/*! 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} Q9P Find the disintegration energy Q... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the disintegration energy Q for the decay of 49Vby Kelectron capture (see Problem 54). The needed data are mv=48.94852u,mn=48.94787uand .

Short Answer

Expert verified

The disintegration energy for the decay of vanadium atom is 600 keV .

Step by step solution

01

Identification of given data

The mass of the vanadium atom is mv=48.94852u

The mass of the Titanium atom is mTi=48.94787u

The energy for K electron capture is Ek=5.47keV

02

Concept Introduction

The disintegration energy for an atom is the energy necessary for breaking the nucleus of the atom.

The disintegration energy for the decay of vanadium atom is given as:

E=(mv-mTi)c2-Ek…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(1)

Here, is the speed of light.

03

Determination of disintegration energy for the decay of vanadium atom

Substitute all the values in the above equation (1), and we get.

E=(48.94852u-48.94787u)c2-5.47keV=0.00065u(c2)1c2931.5×103keVu-5.47keV=600keV

Therefore, the disintegration energy for the decay of vanadium atom is 100 keV.

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

The magic nucleon numbers for nuclei are given in Module 42-8 as 2, 8, 20, 28, 50, 82, and 126. Are nuclides magic (that is, especially stable) when (a) only the mass number A, (b) only the atomic number Z, (c) only the neutron number N, or (d) either Zor N(or both) is equal to one of these numbers? Pick all correct phrases.

Figure 42-16 gives the activities of three radioactive samples versus time. Rank the samples according to their (a) half-life and (b) disintegration constant, greatest first. (Hint:For (a), use a straightedge on the graph.)

Two radioactive materials that alpha decay,U238and T232h, and one that beta decaysK40, are sufficiently abundant in granite to contribute significantly to the heating of Earth through the decay energy produced. The alpha-decay isotopes give rise to decay chains that stop when stable lead isotopes are formed. The isotopeK40has single beta decay. (Assume this is the only possible decay of that isotope.) Here is the information:

In the table Qis the totalenergy released in the decay of one parent nucleus to the finalstable endpoint and fis the abundance of the isotope in kilograms per kilogram of granite;means parts per million. (a) Show that these materials produce energy as heat at the rate of1.0×10-8Wfor each kilogram of granite. (b) Assuming that there is2.7×1022kgof granite in a 20-km-thick spherical shell at the surface of Earth, estimate the power of this decay process over all of Earth. Compare this power with the total solar power intercepted by Earth,1.7×1017W1.

A rock recovered from far underground is found to contain 0.86 mg of U238, 0.15 mg ofP206b , and 1.6 mg ofA40r . How muchK40 will it likely contain? Assume thatK40 decays to onlyA40r with a half-life of1.25×109y . Also assume thatU238 has a half-life of4.47×109y .

Make a nuclidic chart similar to Fig. 42-6 for the 25 nuclides T118-122e, S117-121b, S116-120n, ln115-19, and C114-118d. Draw in and label (a) all isobaric (constant A) lines and (b) all lines of constant neutron excess, defined as N - Z.

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.