physics : class12 Nuclear Stability and Mass Defect : Mass Defect : Atoms and Nuclei MCQ Questions and Answers
Practice important
Nuclear Stability and Mass Defect
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. The atomic mass of a nucleus with Z = 26 and A = 56 is 55.9349 u. Using m_H = 1.007825 u, m_n = 1.008665 u, and 1 u = 931.5 MeV/c^2, what are its mass defect and binding energy per nucleon, respectively?
A.
0.5285 u and 8.79 MeV
B.
0.5285 u and 9.44 MeV
C.
0.4750 u and 7.91 MeV
D.
0.6032 u and 10.05 MeV
Correct Answer:
0.5285 u and 8.79 MeV
Explanation:
The mass defect is Δm = 26(1.007825) + 30(1.008665) − 55.9349 = 0.5285 u. Thus, the total binding energy is 0.5285 × 931.5 ≈ 492.3 MeV, and the binding energy per nucleon is 492.3/56 ≈ 8.79 MeV.
2. Two hypothetical nuclei have the following properties: nucleus X has A = 120 and total binding energy 1000 MeV, while nucleus Y has A = 118 and total binding energy 1005 MeV. Which conclusion about their nuclear stability is most justified?
A.
X is more stable because its total binding energy is only slightly lower
B.
Y is more stable because its binding energy per nucleon is larger
C.
Both have equal stability because their total binding energies are nearly equal
D.
X is more stable because it contains more nucleons
Correct Answer:
Y is more stable because its binding energy per nucleon is larger
Explanation:
Stability is compared using binding energy per nucleon, not total binding energy. For X, BE/A = 1000/120 = 8.33 MeV, whereas for Y, BE/A = 1005/118 ≈ 8.52 MeV. Hence Y is more tightly bound per nucleon.
3. A hypothetical nucleus with A = 240 has a binding energy of 8.20 MeV per nucleon. Taking 1 u = 931.5 MeV/c^2, what is its mass defect approximately?
A.
1.96 u
B.
2.11 u
C.
2.46 u
D.
0.0088 u
Correct Answer:
2.11 u
Explanation:
The total binding energy is 240 × 8.20 = 1968 MeV. Therefore, Δm = BE/(931.5 MeV/u) = 1968/931.5 ≈ 2.11 u.
4. The atomic masses of 14C and 14N are 14.003242 u and 14.003074 u, respectively. Using 1 u = 931.5 MeV/c^2, what does the positive mass difference imply for the process 14C → 14N + e− + antineutrino?
A.
It is energetically allowed with Q ≈ 0.157 MeV
B.
It is energetically forbidden because nitrogen has a smaller atomic mass
C.
It is allowed with Q ≈ 1.57 MeV only if a positron is emitted
D.
It has Q = 0 because both nuclei have the same mass number
Correct Answer:
It is energetically allowed with Q ≈ 0.157 MeV
Explanation:
For beta-minus decay using atomic masses, Q = [M(14C) − M(14N)]c^2 = (0.000168 u)(931.5 MeV/u) ≈ 0.157 MeV. Since Q is positive, the decay is energetically allowed and moves the isobar toward greater stability.
5. Using m_H = 1.007825 u, m_n = 1.008665 u, the atomic mass of deuterium = 2.014102 u, and 1 u = 931.5 MeV/c^2, what is the binding energy of the deuteron?
A.
0.002388 MeV
B.
1.116 MeV
C.
2.225 MeV
D.
4.450 MeV
Correct Answer:
2.225 MeV
Explanation:
The mass defect is Δm = m_H + m_n − m_D = 1.007825 + 1.008665 − 2.014102 = 0.002388 u. Hence BE = 0.002388 × 931.5 ≈ 2.225 MeV.
6. Nucleus P has mass number 100 and binding energy per nucleon 8.60 MeV, while nucleus Q has mass number 200 and binding energy per nucleon 7.90 MeV. Which statement correctly compares their total mass defects?
A.
P has the larger mass defect because its binding energy per nucleon is larger
B.
Q has the larger mass defect because its total binding energy is larger
C.
Both have equal mass defects because their average nucleon masses are similar
D.
The mass defects cannot be compared without knowing their atomic numbers
Correct Answer:
Q has the larger mass defect because its total binding energy is larger
Explanation:
P has total binding energy 100 × 8.60 = 860 MeV, whereas Q has 200 × 7.90 = 1580 MeV. Since mass defect is proportional to total binding energy, Q has the larger mass defect despite its lower binding energy per nucleon.
7. Consider the fission reaction 235U + n → 92Kr + 141Ba + 3n. Given atomic masses m(235U) = 235.043924 u, m(92Kr) = 91.92616 u, m(141Ba) = 140.9144 u, and m_n = 1.008665 u, what is the approximate energy released?
A.
17.3 MeV
B.
93.6 MeV
C.
173.3 MeV
D.
931.5 MeV
Correct Answer:
173.3 MeV
Explanation:
The initial mass is 235.043924 + 1.008665 = 236.052589 u. The final mass is 91.92616 + 140.9144 + 3(1.008665) = 235.866555 u. Thus Δm = 0.186034 u and released energy is 0.186034 × 931.5 ≈ 173.3 MeV, reflecting the increase in average binding energy per nucleon.
8. Two isobars have A = 60. Their atomic masses are M1 = 59.9308 u for Z = 27 and M2 = 59.9302 u for Z = 28. Neglecting electron-binding energies, which conclusion follows from these data?
A.
The Z = 27 isobar is more stable because it has fewer protons
B.
The Z = 28 isobar is more stable because it has the smaller mass and larger mass defect
C.
Both are equally stable because their mass numbers are identical
D.
The Z = 27 isobar must undergo alpha decay because its mass is larger
Correct Answer:
The Z = 28 isobar is more stable because it has the smaller mass and larger mass defect
Explanation:
For isobars, the lower-mass state has the greater total binding energy and mass defect. Since M2 < M1, the Z = 28 isobar is more tightly bound and is energetically favored against beta conversion toward the other isobar.
9. A nucleus has 200 nucleons and a binding energy per nucleon of 7.50 MeV. Another nucleus formed from the same total number of nucleons has a binding energy per nucleon of 8.10 MeV. If the transformation is energetically possible, what mass defect is converted into energy?
A.
0.129 u
B.
0.645 u
C.
1.29 u
D.
120 u
Correct Answer:
0.129 u
Explanation:
The increase in total binding energy is 200(8.10 − 7.50) = 120 MeV. The corresponding mass conversion is Δm = 120/931.5 ≈ 0.129 u. The product is more stable because it has the greater binding energy per nucleon.
10. A nucleus has a mass defect of 1.20 u and contains 150 nucleons. Another nucleus has a mass defect of 1.00 u and contains 100 nucleons. Which statement is correct regarding their relative stability?
A.
The first nucleus is more stable because its total mass defect is larger
B.
The second nucleus is more stable because its mass defect per nucleon is larger
C.
Both have equal stability because their mass defects are of the same order
D.
Stability cannot be compared using mass defect under any circumstance
Correct Answer:
The second nucleus is more stable because its mass defect per nucleon is larger
Explanation:
The first nucleus has mass defect per nucleon 1.20/150 = 0.0080 u, while the second has 1.00/100 = 0.0100 u. Since binding energy per nucleon is proportional to mass defect per nucleon, the second nucleus is more tightly bound and generally more stable.
Short Questions and Answers
Practice short questions and answers related to
Nuclear Stability and Mass Defect
.
Short questions are currently unavailable for this page.