Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Question
Chapter 45, Problem 45SP

(a)

To determine

The atomic number and mass number of the nucleus UX1.

(a)

Expert Solution
Check Mark

Answer to Problem 45SP

Solution:

90, 234

Explanation of Solution

Given data:

By natural radioactivity, U238 emits an alpha particle to form a nucleusUX1.

The heavy nucleus UX1 emits a beta particle to form the resultant nucleusUX2.

Formula used:

The conservation of mass is applied in the reaction PZ1A1QZ2A2+RZ3A3 by equating the atomic mass on the left side with the atomic mass on the right side and by equating the atomic number on the left side with the sum of atomic numbers on the right side.

Z1=Z2+Z3A1=A2+A3

Here, Z1, Z2, and Z3 represent the atomic numbers of P, Q, and R, respectively. Here, A1, A2, and A3 represent the atomic masses of P, Q, and R, respectively.

The reaction, when an alpha particle is emitted from a nucleus, is written as

PZADyx+α24

Here, α24 represents the nucleus of an alpha particle, P is the parent nucleus with atomic number Z and atomic mass A, and D is the daughter nucleus with atomic number y and mass number x.

Explanation:

Consider the given nucleus U238. The symbol U indicates that this is a uranium isotope. Write the value of the atomic number of uranium from the periodic table. The atomic number of uranium is 92. The complete representation of the nucleus is U92238.

Compare the nucleus U92238 with the parent nucleus PZ1A1. Therefore,

Z1=92A1=238

Write the standard expression for the emission of an alpha particle from a nucleus as

PZADyx1+α24

Here, x and y indicate the mass number and atomic number of the nucleus UX1, respectively.

The parent nucleus for the reaction is U92238 and the daughter nucleus formed is UX1. Therefore, substitute U for P, 238 for A1, 92 for N1, and UX1 for D1.

U92238UyxX1+α24…… (1)

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (1):

238=x+4x=2384x=234

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (1):

92=y+2y=922y=90

The value of x is calculated as 234 and the value of y is calculated as 90.

Rewrite equation (1) as

U92238UyxX1+α24

Substitute 234 for x and 90 for y

U92238U90234X1+α24

Conclusion:

The complete equation, after filling the values of x and y, is U92238D902341+α24. The daughter nucleus is represented as U90234X1, which indicates that the atomic number of the daughter nucleus UX1 is 90 and the mass number of UX1 is 234.

(b)

To determine

The atomic number and mass number of the nucleus UX2.

(b)

Expert Solution
Check Mark

Answer to Problem 45SP

Solution:

91, 234

Explanation of Solution

Given data:

By natural radioactivity, U238 emits an alpha particle to form a nucleus UX1.

The heavy nucleus UX1 emits a beta particle to form the resultant nucleus UX2.

Formula used:

The conservation of mass is applied in the reaction PZ1A1QZ2A2+RZ3A3 by equating the atomic mass on the left side with the atomic mass on the right side and by equating the atomic number on the left side with the sum of atomic numbers on the right side.

Z1=Z2+Z3A1=A2+A3

Here, Z1, Z2, and Z3 represent the atomic numbers of P, Q, and R, respectively. Here, A1, A2, and A3 represent the atomic masses of P, Q, and R, respectively.

The emission of a β particle corresponds to the emission of an electron. The chemical reaction for β emission from a parent nucleus to form a daughter nucleus is expressed as

PZADba+e10

Here, P is the parent nucleus, A and Z are the atomic mass and atomic number of the parent nucleus, respectively, D is the daughter nucleus, a and b are the atomic mass and atomic number of daughter nucleus, respectively, and e10 is the beta particle emitted.

Explanation:

Write a standard expression for the emission of abeta particle from the nucleus UX1 to form the nucleus UX2 as

UyxX1UbaX2+e10

Here, a and b indicate the mass number and atomic number of the nucleus UX2, respectively.

From part (a), the parent nucleus for this reaction is U90234X1 and the daughter nucleus formed is UX2. Therefore, substitute 234 for x and 90 for y

U90234X1UbaX2+e10 …… (2)

Apply the conservation of mass by equating the atomic masses on both sides of chemical equation (2):

234=a+0a=234

Apply the conservation of mass by equating the atomic numbers on both sides of chemical equation (2):

90=b1b=90+1b=91

The value of a is calculated as 234 and the value of b is calculated as 91.

Rewrite equation (2) as

U90234X1UbaX2+e10

Substitute 234 for a and 90 for y

U90234X1U91234X2+e10

Conclusion:

The complete equation, after filling the values of a and b, is U90234X1U91234X2+e10. The daughter nucleus is represented as U91234X2, which indicates that the atomic number of the daughter nucleus UX2 is 91 and the mass number of UX2 is 234.

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