Answer:

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 (
)
with this expression we can find the closest approach distance (r)
There are 8 electrons in the third energy level of Calcium atom.
Given: Heat Qout means useful work = 2800 J
Heat Qin = 8900 J
Required; Efficiency = ?
Formula: Efficiency = Qout/Qin = x 100%
= 2800 J/8900 J = 0.3146 X 100 %
Efficiency = 31.46%
the answer could be (very basic) since options arent given
There is half the force that there was before it was split in half