🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 23 Easy Difficulty # The nucleus of$^{214} \mathrm{Po}$decays radioactively by emitting an alpha particle (mass$6.65 \times 10^{-27} \mathrm{kg}$) with kinetic energy$1.23 \times 10^{-12} \mathrm{J},$as measured in the laboratory reference frame Assuming that the Po was initially at rest in this frame, find the recoil velocity of the nuclens that remains after the decay. ### Answer ##$v=3.67 \times 10^{5} \mathrm{m} / \mathrm{s}\$

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Moment, Impulse, and Collisions

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

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### Video Transcript

in this question of polonium nuclear's decay. Radioactively by a meeting in off a particle that alpha particle has a mass off 6.65 times stand to minus 27 kg and magnetic energy off 1.23 times. Stand to minus 12 Jews. All we have to do in this question is evaluate what is the recoil velocity off the remaining nuclear's? For that, we can use the law off conservation of momentum to begin with. So the initial momentum p zero and I'm talking here about the net momentum. So let me right on em here should be equal to the final net momentum. Let me call this P M. The initial net Momenta musicals to zero because at the beginning the wall no clues. Wasat rest. Then we have zero being equals two at the end. We have the alpha particle moving in one direction, and the nuclear is moving in the other direction. So by choosing these reference frame, we can say the following. After the decay, the alpha particle moves to the right with a momentum given by the mass off that alpha particle times the velocity off that alpha particle and at the same time, the nucleus moves to the other direction with the velocity that I'll call V. Then the momentum off the nuclear's after the decay is given by minus the mass off the nucleus times the final velocity. Then we stole for V. By doing that, we get the following V, which is the final velocity of the nucleus is given by the mass off the alpha particle times the velocity off the alpha particle divided by the final mass off the nucleus. Now we have to evaluate what is the velocity off the alpha particle? For that, we can use the information about the kinetic energy. We know that the kinetic energy off the alpha particle is given by one half the mass of the alpha particle times the velocity of the alpha particle squared. Now we solve this expression for the velocity off the alpha particle. It goes as follows. So we offer squared is given by two times the kinetic energy off the alpha particle divided by the mass off the particle. So the velocity off that alpha particle is given by the square root off three times the kinetic energy divided by the mass and I know that the velocity is positive in my reference frame, since the velocity is pointing to the right and so is my reference frame. Then we can plug in this expression for the velocity off the alpha particle, the expression for the velocity off the resulting nucleus. By doing that to get the following the velocity of the nucleus after the decay V is given by the mass off the alpha particle divided by the mass off the nucleus times the square root off two times the kinetic energy off the awful particle divided by the mass off the off particle. Now we have to evaluate what is the mass off that nucleus after the decay. We know that before the decay we had under clues off 214 polonium. Then after the decay, what we get is the following. Now we have 210 polonium, plus an Alfa particle which has an atomic mass of four since it is composed by two protons and two neutrons. The reform, the mass of the resulting polonium nucleus M, is given by 210 units off atomic mass. Now we have to convert from units of atomic Mass 2 kg. This is easy to do if you remember that each unit off atomic mass is the same as 1.66 times 10 to minus 27 kg. If you remember that, then you also know that the mass is given by 210 times 1.66 times tend to minus 27 kg. Now all you have to do is plugging everything that we had discovered into the equation for the velocity. So we is given by the mass off that off a particle in kilograms, which is 6.65 time stand to minus 27 divided by the mass off the polonium nuclear after the decay, which is 210 times 1.66 times 10 to minus 27 times the square root off two times the kinetic energy off the off a particle 1.23 times 10 to minus 12 divided by the mass off that off a particle 6.65 times stem to minus 27 and then this results in a velocity of approximately three 0.67 times, 10 to the fifth meters per second. And this is the answer to this question. Of course, this velocity points to the left

Brazilian Center for Research in Physics

#### Topics

Moment, Impulse, and Collisions

##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp