Calculating Recoil Velocity in Po^214 Radioactive Decay

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SUMMARY

The discussion focuses on calculating the recoil velocity of the Po^214 nucleus after it emits an alpha particle with a kinetic energy of 1.16×10−12 J. The initial approach utilized the kinetic energy formula, leading to a recoil velocity of 1.92×10^7 m/s. The confusion arose regarding the mass of the daughter nucleus, which is 3.50×10^-25 kg, necessary for applying conservation of momentum to find the final velocity of the nucleus. The solution confirms that the mass of the daughter nucleus is essential for accurate calculations in radioactive decay scenarios.

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Homework Statement



The nucleus of Po^214 decays radioactively by emitting an alpha particle (mass 6.65×10−27 kg) with kinetic energy 1.16×10−12 J, as measured in the laboratory reference frame.

a) Assuming that the Po was initially at rest in this frame, find the magnitude of the recoil velocity of the nucleus that remains after the decay.


Homework Equations





The Attempt at a Solution



My approach to the problem was using the fact that K1=K2

I then solved for Vf by doing the following

k1=1/2mvf2

squareroot(2k1/m)=vf

squareroot[2(1.23*10^-12j)/(6.65*10^-27kg)]= 1.92*10^7

What I am confused about is my books solution manual does something else as well they use the following formula after they acquire the value above: mfvf+mnvn=0

They solve for vn and get

vn=[(6.65*10^-27)/(3.50*10^-25kg)]*1.92*10^7=3.65*10^5


I don't understand how they get 3.50*10^-25, and why its necessary to solve for vn.

Thank you
 
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it would seem that 3.50*10-25kg is the mass of the daughter nucleus, which is then nucleus left behind when the alpha particle is emitted. Did they not give you this information? Do you have any way of determining the mass of the daughter nucleus?
 

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