Distance of Closest Approach of Proton & Alpha Particle

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SUMMARY

The discussion centers on calculating the distance of closest approach between a proton and an alpha particle, both initially traveling at 3.9×10−3c. The user initially calculated the distance as 1.7×10−13 m using the equation k(Qp)(Qa)/r = (0.5(Mp)V2) + (0.5(Ma)V2), but the correct answer is 1.3×10−13 m. The error stems from the frame of reference used in the calculations, specifically whether the particles are considered to be at rest in the lab frame at the point of closest approach.

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  • Understanding of electric potential energy and kinetic energy.
  • Familiarity with the concept of charge and mass of subatomic particles, specifically protons and alpha particles.
  • Knowledge of the equation for distance of closest approach in nuclear physics.
  • Basic understanding of reference frames in physics.
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  • Review the derivation of the distance of closest approach formula in nuclear physics.
  • Study the implications of different reference frames on particle interactions.
  • Learn about the properties of alpha particles and their interactions with protons.
  • Explore the concept of conservation of energy in particle collisions.
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Homework Statement


An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.
A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10^−3 c when they were far away.Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.

What is the distance of closest approach between the proton and the alpha particle?

Homework Equations



potential energy and electric potential energy.

The Attempt at a Solution


following is a simplified attempt at solution

k(Q_p)(Q_a)/r=(.5(M_p)V^2)+(.5(M_a)V^2)
and solved for r and got 1.7X10^-13 m
but i keep getting it wrong could some 1 tell me where i am going wrong
the answer is 1.3X10^-13m
 
Last edited:
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seto6 said:

Homework Statement


An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.
A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10^−3 c when they were far away.Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.

What is the distance of closest approach between the proton and the alpha particle?

Homework Equations



potential energy and electric potential energy.

The Attempt at a Solution


following is a simplified attempt at solution

k(Q_p)(Q_a)/r=(.5(M_p)V^2)+(.5(M_a)V^2)
and solved for r and got 1.7X10^-13 m
but i keep getting it wrong could some 1 tell me where i am going wrong
the answer is 1.3X10^-13m
The problem is the frame of reference you are using.

At the point of closest approach, are the two particles stopped in the lab frame? Why, or why not?

AM
 

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