## a. {5}^{10} B+_{2}^{4} H e \rightarrow \frac{13}{6} C+_{1}^{1} Hb. _{6}^{13} C+_{1}^{1} H \rightarrow_{5}^{10} B+_{2}^{4} H e

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in number 39. We're told to basically find the missing piece of each reaction. We're told that um, Boron 10 is struck with an Alfa particle. So boring. 10. So, um, struck with an Alfa particle. So that's for you to basically a helium nucleus. And that produces a proton. And what, remember, a proton has a massive one and one proton. So here, I'm just gonna go across the top with my masses. So on the side of have temples for 14 over here, I have one. So I need 13 more and for on the bottom from a mess, huh? My protons. I have 567 on this side. And here I only one on this side. So I need six more. Look on the periodic table and I see six protons is carbon. So this is carbon 13 for part B. They're telling me the products. So what's produced is an Alfa particle. Oh, and this is what I'm trying to find their asking me. What else is produced of particles? Product? They're produced when carbon 13 struck by a proton. So carbon 13 struck by a proton. So I'm going across on top. I have 13. 14 on the sign here I have for my alpha particles. I need 10 more. And down here I have 67 So here I need seven and 500. Periodic table is boring. So basically, these reactions were opposites of each other.

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