Which Oxygen Isotope Undergoes Mirror Nuclei Decay?

In summary, the homework statement asks which isotope of oxygen decays into mirror nuclei. TheAttempt at a Solution suggests that the energy balance may give the answer. However, it is not clear which isotope should be chosen for experimentation.
  • #1
skrat
748
8

Homework Statement


Using semi empirical mass formula calculate which isotope of oxygen (##_8O##) decays into mirror nuclei. Is it ##\beta +## or ##\beta -##?

Homework Equations


$$W(A,Z)=w_0A-w_1A^{2/3}-w_2\frac{Z^2}{A^{1/3}}-w_3\frac{(A-2Z)^2}{A}-w_4A^{-3/4}\delta(A,Z)$$ where $$w_0=15.8MeV$$ $$w_1=17.8MeV$$ $$w_2=0.71MeV$$ $$w_3=23.7MeV$$ and $$w_4=11.2MeV$$

The Attempt at a Solution


[/B]
Hmmm, firstly: $$\beta ^+ : \qquad ^A_8O \rightarrow ^A_7X+e^++\nu _e$$ and $$\beta ^- : \qquad ^A_8O \rightarrow ^A_9Y+e^-+\bar{\nu _e}$$ I don't really know what to do here but... My idea was that maybe the energy balance would give me the answer?...

Something like:
For ##\beta ^+##: $$M_Oc^2=M_xc^2+m_ec^2$$
$$8m_pc^2+(A-8)m_nc^2-W(A,8)-(7m_pc^2+(A-7)m_nc^2-W(A,7)+m_ec^2)=C$$ Now my idea was if that ##C>0## than the decay is not possible. But... I amnot sure this is the right way, since I don't have the number ##A##.

Does anybody have a hint on what to do here? :/
 
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  • #2
You want to show (as you stated) that the mass of the left side of the decay equation is greater than the mass on the right side. You must choose the value of A for known isotopes of oxygen and try each one to see what happens.
 
  • #3
gleem said:
You want to show (as you stated) that the mass of the left side of the decay equation is greater than the mass on the right side. You must choose the value of A for known isotopes of oxygen and try each one to see what happens.

Aha, so there isn't any mysterious way to find that A. I just have to guess it and look for the right one?
 
  • #4
Yep. Common isotopes of oxygen are 14,15,16,17,18,19, 20 .three are stable. There are a few more with Half-Lifes less than 1 sec
 

FAQ: Which Oxygen Isotope Undergoes Mirror Nuclei Decay?

What is the Semi-Empirical Mass Formula?

The Semi-Empirical Mass Formula (SEMF) is a mathematical formula used to calculate the approximate nuclear binding energy of an atomic nucleus. It takes into account the number of protons and neutrons in the nucleus, as well as other factors such as the pairing effect and the shell correction.

What is the significance of the Semi-Empirical Mass Formula?

The SEMF is significant because it helps us understand the stability of atomic nuclei and the process of nuclear fusion and fission. It also allows us to predict the binding energy and therefore the stability of new, undiscovered nuclei.

How accurate is the Semi-Empirical Mass Formula?

The SEMF is an approximation and does not account for all factors that contribute to the binding energy of a nucleus. Therefore, it is not always accurate and may have significant discrepancies with experimental values for certain nuclei. However, it is generally considered to be a good estimate for the binding energy of most nuclei.

Can the Semi-Empirical Mass Formula be applied to all nuclei?

No, the SEMF is only applicable to stable nuclei and cannot be used for highly unstable or exotic nuclei. It is also limited in its accuracy for very heavy or very light nuclei.

Who developed the Semi-Empirical Mass Formula?

The SEMF was developed in the 1930s by German physicist Carl Friedrich von Weizsäcker and was later refined by several other scientists, including Hans Bethe and Robert Bacher. It continues to be used and modified by scientists today in the study of nuclear physics.

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