To solve this problem, let's break down the information provided:

1. Uniform Circular Motion: The proton is undergoing uniform circular motion, meaning it travels in a circular path at constant speed, and the acceleration is always directed toward the center of the circle (centripetal acceleration).

2. Given Data:

   • Time $t_1 = 3.25$ ns (nanoseconds)
   • Position at $t_1 = \langle 50.0 \, \text{cm}, 50.0 \, \text{cm} \rangle$
   • Velocity at $t_1 = -(4.29 \times 10^7) \, \text{m/s} \, \hat{i}$
   • Acceleration at $t_1$ is along the $+y$-axis.
   • Time $t_2 = 58.2$ ns
   • Velocity at $t_2 = -(4.29 \times 10^7) \, \text{m/s} \, \hat{i}$
   • Acceleration at $t_2$ is along the $-x$-axis.
   • The time difference $\Delta t = t_2 - t_1 = 58.2 \, \text{ns} - 3.25 \, \text{ns} = 54.95 \, \text{ns}$, which is less than one period of revolution.

Let's proceed step by step to answer each part:

A. Period of Revolution of the Electron

1. Calculate the Fraction of the Circle Traveled:

   Since the velocity of the electron at $t_1$ and $t_2$ is in the $-\hat{i}$ direction but the acceleration changes from $+y$ to $-x$, this suggests that the electron travels a quarter (1/4) of a full circle between $t_1$ and $t_2$.

2. Determine the Period of the Circular Motion:

   The time to complete a quarter-circle is $\Delta t = 54.95 \, \text{ns}$.

   Therefore, the full period $T$ (time to complete a full circle) is: $T = 4 \times \Delta t = 4 \times 54.95 \, \text{ns} = 219.8 \, \text{ns}$

B. Center of the Circular Path

3. Finding the Center:

   At $t_1$, the position of the electron is $(50.0, 50.0)$ cm. The acceleration is directed along the $+y$-axis, which means the center of the circular path must be to the left (in the negative $x$-direction) of this point because the centripetal acceleration always points toward the center.

   At $t_2$, the acceleration is along the $-x$-axis, meaning the center is below this point (negative $y$-direction).

   Since the path forms a circle and we know the acceleration directions at two points, the center of the circular path must be at the same distance from both positions but offset by the radius.

   Given that at time $t_1$, the acceleration is along the $+y$-axis, the center must lie on the line where $x = 50.0 \, \text{cm}$. Similarly, since at time $t_2$ the acceleration is along the $-x$-axis, the center must lie on the line where $y = 50.0 \, \text{cm}$. Therefore, the center of the circle is at: $\text{Center} = (0, 0) \, \text{cm}$

C. Centripetal Acceleration of the