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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Textbook Question
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Chapter 45, Problem 48SP

Complete the notations for the following processes.

(a) 24 Mg ( d , p ) ?

(b) 26 Mg ( d , p ) ?

(c) 40 Ar ( α , p ) ?

(d) 12 C ( d , n ) ?

(e) 130 Te ( d , 2 n ) ?

(f) 55 Mn ( n , γ ) ?

(g) 59 Co ( n , α ) ?

(a)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process for magnesium, if the alpha particles are emitted in the decay process as, 24Mg(d,α).

Answer to Problem 48SP

Solution:

24Mg(d,α)N22a

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h

Explanation:

Write the decay equation for 24Mg(d,α)Xa to complete the notation of the decay. Here particle d is the deuteron (12H), and α stands for the alpha particle.

1224Mg+12H baX+24α

Here, X is the element of the atomic mass number a and the number of protons is b.

The sum of the mass number of the reactants and products are equal. Hence,

24+2=a+4

Solve for a.

a=22

The sum of the atomic number of the reactants and products are equal. Hence,

12+1=b+2

Solve for b.

b=11

The sodium element has atomic number 11 and mass number 22 .

Thus, rewrite the notation.

24Mg(d,α)Xa

Substitute Na for X and 22 for a

24Mg(d,α)N22a.

Conclusion:

Therefore, the notations of the process are 24Mg(d,α)N22a.

(b)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process for magnesium of the mass number 26, if the proton is emitted in the decay process.

Answer to Problem 48SP

Solution:

26Mg(d,α)M27g

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h

Explanation:

Write the decay equation for 26Mg(d,p)Xa to complete the notation of the decay. Here particle d is the deuteron (12H), and p stands for the proton.

1226Mg+12H baX+11H

Here, X is the element of the atomic mass number a and thenumber of protons is b.

The sum of the mass number of the reactants and products are equal. Hence,

26+2=a+1

Solve for a.

a=27

The sum of the atomic number of the reactants and products are equal. Hence,

12+1=b+1

Solve for b.

b=12

The magnesium element has an atomic number thatequals to 12.

Rewrite the notation.

26Mg(d,p)Xa

Substitute Mg for X and 27 for a

26Mg(d,α)M27g.

Conclusion:

Therefore, the notations of the process are 26Mg(d,α)M27g.

(c)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process for argon of mass number equals to 40 given as 40Ar(α,p).

Answer to Problem 48SP

Solution:

40Ar(α,p)K43

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h

Explanation:

Write the decay equation for 40Ar(α,p)Xa to complete the notation of the decay. Here, particle α stands for the alpha particle and p stands for the proton.

1840Ar+24α baX+11H

Here, X is the element of the atomic mass number a and the number of protons is b.

The sum of the mass number of the reactants and products are equal. Hence,

40+4=a+1

Solve for a.

a=43

The sum of the atomic number of the reactants and products are equal. Hence,

18+2=b+1

Solve for b.

b=19

The potassium element has atomic number 19 and mass number 43 .

Rewrite the notation.

40Ar(α,p)Xa

Substitute K for X and 43 for a

40Ar(α,p)K43

Conclusion:

Therefore, the notations of the process are 40Ar(α,p)K43.

(d)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process forcarbon of the mass 12 given as 12C(d,n), if the neutron is ejected.

Answer to Problem 48SP

Solution:

12C(d,n)N13

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h

Explanation:

Write the decay equation for 12C(d,n)Xa to complete the notation of the decay. Here, particle d stands for the deuteron (12H) and n stands for the neutron.

612C +12H baX+01n

Here, X is the element of the atomic mass number a and the number of protons is b.

The sum of the mass number of the reactants and products are equal. Hence,

12+2=a+1

Solve for a.

a=13

The sum of the atomic number of the reactants and products are equal. Hence,

6+1=b+0

Solve for b.

b=7

The nitrogen element has atomic number 7 and mass number 13 .

Rewrite the notation.

12C(d,n)Xa

Substitute N for X and 13 for a

12C(d,n)N13

Conclusion:

Therefore, the notations of the process are 12C(d,n)N13.

(e)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process for tellurium Te of mass number 130 given as 130Te(d,2n), if two neutrons are ejected.

Answer to Problem 48SP

Solution:

130Te(d,2n)I130

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h

Explanation:

Write the decay equation for 24Te(d,2n)Xa to complete the notation of the decay. Here, particle d stands for the deuteron (12H), and n stands for the neutron.

52130Te+12H baX+201n

The sum of the mass number of the reactants and products are equal. Hence,

130+2=a+2

Solve for a.

a=130

The sum of the atomic number of the reactants and products are equal. Hence,

52+1=b+2(0)

Solve for b.

b=53

The iodine element has atomic number 53 and mass number 130 .

Rewrite the notation.

130Te(d,2n)Xa

Substitute I for X and 130 for a

130Te(d,2n)I130

Conclusion:

Therefore, the notations of the process are 130Te(d,2n)I130.

(f)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process formanganese of mass number 55 given by, if gamma particle is separated in the decay process.

Answer to Problem 48SP

Solution:

55Mn(n,γ)M56n

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h.

Explanation:

Write the decay equation for 55Mn(n,γ)Xa to complete the notation of the decay. Here, particle n stands for the neutron, and γ stands for the gamma particle.

2555Mn+01n baX+γ

The sum of the mass number of the reactants and products are equal. Hence,

55+1=a

Solve for a.

a=56

The sum of the atomic number of the reactants and products are equal. Hence,

25=bb=25

The manganese element has atomic number 25 and mass number 55 .

Rewrite the notation.

55Mn(n,γ)Xa

Substitute Mn for X and 56 for a

55Mn(n,γ)M56n

Conclusion:

Therefore, the notations of the process are 55Mn(n,γ)M56n.

(g)

Expert Solution
Check Mark
To determine

To complete: The notation of the decay process forcobalt of mass number 59 given by 59Co(n,α), if alpha particle is separated in the decay process.

Answer to Problem 48SP

Solution:

59Co(n,α)M56n

Explanation of Solution

Introduction:

The decay process equation is written as,

baA+Dfe dcB+Ehg

The notation for decay process is written as baA(D,E)fcB, where Dfe and Ehg are the light bombarding and the light product particle, respectively.

The atomic number and mass number in the decay equation of the reactants and products are conserved.

That is, a+e=c+g and b+f=d+h.

Explanation:

Write the decay equation for 59Co(n,α)Xa to complete the notation of the decay. Here, particle n stands for the neutron, and α stands for the alpha particle.

2759Co+01 baX+ 24α

The sum of the mass number of the reactants and products are equal. Hence,

59+1=a+4

Solve for a.

a=56

The sum of the atomic number of the reactants and products are equal. Hence,

27+0=b+2

Solve for b.

b=25

The manganese element has atomic number 25 and mass number 56 .

Rewrite the notation.

59Co(n,α)Xa

Substitute Mn for X and 56 for a

59Co(n,α)M56n

Conclusion:

Therefore, the notations of the process are 59Co(n,α)M56n.

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