/*! 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} Q15PE Find the mass of \(^{239}{\rm{Pu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the mass of \(^{239}{\rm{Pu}}\) that has an activity of\(1.00\,\mu {\rm{Ci}}\).

Short Answer

Expert verified

The mass of \(^{239}{\rm{Pu}}\) that has an activity of\(1.00\,\mu {\rm{Ci}}\) is \(17.2\,\mu {\rm{g}}\).

Step by step solution

01

Definition of mass

Mass is a unit of measurement for the amount of matter in a body.

02

use the radioactive decay law

Consider the given problem and solve.

From appendix A, read the atomic mass number of \({\;^{259}}{\rm{Pu}}\) :

\({A_{{\rm{Pu}}}} = 259\)

From appendix B, read the half life time of \(^{259}{\rm{Pu}}\):

\({t_{1/2}} = 2.41 \cdot {10^4}{\rm{y}}\)

Using the radioactive decay law,

\(R = \frac{{N\ln 2}}{{{t_{1/2}}}}\)

Hence, the fact: \(1\,{\rm{Ci}} = 3.70 \cdot {10^{10}}\,\;{\rm{Bq}}\).

03

Find the mass

Let us solve the given problem.

We obtain,

\(\begin{aligned} N &= \frac{{R{t_{1/2}}}}{{\ln 2}}\\ &= \frac{{1.00 \times {{10}^6} \times 3.70 \times {{10}^{10}}\;{{\rm{s}}^1} \times 2.41 \times {{10}^4} \times 360 \times 24 \times 3600\;{\rm{s}}}}{{0.693}}\\ &= 4.00 \times {10^{16}}\end{aligned}\)

Mass of Plutonium:

\(\begin{aligned} m &= NAu\\ &= 4.00 \times {10^{16}} \times 259 \times 1.66 \times {10^{27}}\;{\rm{kg}}\\ &= 17.2\,\mu {\rm{g}}\end{aligned}\)

Therefore, mass of Plutonium is \(17.2\,\mu {\rm{g}}\).

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 91Ó°ÊÓ!

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

Breeding plutonium produces energy even before any plutonium is fissioned. (The primary purpose of the four nuclear reactors at Chernobyl was breeding plutonium for weapons. Electrical power was a by-product used by the civilian population.) Calculate the energy produced in each of the reactions listed for plutonium breeding just following Example 32.4. The pertinent masses are \(m\left( {{\rm{ }}239{\rm{ U}}} \right){\rm{ }} = {\rm{ }}239.054289{\rm{ u }},{\rm{ }}m\left( {{\rm{ }}239{\rm{ Np}}} \right){\rm{ }} = {\rm{ }}239.052932{\rm{ u }},{\rm{ and }}m\left( {{\rm{ }}239{\rm{ Pu}}} \right){\rm{ }} = {\rm{ }}239.052157{\rm{ u}}\)

A large power reactor that has been in operation for some months is turned off, but residual activity in the core still produces 150 MW of power. If the average energy per decay of the fission products is 1.00 MeV, what is the core activity in curies?

The energy produced by the fusion of a \(1.00 - kg\) mixture of deuterium and tritium was found in Example Calculating Energy and Power from Fusion. Approximately how many kilograms would be required to supply the annual energy use in the United States?

Are some types of cancer more sensitive to radiation than others? If so, what makes them more sensitive?

(a) Two annihilation \(\gamma \) rays in a PET scan originate at the same point and travel to detectors on either side of the patient. If the point of origin is \({\rm{9}}{\rm{.00\;cm}}\)closer to one of the detectors, what is the difference in arrival times of the photons? (This could be used to give position information, but the time difference is small enough to make it difficult.)

(b) How accurately would you need to be able to measure arrival time differences to get a position resolution of \({\rm{1}}{\rm{.00\;mm}}\)?

See all solutions

Recommended explanations on Physics 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.