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$\mathrm{A}^{232} \mathrm{Th}$ (thorium) nucleus at rest decays to a $^{228} \mathrm{Ra}$ (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is $6.54 \times 10^{-13} \mathrm{J}$ . An alpha particle has 1.76$\%$ of the mass of a 28 Ra nucleus. Calculate the kinetic energy of (a) the recoiling 28 Ra nucleus and $(b)$ the alpha particle.

$K_{B}=0.11 \times 10^{-13} \mathrm{J}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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So it's exercise. I have a theory of nucleus that is initially at rest. And this nucleus the case in the radium nucleus by emitting a Alfa particle. Okay. And we know that the kinetic energy off the off the radium nucleus, plus the kinetic energy off the alpha particle, is equal to 6.54 time sent to the minus 13 jewels. We also know the mass off the alpha particle. And we also know the mass off the radio. Luckless. So in question A. We want to find what is going to be the kinetic energy off the Raiders Nuclear radium nucleus. Sorry. Okay, so we have that there is no external force acting during the decay off the theory, a nucleus. So we must have a conservation off energy in the before scenario. And when the Sorry, the conservation off Linder moment. When the before scenario with the in a moment, um, off the after scenario. So we have that p zero only zero Izzy before scenario has to be equal to p after. Okay, which I'll call PF. And this means that zero since since the velocity off the thorium nuclear is equal to zero so we have that zero has to be equal to the mass off the radium nucleus times the velocity off the radium nucleus. This'll minus the velocity off the alpha particle times the mass of the alpha particles. So we have a relation between the velocity off the radio nucleus and the philosophy of the alpha particle. So let me write it. Here we have that we Alfa is going to be equal to the mass off the radium nucleus over the mass, off the other particle times the philosophy off the radium. Luckless. Okay, since we want to find the kinetic energy, we can write velocity in terms off the kind of energy if way Remember that the definition of kinetic energy so is equal to the massive particle times its velocity over to sorry velocity squared. So we have that V is equal to to the square root off two times the kinetic energy over the mass of the particle. So we have that. Um, yeah. Um so we have that the square root off two k alfa over an Alfa has to be equal to the square root off to K radium over the mass off the radium times the mess off the radium over the mass off the awful particle. So we can, uh, So we can square both sides of the equation so we can get rid off the square root and we have a relation and find the relation between K Alfa and K R A. So we have that. Okay, Alfa is equal to We'll also be able to get rid of this. So this is equal to key. So sorry, K r a, uh times am Alfa over em are a Times M are a over m Alfa squared so you can get rid off the squared by dividing a male fell by m Alfa m m A by m a. Sorry, uh, we only get rid off the squared so we get rid off this on, we're left with M R a over M Alfa. Okay, so we have this relation and we know that. Okay, Alfa plus K r. A has to be equal to 6.54 times 10 to the minus 13 jewels. So we just re stewed the expression we found for K Alfa in terms off K. R s. So we have that. Okay, Are eight times m r a over m alfa plus skate are a I am alpha. Okay, so we can rewrite this by putting a k r a in evidence, and we can perform this. Uh, sorry. There is no over m Alfa here. Sorry. So we can perform the, uh, this addition. So m r a over, um, Alpha plus one and this equal to six point 54 times, then to the miners. 13 jewels. So we only super suit the values we know. So we know m r a and m Alfa, and we can isolate k r A. So we have k r a has to be equal to 6.54 times sent to the minus 13 over now, uh, one plus the mass off the radium nuclear. So 6.65. Sorry. Um, 3.77 times, 10 to the minus 25 kg over the best of the alpha particle, which is 6.65 time. Stand to the minus 27. Okay. And performing this American population, we find the kinetic energy off the radium Nicholas to be equal to 0.11 times 10 to the minus 30 Angels. Okay, Now questions be asked that she calculate what is going to be the Connecticut energy off the awful particle. So we can just be stood what we found in question A. And we have that K Alfa is equal to the total kinetic energy minus key R k r a and have that This is equal to 6.54 times 10 to the minus 13 minus 0.11 times 10 to the minus 13 and you find the kinetic energy off the off. A particle is going to be equal. Thio six point 42 times, 10 to the minus 13 jewels. So some something like this and this is the final answer off the exercise.

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