The following table gives values for the mass defect ∆m for four hypothetical nuclei: A, B, C, and D, isotopes of an element.
ABCD Mass defect, ∆m(×10^-29 kg)
A. B. C. D
-6.0 -2.0 0 +6.0
Which statement is true regarding the stability of these nuclei?
(a)Nucleus D is the most stable, and nucleus A is the least stable.
(b)Nucleus C is stable, whereas nuclei A, B, and D are not.
(c)Nucleus A is the most stable, and nucleus D is not stable.
(d)Nuclei A and B are stable, but nucleus B is more stable than nucleus A.
I'm confused!! Please help
Which part do you get confused on?
Just wondering.
To determine the stability of the nuclei, we need to consider the mass defect. The mass defect is the difference between the predicted mass and the actual mass of a nucleus.
A positive mass defect indicates that the actual mass of the nucleus is less than the predicted mass. This means that the nucleus is less stable, as there is excess energy present.
A negative mass defect indicates that the actual mass of the nucleus is greater than the predicted mass. This indicates that energy has been released during the formation of the nucleus, making it more stable.
The table shows that nuclei A and B have negative mass defects (-6.0 and -2.0 respectively), while nucleus D has a positive mass defect (+6.0). Nucleus C has a mass defect of zero.
From this, we can determine that nuclei A and B are more stable (having negative mass defects) compared to nucleus D (having a positive mass defect). Nucleus C, having a mass defect of zero, is also stable.
Therefore, the correct statement is:
(d) Nuclei A and B are stable, but nucleus B is more stable than nucleus A.
To determine the stability of the nuclei, we need to analyze the mass defect values. The mass defect represents the difference between the actual mass of a nucleus and the sum of the masses of its individual nucleons (protons and neutrons).
In general, more stable nuclei tend to have a larger positive mass defect. This is because some mass is converted into binding energy during the formation of a nucleus, according to Einstein's mass-energy equivalence principle (E=mc^2).
Let's analyze the given values for the mass defect:
∆m(A) = -6.0 × 10^-29 kg
∆m(B) = -2.0 × 10^-29 kg
∆m(C) = 0
∆m(D) = +6.0 × 10^-29 kg
From the values, we can see that nuclei A and B have negative mass defects. This indicates that they have a larger mass than the sum of their constituent nucleons, implying that they are unstable.
Nucleus C has a mass defect of zero, suggesting that it is at the border of stability. It is not necessarily unstable, but it is not as stable as a nucleus with a positive mass defect.
Finally, nucleus D has a positive mass defect, indicating that it is the most stable among the four.
Based on this analysis, the correct statement regarding the stability of these nuclei is:
(a) Nucleus D is the most stable, and nucleus A is the least stable.