Among the three isobars which has the highest neutron-proton ratio.

Question

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Answer 1

Question

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

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Answer 2

Question

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

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Answer 3

Question

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

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Answer 4

Question

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

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Answer 5

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

Answer 6

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Answer 7

Chapter 13, Problem 13P

(a)

To determine

Among the three isobars which has the highest neutron-proton ratio.

Answer 8

(a)

Expert Solution

Answer to Problem 13P

Among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for neutron to proton ratio.

$neutron to proton ratio=A−ZZ$ (I)

Here, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $55$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $55139Cs$ .

$neutron to proton ratio=139−5555=1.53$

Substitute $57$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $57139La$ .

$neutron to proton ratio=139−5757=1.43$

Substitute $59$ for $Z$ , and $139$ for $A$ in equation (I), to find neutron to proton ratio of $59139Pr$ .

$neutron to proton ratio=139−5959=1.35$

Therefore, among the three isobars $55139Cs$ has the highest neutron-proton ratio equal to $1.53_$ .

Answer 13

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Answer 14

(b)

To determine

Among the three isobars which has the highest binding energy per nucleon.

Answer 15

(b)

Expert Solution

Answer to Problem 13P

Among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for binding energy per nucleon.

$EbA=1A[C1A−C2A2/3−C3(Z(Z−1))A−1/3−C4(N−Z)2A]$

Here, $EbA$ is the binding energy per nucleon, $A$ is the mass number, and $Z$ is the atomic number.

Conclusion:

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $80$ for $N$ in equation (I), to find $EbA$ for $59139Pr$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(58)(139)1/3−23.6(80−59)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(21)2139]=1160.8 MeV139=8.351 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $57139La$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(82−57)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(25)2139]=1161.1 MeV139=8.355 MeV$

Substitute $15.7$ for $C1$ , $17.8$ for $C2$ , $0.71$ for $C3$ , $23.6$ for $C4$ , $139$ for $A$ , $59$ for $Z$ and $82$ for $N$ in equation (I), to find $EbA$ for $55139Cs$ .

$EbA=1139[(15.7)(139)−(17.8)(139)2/3−0.71(55)(54)(139)1/3−23.6(84−55)2139]1139[(15.7)(139)−(17.8)(139)2/3−0.71(59)(139)1/3−23.6(29)2139]=1154.9 MeV139=8.308 MeV$

Therefore, among the three isobars $57139La$ has the highest binding energy per nucleon equal to $8.353 MeV_$ .

Answer 20

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .

Conclusion:

Therefore, among the two radioactive nuclei $139Cs$ is more heavier.

Answer 21

(c)

To determine

Among the two radioactive nuclei $(139Pr and139Cs)$ which one is heavier.

Answer 22

(c)

Expert Solution

Answer to Problem 13P

Among the two radioactive nuclei $139Cs$ is more heavier.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The mass of neutron is greater than the mass of proton , thus the atom with more number of nucleon $(N)$ and smallest atomic number $(Z)$ is more heavier.

Therefore $55139Cs$ is more heavier having a mass of $138.913 u$ .