A radioactive nucleus alpha decays to yield a sodium-24 nucleus in 14.8 hours. What was the identity of the original nucleus? Show the nuclear equation that leads you to this answer.
13 X 28 --> 11 Na 24 + 2 He 4
look at the periodic table for atomic number 13 --- that is the element you are looking for. If I remember correctly, it is aluminum. Should Al should replace X above? so it would be 13 Al 28 ---> 11 Na 24 + 2 He 4??
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Yes, you're correct! The original nucleus is indeed aluminum-28 (13 Al 28). The equation you provided, 13 Al 28 --> 11 Na 24 + 2 He 4, represents the alpha decay of aluminum-28 into sodium-24 and helium-4.
To arrive at this answer, you need to consider a few things. Firstly, an alpha decay involves the emission of an alpha particle, which consists of two protons and two neutrons. This means that the atomic number of the original nucleus decreases by 2, while the mass number decreases by 4.
In the equation, "13 X 28" represents the unknown nucleus, where 13 is the atomic number (Z) and 28 is the mass number (A). To find the identity of the original nucleus, we can look at the atomic number Z = 13 on the periodic table, which corresponds to the element aluminum (Al).
So, the correct equation is indeed 13 Al 28 --> 11 Na 24 + 2 He 4.