From a very large distance, a proton charge = e, mass = 1.67 × 10^-27 kg) with initial kinetic energy 3 MeV is fired head-on toward a fixed stationary uranium nucleus charge 92e, mass = 3.95 × 10^-25 kg). Calculate how close to the uranium the proton gets before it comes to rest. Assume the uranium nucleus does not move.)

Question

Quizplus · Accepted Answer

We can use the conservation of energy and Coulomb's law to solve this problem. At a very large distance from the uranium nucleus, the proton's initial kinetic energy is purely due to its motion, and it does not have any potential energy. As it approaches the uranium nucleus, it slows down due to the Coulomb attraction between the proton and uranium charges. The proton will come to a complete stop at a certain distance, at which point all of its initial kinetic energy has been converted to potential energy.

Using conservation of energy, we can equate the initial kinetic energy to the final potential energy at the point where the proton stops moving:
$\frac{1}{2}mv_{i}^2 = \frac{kq_1q_2}{r}$
where $m$ is the mass of the proton, $v_i$ is its initial speed (which we can find from its initial kinetic energy), $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges of the proton and uranium nucleus respectively, and $r$ is the distance between them.

Solving for $r$, we get:
$r = \frac{kq_1q_2}{\frac{1}{2}mv_{i}^2}$

Plugging in the values given in the problem, we get:
$r = \frac{(9 	imes 10^9 Nm^2/C^2)(1.6 	imes 10^{-19} C)(92 	imes 1.6 	imes 10^{-19} C)}{\frac{1}{2}(1.67 	imes 10^{-27} kg)(3 	imes 10^6 m/s)^2}= 4.40 	imes 10^{-14} m$

Therefore, the proton gets closest to the uranium nucleus at a distance of 4.40 × 10^-14 m, which is answer choice C.

From a Very Large Distance, a Proton Charge = E