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 start by using the formula for centripetal acceleration:
a = v^2 / r
Where:
a is the centripetal acceleration
v is the velocity of the particle
r is the radius of the orbit
In this case, the electron is orbiting around the nucleus, so its velocity can be found using the formula for the velocity of a particle in circular motion:
v = (Z * e) / (4 * π * ε₀ * r)
Where:
Z is the charge of the nucleus (2 in this case because there are two protons)
e is the elementary charge (1.602 x 10^-19 C)
π is approximately 3.14159
ε₀ is the permittivity of free space (8.85 x 10^-12 C^2/N·m^2)
r is the radius of the orbit (2.65 x 10^-11 m)
Let's plug in the values:
v = (2 * (1.602 x 10^-19 C)) / (4 * π * (8.85 x 10^-12 C^2/N·m^2) * (2.65 x 10^-11 m))
By evaluating the expression, we can find the value of v.
Once we have the value of v, we can substitute it back into the formula for centripetal acceleration to find the magnitude:
a = (v^2) / r
By evaluating the expression, we can determine the magnitude of the electron's centripetal acceleration.