Many transuranium elements, such as plutonium-232, have very short half-lives. (For \({ }^{232} \mathrm{Pu}\), the half-life is 36 minutes.) However, some, like protactinium-231 (half-life \(=3.34 \times\) \(10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \({ }^{232}\) Pu nuclei and 1 mole of \({ }^{231}\) Pa nuclei are each formed from their respective number of protons and neutrons.

Short Answer

Expert verified
The change in energy when 1 mole of ${ }^{232}\mathrm{Pu}$ nuclei are formed is 2.16049 x 10^{-10} J, and the change in energy when 1 mole of ${ }^{231}\mathrm{Pa}$ nuclei are formed is 2.75236 x 10^{-10} J.

Step by step solution

01

Calculate mass difference for \({ }^{232}\mathrm{Pu}\) and \({ }^{231}\mathrm{Pa}\)

We'll calculate mass difference for the two nuclei separately. To do this, we'll first calculate the total mass of protons and neutrons and then subtract the mass of the nucleus. For \({ }^{232}\mathrm{Pu}\): Number of protons = 94 Number of neutrons = 232 - 94 = 138 Mass of 1 proton = 1.00728 amu Mass of 1 neutron = 1.00867 amu Total mass of protons and neutrons = 94 x 1.00728 amu + 138 x 1.00867 amu = 94.28544 amu + 139.19646 amu = 233.4819 amu Mass of 1 mole of \({ }^{232}\mathrm{Pu}\) nucleus = 232.037 amu (given) Mass difference = 233.4819 amu - 232.037 amu = 1.4449 amu For \({ }^{231}\mathrm{Pa}\): Number of protons = 91 Number of neutrons = 231 - 91 = 140 Total mass of protons and neutrons = 91 x 1.00728 amu + 140 x 1.00867 amu = 91.66348 amu + 141.2148 amu = 232.87828 amu Mass of 1 mole of \({ }^{231}\mathrm{Pa}\) nucleus = 231.036 amu (given) Mass difference = 232.87828 amu - 231.036 amu = 1.84228 amu
02

Convert mass difference to energy

We will use the Einstein's mass-energy equivalence formula, assuming \(c^2 = 9 \times 10^{16} m^2/s^2\) to convert the mass difference (in amu) to energy difference (in Joules). 1 amu = 1.66 x 10^{-27} kg For \({ }^{232}\mathrm{Pu}\): Mass difference = 1.4449 amu = 1.4449 x 1.66 x 10^{-27} kg = 2.40054 x 10^{-27} kg Energy difference = \(E = mc^2\) = 2.40054 x 10^{-27} kg x 9 x 10^{16} m^2/s^2 = 2.16049 x 10^{-10} J For \({ }^{231}\mathrm{Pa}\): Mass difference = 1.84228 amu = 1.84228 x 1.66 x 10^{-27} kg = 3.05818 x 10^{-27} kg Energy difference = \(E = mc^2\) = 3.05818 x 10^{-27} kg x 9 x 10^{16} m^2/s^2 = 2.75236 x 10^{-10} J These are the energy changes for 1 mole of \({ }^{232}\mathrm{Pu}\) and \({ }^{231}\mathrm{Pa}\) nuclei, respectively.

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

Much of the research on controlled fusion focuses on the problem of how to contain the reacting material. Magnetic fields appear to be the most promising mode of containment. Why is containment such a problem? Why must one resort to magnetic fields for containment?

A \(0.10-\mathrm{cm}^{3}\) sample of a solution containing a radioactive nuclide \(\left(5.0 \times 10^{3}\right.\) counts per minute per milliliter) is injected into a rat. Several minutes later \(1.0 \mathrm{~cm}^{3}\) blood is removed. The blood shows 48 counts per minute of radioactivity. Calculate the volume of blood in the rat. What assumptions must be made in performing this calculation?

A certain radioactive nuclide has a half-life of \(3.00\) hours. a. Calculate the rate constant in \(\mathrm{s}^{-1}\) for this nuclide. b. Calculate the decay rate in decays/s for \(1.000\) mole of this nuclide.

To determine the \(K_{\text {sp }}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), a chemist obtained a solid sample of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in which some of the iodine is present as radioactive \({ }^{131} \mathrm{I}\). The count rate of the \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) 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 into some water, and the solid is allowed to come to equilibrium with its respective ions. A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\). \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\)

The sun radiates \(3.9 \times 10^{23} \mathrm{~J}\) of energy into space every \(\mathrm{sec}-\) ond. What is the rate at which mass is lost from the 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