysis (reaction with water). (a) Outline an experiment (using radioactivity) that can be used to determine where the cleavage results. (b) Describe the results that would lead you to conclude that cleavage results at a, (c) Describe the results that would lead you to conclude that cleavage results at \(\mathbf{b}\). Results show that the cleavage occurs at \(\mathrm{b}\).

A rock from an archaeological dig was found to contain \(0.255 \mathrm{~g}\) of Pb-206 per gram of U-238. Assume that the rock did not contain any Pb-206 at the time of its formation and that U-238 decayed only to Pb-206. How old is the rock? (For \(\mathrm{U}-238, t_{1 / 2}=4.5 \times 10^{9} \mathrm{y}\).)

It is possible to estimate the activation energy for fusion by calculating the energy required to bring two deuterons close enough to one another to form an alpha particle. This energy can be obtained by using Coulomb's law in the form \(E=8.99 \times 10^{9} q_{1} q_{2} / r\), where \(q_{1}\) and \(q_{2}\) are the charges of the deuterons \(\left(1.60 \times 10^{-19} \mathrm{C}\right), r\) is the radius of the He nucleus, about \(2 \times 10^{-15} \mathrm{~m}\), and \(E\) is the energy in joules. (a) Estimate \(E\) in joules per alpha particle. (b) Using the equation \(E=m v^{2} / 2\), estimate the velocity (meters per second) each deuteron must have if a collision between the two of them is to supply the activation energy for fusion \((m\) is the mass of the deuteron in kilograms).

Carbon from a cypress beam obtained from the tomb of an ancient Egyptian king gave \(9.2\) disintegrations/minute of \(C-14\) per gram of carbon. Carbon from living material gives \(15.3\) disintegrations/min of C-14 per gram of carbon. Carbon-14 has a half-life of 5730 years. How old is the beam?

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