What particle (a particle, electron, or positron) is emitted in t...

Question

What particle (a particle, electron, or positron) is emitted in the following radioactive decays? (a) 27 14SiS 27 13 Al; (b) 238 92U S 234 90Th; (c) 74 33As S 7434Se.

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to complete the decay equation for the given unstable new client by indicating the type of release subatomic particles. We were given three different decay equation. And the options given listed the different type of release subatomic particles for each radioactive decay equations. The way we want to tackle this problem is to realize that for each radioactive reaction, the total charge and the nucleon number should be conserved. So let A Z X be the unknown emitted particle for each decay equation. So we have a Z X and for radioactive equation one, we have 50 25 M which will decay into 50 24 C R plus A Z X looking at the A, we will have a on the left side which is going to be 50 minus A on the right side which is going to be 50 as well. So then we know our A four A Z X released subatomic particles will equals to zero. Looking at the Z, we will have 25 on the left side and 24 on the right side. So we know that the A, the Z for our A Z X subatomic particle is going to be equal to one or plus one. So in this case, we know that X will have a positive charge of plus one electron. Therefore, X will be a positron. So then the complete radioactive decay, we will have 50 25 M decays into 50 C R plus a positron which is zero and plus one E just like. So, so the answer for the first part is going to be zero plus one E positron moving on to the second part. The second part, we will have 210 84 P O which will decay to 206 82 PB plus A Z X. Looking at the a balance, we have 210 on the left side and 206 on the right side. So our A for A Z X will be four and then the Z on the left side, we have 84 on the right side, we have 82. So the Z for A Z X is going to be two. So then we will have a four and two particle which in this case, we know that that will be a double positive charge plus two electron with a mass of four U. Therefore, X will actually be an alpha particle which will be helium. So in this case, X will have a double positive plus two charge and a mass of four U. So that will be helium. So the complete um decay equation will then be equal to 210 84 P O decays into 206 82 B B plus four, two helium just like. So, so helium alpha particle lastly moving on to the last part or the third part, we have the radioactive equation of 1 21 50 S N decays into 1 21 uh 51 as B plus A Z X. And looking at the A balance, we have 1 21 on the left side and 1 21 on the right side, which will be equals to zero for the E Z X. And then looking at the C Z, we have 50 on the left side minus 51 on the right side, which will come out to be minus one. So then we know that X will have a negative charge minus one electron. Therefore, X will be a, an electron. So to complete the equation, radioactive decay will be equal to 1 21 50 as N equals to uh or deca it to 1 21 51 S B plus zero minus one electron just like. So, so there we have it. The first uh subatomic particle is going to be a zero plus one positron or zero plus one electron. And then the second one is going to be 42 helium alpha particle, four plus two helium alpha particle. And then the last part is going to B or the third part is going to be an electron zero minus zero, negative one electron. And all of this will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.

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What particle (a particle, electron, or positron) is emitted in the following radioactive decays? (a) 27 14SiS 27 13 Al; (b) 238 92U S 234 90Th; (c) 74 33As S 7434Se.

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