Write a partial decay series for Po-214 undergoing the following sequential decays: á, â, á, â. Use e for an electron. Use the format +1 or -1 for charge.
a)a decay
b)â decay
c)a decay
d)â decay
oops b and c should be b decay and in the directions its a,b,a,b
I cant read your post.
To write a partial decay series for Po-214, we need to know the initial element and the type of decay for each step. Here is how you can determine the type of decay for each step:
1. α decay (α): In α decay, the atomic number decreases by 2 and the mass number decreases by 4. This means that an α particle, which consists of 2 protons and 2 neutrons, is emitted from the nucleus.
2. β decay (β): In β decay, a neutron is converted into a proton or a proton is converted into a neutron, and an electron (e-) or a positron (e+) is emitted.
Based on the given sequence of decays (á, â, á, â), we can determine the partial decay series as follows:
a) α decay (α): The initial element is Po-214. In α decay, the atomic number decreases by 2, so the resulting element is Rn-210 (Polonium-210).
b) β decay (β): The resulting element Rn-210 undergoes β decay. Since we have â (á followed by â), it indicates a double β decay. This means that two β emissions occur in this step. Let's denote β1 and β2.
c) α decay (α): The resulting element after the double β decay is an isotope of Pb (Lead), specifically Pb-210. In α decay, the atomic number decreases by 2, so the resulting element is Bi-206 (Bismuth-206).
d) β decay (β): The resulting element Bi-206 undergoes β decay. Again, we have â (á followed by â), indicating a double β decay. Let's denote β3 and β4.
Putting it all together, the partial decay series for Po-214 undergoing the given sequential decays is:
Po-214 → α (α1) → Rn-210 + β1 + β2 → α (α2) → Pb-210 + β3 + β4 → Bi-206
Please note that this is only a partial decay series since we only listed the decays indicated by the given sequence. The complete decay series would include further nuclear decays until reaching a stable or nearly stable element.