Understanding Nuclear Decay: The Role of Beta Emission in 27-Al Disintegration

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SUMMARY

The discussion centers on the nuclear decay of 27-Al, specifically addressing its transformation into 29-Al through beta emission. The consensus is that beta decay is the correct answer, as it results in an increase in atomic number without altering the mass number. However, there is a critical note that 27-Al is a stable isotope and does not undergo decay, raising questions about the validity of the question posed. The confusion stems from the misunderstanding of decay processes and the stability of isotopes.

PREREQUISITES
  • Understanding of nuclear decay processes, specifically beta decay.
  • Familiarity with atomic structure, including atomic number and mass number.
  • Knowledge of isotopes and their stability, particularly in relation to aluminum.
  • Basic grasp of particle physics, including the roles of alpha particles, beta particles, positrons, and protons in nuclear reactions.
NEXT STEPS
  • Research the principles of beta decay and its implications in nuclear physics.
  • Study the stability of isotopes, focusing on aluminum isotopes and their decay characteristics.
  • Explore the differences between alpha, beta, and positron emissions in nuclear reactions.
  • Investigate the role of neutrons in nuclear stability and decay processes.
USEFUL FOR

Students of nuclear physics, educators teaching atomic structure, and anyone interested in the principles of nuclear decay and isotope stability.

krateesh
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Homework Statement


27-Al disintegates to form 29-Al, by what emission this is possible?

a)alpha b)beta c)positron d)proton
answer given- beta

Homework Equations


beta results in atomic no. increase,alpha results in both mass &no. decrease,positron results in decrease in no.,proton results in decrease by 1 in both.


The Attempt at a Solution


the answer i think given is wrong as we want a paricle like neutron to make it happen (which is not in the options)

are answers and options wrong or just my basics are being tested?
 
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This is really late, but for what it's worth, your thinking is correct and the question is wrong (besides, Al-27 is the stable isotope - it does not decay).
 

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