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In certain radioactive beta decay processes (more about ese in Chapter $30,$ the beta particle (an electron) leaves the omic nucleus with a speed of 99.95$\%$ the speed of light relave to to the decaying nucleus. If this nucleus is moving at 5.00$\%$ the speed of light, find the speed of the emitted electron relative to the laboratory reference frame if the electron isemitted (a) in the same direction that the nucleus is moving, (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find the kinetic energy of theelectron as measured in (i) the laboratory frame and (ii) thereference frame of the decaying nucleus.

$$6750 \times 10^{-15} \mathrm{J}$$$$740 \times 10^{-15} \mathrm{J}$$$$2511 \times 10^{-15} \mathrm{J}$$

Physics 101 Mechanics

Chapter 27

Relativity

Gravitation

Cornell University

Simon Fraser University

Hope College

Lectures

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all right. In part a of this problem. We have a nucleus that's moving as you're a 0.75 si, and an electron is admitted in the same direction that the nucleus is going in at a speed of 0.9995 c relative to the nucleus. So we want the speed of the electron in the lab frame. And, as usual, we use primed coordinates and to know the moving friend. So our velocity edition formula looks like this. You is the speed of the nucleus as your 0.75 c and be prime is the speed of the electron from the perspective of the nucleus. So if we want the it's simply 0.7 C plus 0.9995 scene over one, plus the product of these two in Part B. We're asked for the velocity of the electron if it's moving in the opposite direction. In that case, we simply replace on the prime with its negative, so this will look like V is equal to oh, technically, I wrote thes swapped, although it doesn't change the answer, the in part B, it's going to be negative. 0.9995 plus 0.75 si and I've written it and perhaps an awkward order just to match the original equation. And this now becomes a minus. And in part, C were asked about the kinetic energy. In each case, the kinetic energy is given by I'm C squared gamma minus one, and gamma is one over square of one minus. V squared over C squared. So, um, the mass of an electron is 9.11 times 10 to the negative 31 killer grounds and, um, in the frame of the nucleus, it's not going to matter which direction the electron is going in. Um, because from the perspective of the nucleus, it'll have the same speed either way. And so to find the chaotic energy of the Atran on in the rest frame of the nucleus than you would just substitute, um, the is equal to 0.9995 c in this expression. And if you want to find the kinetic energy in the lab frame now, the cannon energies, they're going to be different. And, um, if you want to know the kinetic energy when the electron is moving in the same direction is the nucleus. You would substitute into this equation the value of the U got in part A. And if you want to know the kind of energy when the electrons moving in the opposite direction, you take the value of the that you got in part being you substituted and for gamma here.

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