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dimulka [17.4K]
3 years ago
8

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

r which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get
Physics
1 answer:
Klio2033 [76]3 years ago
3 0

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

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