Determining # of Beta Decays from Alpha Decay

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SUMMARY

The discussion focuses on calculating the number of beta decays resulting from a series of alpha decays in a nuclear reaction involving Uranium-238. It establishes that 8 alpha decays occur, reducing the mass from 238 to 206. Following these alpha decays, the remaining number of protons is 76, leading to the conclusion that 6 beta decays will occur to stabilize the nucleus.

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Homework Statement
Consider a nuclear process in which U-238 undergoes several Alpha and Beta decays and becomes Pb-206. How many Alpha and Beta decays occur in this process?
Relevant Equations
.
Alpha particle has two protons and neutrons so mass = 4.

4x = 238 -206.

x = 1/4(238-206) = 8 Alpha decays.

How would I determine the number of Beta decays? Since the mass doesn't change here.
 

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How many protons are left if you do all the alpha decays first?
 
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76. So there will be 6 Beta decays.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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