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 two naturally occurring electrons is removed. The radius of the orbit is 2.65 10-11 m. Determine the magnitude of the electron's centripetal acceleration.
To determine the magnitude of the electron's centripetal acceleration, we can use the formula:
ac = v^2 / r
Where ac is the centripetal acceleration, v is the velocity of the electron, and r is the radius of the orbit.
Since we know the radius of the orbit, we can calculate the velocity of the electron using the formula for the centripetal force:
Fc = (mv^2) / r
Where Fc is the force acting on the electron, m is the mass of the electron, and r is the radius of the orbit.
In this case, the force acting on the electron is the electrostatic force of attraction between the electron and the two protons in the nucleus. This force is given by:
Fe = k * (|q1*q2|) / r^2
Where Fe is the electrostatic force, k is the electrostatic constant (9 × 10^9 N⋅m^2/C^2), q1 and q2 are the charges of the protons (+2e each), and r is the radius of the orbit.
Since the centripetal force is equal to the electrostatic force, we can equate the two equations:
(mv^2) / r = k * (|q1*q2|) / r^2
Rearranging the equation:
v^2 = (k * |q1*q2| * r) / m
Now we can substitute the known values:
m = mass of the electron = 9.11 × 10^-31 kg
r = radius of the orbit = 2.65 × 10^-11 m
q1 = q2 = charge of a proton = +1.6 × 10^-19 C
k = electrostatic constant = 9 × 10^9 N⋅m^2/C^2
Plugging in the values:
v^2 = (9 × 10^9 N⋅m^2/C^2 * (1.6 × 10^-19 C)^2 * 2.65 × 10^-11 m) / 9.11 × 10^-31 kg
Simplifying:
v^2 ≈ 2.6106 × 10^19 m^2/s^2
Finally, we can use this velocity to calculate the centripetal acceleration:
ac = v^2 / r = (2.6106 × 10^19 m^2/s^2) / (2.65 × 10^-11 m)
Evaluating this expression, the magnitude of the electron's centripetal acceleration is approximately 9.846 x 10^28 m/s^2.