Minimum Distance of Alpha Particle to Radium Core?

The minimum distance an alpha particle with kinetic energy of 4.77 MeV can approach a radium core 226Ra88 is equal to half the radius of the nucleus. This can be determined using conservation of energy and the equation for Coulomb potential energy. The book's answer may differ due to different assumptions or approximations used in the calculation.
  • #1
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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1. What is electrostatics?

Electrostatics is the branch of physics that deals with the study of stationary electric charges and the forces they exert on each other.

2. What is an electrostatics problem with energy?

An electrostatics problem with energy involves determining the amount of energy associated with a system of static electric charges or the amount of work required to move a charge within an electric field.

3. How is the energy in an electrostatics problem calculated?

The energy in an electrostatics problem is calculated using the formula U = k(q1q2)/r, where U is the electric potential energy, k is the Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

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