Alpha Particle Approaching a Gold Nucleus

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SUMMARY

The discussion centers on calculating the closest approach distance between an alpha particle and a gold nucleus, utilizing Coulomb's law and principles of conservation of momentum and energy. The alpha particle, with a velocity of 3.5 x 106 m/s, interacts with a gold nucleus (atomic number 79, mass number 197). The calculations involve determining the momentum (p = mv) and kinetic energy (KE = 1/2mv2) of the alpha particle, and the need to account for both conservation of momentum and energy to accurately find the distance of closest approach.

PREREQUISITES
  • Coulomb's Law
  • Conservation of Momentum
  • Conservation of Energy
  • Kinetic Energy Calculations
NEXT STEPS
  • Study the implications of Coulomb's Law in nuclear physics.
  • Learn about the conservation of momentum in particle collisions.
  • Explore energy potential equations in electrostatic interactions.
  • Investigate the behavior of alpha particles in nuclear reactions.
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This discussion is beneficial for physics students, educators, and anyone interested in nuclear physics, particularly those studying particle interactions and energy conservation principles.

FelaKuti
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Homework Statement


An alpha-particle with velocity 3.5 x 106 strikes a block of gold, atomic number 79 and mass number 197. Find the distance of the closest possible approach between the alpha particle and a gold nucelus, assuming Coulomb's law holds over such distances. Assume mp = mn = 1.67 x 10-27

Homework Equations


p = mv
KE = 1/2mv2
EPE = kq1q2/r2

The Attempt at a Solution



Considering conversation of momentum and conservation of energy I worked out p = mv = (4 x 1.67 x 10-27)(3.5 x 106) = 2.3 x 10-20.

The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = mauv' + malpha v'
v' = p/mau + mualpha
v' = 2.3 x 10-20 / 201 x 1.67 x 10-27
v' = 6.4 x 104

So the change in KE in KEv minus KEv'.

At this point I believe I've gone wrong as I've calculated the change in KE to be negligible yet the online question pp I'm using tells me to consider conversation of momentum (here). As I understand the remainder of the question is simply equate change KE to change in EPE, rearrange to find r and plug in values.
 
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The gold nucleus is to be assumed as a field/potential generator and hold firm in its initial place. In this case, only conservation of energy is needed. Start from infinite distance where the alpha particle is moving with the given velocity in zero potential. Then the total energy in the closest distance will give you the answer.
 
blue_leaf77 said:
The gold nucleus is to be assumed as a field/potential generator and hold firm in its initial place.
I don't believe that's true in this instance. If you check the question and hints at the website you'll find that the use of conservation of momentum is recommended in addition to conservation of energy.

@FelaKuti , check your calculation for v'. Your method's okay but the numerical result looks off; maybe a finger problem on the calculator.
 
I see. I only read the first hint where no mention of momentum conservation is made. So yes the momentum conservation is needed and the gold nucleus is also moving.
 

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