emission question

**prasum** · March 24th 2011, 05:53 AM

if 13 Al 27 isotope is stable which emission can 13 Al 29 undergo

**e^(i*pi)** · March 24th 2011, 06:40 AM

For future reference you don't need to give us the atomic number since it's implied in the element name. In this case every Al atom has 13 protons - if it didn't it wouldn't be Al!

Back on topic, $^{29}\text{Al}$ needs to lose two atomic mass units (amu) without changing atomic mass number. It's a weird one since to lose 2 amu without changing atomic number implies neutron emission twice which would be $^{29}\text{Al} \rightarrow ^{28}\text{Al} + n \rightarrow ^{27}\text{Al} + n$

Yet google and wolfram say beta minus decay: $^{29}\text{Al} \rightarrow ^{29}\text{Si} + e^- + \bar{\nu_e}$ and $^{29}\text{Si}$ is stable.

I would be inclined to believe google if this is a real world problem since neutron emission usually occurs when there are more than two excess neutrons compared to the stable isotope

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