How many a particles and how many b particles are emitted in the 11-step decay of 235U into 207Pb?

Video transcript

Hello. Everyone in this video are trying to determine the number of alpha particles and beta particles emitted for the decay of 2 37 N. P. 2205 T. I. That occurs in 12 steps. So The reaction that occurs then is again to 37. And the atomic number according to appear on the table is 93 for R. N. P. Elements Going on to 205. Atomic number in according to the Table T. i. So our alpha particle here this is for two alpha. That we decrease the mass number by four and decrease atomic number by two. As for the beta particle here zero negative one E. So increase the atomic number by one. So the number of alpha emissions, let's do these calculations in red here. Again. The number of alpha emissions here. This is equal to the mass number of the parent. New Clyde minus D. Mass number of the daughter new Clyde divided by four. So now plugging in actual values, we have the mass of the number of the N. P minus the mass number. For our T. I divided by four. These are kind of just relating our given situation with this formula here. Now on our next step in this it's going to be adding an actual numerical values. So resetting are given the mass number of N. P. To be to 37 and the mass number of T. I. is 205. We're gonna take the difference of that and divide it by four. Once you put this into a calculator we see here that the image or the number of alpha emissions is then equal to eight. So for all emissions, the atomic number of the product is equal to 93 minus eight times two, which is equal to 93 minus 16, which is equal to 77. Alright, so our first part of the answer for the number of alpha missions is eight. So go ahead and highlight that to signify that that's our answer, we can now go ahead and calculate for the number of beta emissions. So again, we're going from R. N. P. Over two R. T. I. So calculate for the number of beta emissions formula here, is that the atomic number Of Our Product? So TI -77. And there was going to have to go ahead and put in this numerical value of for the atomic number of R. T. I. Which were he said to be 81, 81 -77. Once we put that into the calculator we see that the number of beta emissions that we get, It's going to be equal to four. So that right here is going to be our final and last answer for this problem. Again, we have eight total alpha emissions and four total beta missions