Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

High-mass radionuclides, which may be either alpha or beta emitters, belong to one of four decay chains, depending on whether their mass number A is of the form 4n, 4n+1, 4n+2, or 4n+3, where n is a positive integer. (a) Justify this statement and show that if a nuclide belongs to one of these families, all its decay products belong to the same family. Classify the following nuclides as to family: (b) 235U(c) localid="1661601960557" 236U(d) 238U (e) localid="1661601429038" 239PU (f) localid="1661601438307" 240PU(g) localid="1661601952668" 245PU (h) localid="1661601482780" 246Cm (i) 249Cfand (j) 249Fm.

Short Answer

Expert verified

(a) The mass number of nuclides after the decay of alpha and beta particles belongs to the family of mass numbers 4n,4n+1, 4n+2, and 4n+3.

(b) The U-235 belongs to the 4n+3 family of mass numbers.

(c) The U-236 belongs to the 4n family of mass numbers.

(d) The U-238 belongs to the 4n+2 family of mass numbers.

(e) The Pu-239 belongs to the 4n+3 family of mass numbers.

(f) The Pu-240 belongs to the 4n family of mass numbers.

(g) The Cm-245 belongs to the 4n+1 family of mass numbers.

(h) The Cm-246 belongs to the 4n+2 family of mass numbers.

(i) The Cf-249 belongs to the 4n+1 family of mass numbers.

(j) The Fm-253 belongs to the 4n+1 family of mass numbers.

Step by step solution

01

Radioactive decay

Radioactive decay is the process of decaying protons from the nucleus and variation in the mass number of the atom. The number of protons increases by one unit due to beta decay and no change in the mass number of the atom, while the number of protons changes by two units and mass number by four units for alpha decay.

02

Justification for statement and proof for nuclide belonging to one family of decay(a)

The mass number of nuclides changes by four units in the alpha decay so the mass number of a new nuclide is always of type A - 4n .

Here, A is the mass number of original nuclides and is the number of decayed alpha particles.

The mass number does not change due to the decay of beta particles so the mass number remains unchanged.

The mass number of nuclides for alpha or beta particle decay remains as 4n, 4n+1, 4n+2, and 4n+3.

Therefore, the mass number of nuclides after the decay of alpha and beta particles belongs to the family of mass numbers 4n, 4n+1, 4n+2, and 4n+3.

03

Classification of the family for nuclide(b)

The mass number of U-235 can be expressed as:

If we compare the above-expressed mass number with 4n+3 for n=58 then U-235 belongs to the 4n+3 family of mass numbers.

Therefore, the U-235 belongs to the 4n+3 family of mass numbers.

04

Classification of the family for nuclide(c)

The mass number of U-236 can be expressed as:

A = 236

A = 4(59)

If we compare the above-expressed mass number with 4n for n=59 then U-236 belongs to the 4n family of mass numbers.

Therefore, the U-236 belongs to the 4n family of mass numbers.

05

Classification of the family for nuclide(d)

The mass number of U-238 can be expressed as:

A = 238

A = 4 (59) + 2

If we compare the above-expressed mass number with 4n+2 for n=59 then U-238 belongs to the 4n+2 family of mass numbers.

Therefore, the U-238 belongs to the 4n+2 family of mass numbers.

06

Classification of the family for nuclide(e)

The mass number of Pu-239 can be expressed as:

A = 239

A = 4(59) + 3

If we compare the above-expressed mass number with 4n+3 for n=59 then Pu-239 belongs to the 4n+3 family of mass numbers.

Therefore, the Pu-239 belongs to the 4n+3 family of mass numbers.

07

Classification of the family for nuclide(f)

The mass number of Pu-240 can be expressed as:

A = 240

A = 4(60)

If we compare the above-expressed mass number with 4n for n=60 then Pu-240 belongs to the 4n family of mass numbers.

Therefore, the Pu-240 belongs to the 4n family of mass numbers.

08

Classification of the family for nuclide(g)

The mass number of Cm-245 can be expressed as:

A = 245

A = 4(61) + 1

If we compare the above-expressed mass number with 4n+1 for n=61 then Cm-245 belongs to the 4n+1 family of mass numbers.

Therefore, the Cm-245 belongs to the 4n+1 family of mass numbers.

09

Classification of the family for nuclide(h)

The mass number of Cm-246 can be expressed as:

A = 246

A = 4(61) + 2

If we compare the above-expressed mass number with 4n+2 for n=61 then Cm-246 belongs to the 4n+2 family of mass numbers.

Therefore, the Cm-246 belongs to the 4n+2 family of mass numbers.

10

Classification of the family for nuclide(i)

The mass number of Cf-249 can be expressed as:

A = 249

A = 4(62) + 1

If we compare the above-expressed mass number with 4n+1 for n=62 then Cf-249 belongs to the 4n+1 family of mass numbers.

Therefore, the Cf-249 belongs to the 4n+1 family of mass numbers.

11

Classification of the family for nuclide(j)

The mass number of Fm-253 can be expressed as:

A = 253

A = 4(63) + 1

If we compare the above-expressed mass number with 4n+1 for n=63 then Fm-253 belongs to the 4n+1 family of mass numbers.

Therefore, the Fm-253 belongs to the 4n+1 family of mass numbers.

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

What is the binding energy per nucleon of the europium isotope E63152u? Here are some atomic masses and the neutron mass.

Eu63152151.921742uH11.007825un1.008665u

Nuclear radii may be measured by scattering high energy (high speed) electrons from nuclei. (a) What is the de-Broglie wavelength for 200MeV electrons? (b) Are these electrons suitable probes for this purpose?

What is the binding energy per nucleon of the rutherfordium isotope Rf104259? Here are some atomic masses and the neutron mass.

Rf104259259.10563uH11.007825un1.008665u

A 10.2MeV Li nucleus is shot directly at the center of a Ds nucleus. At what center-to-center distance does the Li momentarily stop, assuming the does not move?

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 of1.0ร—10-8Wfor each kilogram of granite. (b) Assuming that there is2.7ร—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,1.7ร—1017W1.

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.

Sign-up for free