Related questions with answers

If we neglect the neutrino's mass, the total kinetic energy, KK, released in the β\beta^{-}decay (Z,N)(Z, N) \rightarrow (Z+1,N1)+e+νˉ(Z+1, N-1)+e^{-}+\bar{\nu} is given by (17.22). In particular, the criterion that β\beta^{-}decay is possible is just that KK must be positive, or

matom (Z,N)matom (Z+1,N1) (for βdecay) m_{\text {atom }}(Z, N) \geq m_{\text {atom }}(Z+1, N-1) \quad \text { (for } \beta^{-} \text {decay) }

In other words, one has only to compare the atomic masses involved to see if β\beta^{-}decay is possible. (a) Find an expression for the total kinetic energy released in the β+\beta^{+}decay

(Z,N)(Z1,N+1)+e++ν(Z, N) \rightarrow(Z-1, N+1)+e^{+}+\nu

and show that the condition for this process to be possible is

matom (Z,N)matom (Z1,N+1)+2mem_{\text {atom }}(Z, N) \geq m_{\text {atom }}(Z-1, N+1)+2 m_e

 (for β+decay) \text { (for } \beta^{+} \text {decay) }

(b) Do the same for the electron capture

(Z,N)+e(Z1,N+1)+ν(17.85)(Z, N)+e^{-} \rightarrow(Z-1, N+1)+\nu \quad(17.85)

assuming that the original two particles are at rest, and show that the condition for electron capture is the same as (17.84) except that the term 2me2 m_e is missing. (c) The mass of 231U{ }^{231} \mathrm{U} is 231.03626u231.03626 \mathrm{u}, while that of 231 Pa{ }^{231} \mathrm{~Pa} is 231.03588u231.03588 \mathrm{u}. Show that 231U{ }^{231} \mathrm{U} can decay by electron capture but not by β+\beta^{+}emission.

Question

(a) Use the semiempirical binding-energy formula (16.30) to write an expression for the mass of a nucleus in terms of ZZ and AA. (b) Now consider a set of isobars with AA odd. Use your mass formula to show that a plot of mass against ZZ, as in the above figure, should fit a parabola (m=αZ2+βZ+γ)\left(m=\alpha Z^2+\beta Z+\gamma\right). (c) Now consider a set of isobars with AA even. Show that because of the pairing energy, the masses will alternate between two parabolas, one for the nuclei with ZZ even and the other for ZZ odd. (d) Draw the plot of mm against ZZ, and explain why one can have two stable isobars if AA is even. (e) Find two examples of this phenomenon in Appendix D.

Solution

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The semiempirical binding-energy formula is given by

B=avolAasurfA2/3acoulZ2A1/3asym(ZN)2A+εapairA1/2B=a_\text{vol}A-a_\text{surf}A^{2/3}-a_\text{coul}\dfrac{Z^2}{A^{1/3}}-a_\text{sym}\dfrac{(Z-N)^2}{A}+\varepsilon\dfrac{a_\text{pair}}{A^{1/2}}

In this exercise, we use the formula to find the most stable nuclei among a set of isobars.

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