Minimum Distance of Alpha Particle to Radium Core?

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SUMMARY

The discussion centers on calculating the minimum distance an alpha particle (4α2) with a kinetic energy of 4.77 MeV can approach a radium core (226Ra88). The user applied energy conservation principles, leading to the formula for the distance of closest approach, r2=1/[4*∏*ε*Ec/(Z1*Z2*e2)]. However, the calculated result was twice the value found in a reference book, prompting a request for clarification on the discrepancy. The key to resolving this lies in understanding the conditions under which potential energy equals kinetic energy.

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  • Understanding of energy conservation in nuclear physics
  • Familiarity with Coulomb's law and electrostatic potential energy
  • Knowledge of alpha particle properties and behavior
  • Basic calculus for differentiation and solving equations
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  • Review Coulomb's law and its application in nuclear physics
  • Study the concept of potential energy and kinetic energy equivalence
  • Learn about alpha particle scattering and its mathematical models
  • Explore advanced topics in nuclear interactions and energy conservation
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This discussion is beneficial for physics students, nuclear physicists, and educators seeking to deepen their understanding of alpha particle interactions and energy conservation principles in nuclear physics.

anachin6000
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At what minimum distance can a particle 4α2, with kinetic energi Ec=4.77MeV, approach to a radium core 226Ra88.

This is how I solved it:
From energy conservation: ΔEc+ΔEp=0

Ec=[(Z1*Z2*e2)/(4*∏*ε)]*(1/r2-1/r1)

r2=1/[1/r1+4*∏*ε*Ec/(Z1*Z2*e2)] (1)

dr2/dr1=0 ⇔ r1→∞ (2)

From (1) and (2) ⇔ r2=1/[4*∏*ε*Ec/(Z1*Z2*e2)]

Now the problem is that in the book the correct answer is half of mine. Can someone help me to understand why?
 
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