Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which one of the naturally occuring electrons is removed. The radius of the orbit is 2.26 × 10-11 m. Determine the magnitude of the electron's centripetal acceleration.

Electron mass= 9.11X 10^-31 kg
Electron charge= -1.60x10^-19 C
Proton charge= 1.60x10^-19 C

To determine the magnitude of the electron's centripetal acceleration, we can use the formula for centripetal acceleration:

a = v^2 / r

where "a" represents the centripetal acceleration, "v" represents the velocity of the electron, and "r" represents the radius of the orbit.

To find the velocity (v), we can use the formula for the centripetal force:

F = (mv^2) / r

Since the only force acting on the electron is the electrostatic force between the electron and the two protons, we can equate the magnitude of the electrostatic force to the magnitude of the centripetal force:

k * (|e|)(2e) / r^2 = (mv^2) / r

where "k" is the Coulomb constant, "|e|" is the magnitude of the electron charge, "2e" is the magnitude of the charge on the two protons, "m" is the mass of the electron, "v" is the velocity of the electron, and "r" is the radius of the orbit.

Now we can solve for the velocity (v):

k * (|e|)(2e) / r^2 = (mv^2) / r

Simplifying and rearranging, we get:

v^2 = (k * |e| * 2e) / (m * r)

v = √[(k * |e| * 2e) / (m * r)]

Substituting the given values:

v = √[(8.99 × 10^9 N·m^2/C^2 * 1.60 × 10^-19 C * 2 * 1.60 × 10^-19 C) / (9.11 × 10^-31 kg * 2.26 × 10^-11 m)]

v = √[(8.99 × 1.60 × 2.56) / (9.11 * 2.26)]

v = √[36.454 / 41.1586]

v = √0.884

v ≈ 0.940 × 10^5 m/s

Now that we have the velocity, we can determine the magnitude of the electron's centripetal acceleration using the formula:

a = v^2 / r

Substituting the given values:

a = (0.940 × 10^5 m/s)^2 / (2.26 × 10^-11 m)

a ≈ 418 × 10^15 m/s^2

Therefore, the magnitude of the electron's centripetal acceleration is approximately 418 × 10^15 m/s^2.