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 following formula:
a = v² / r
where:
a is the centripetal acceleration,
v is the orbital velocity of the electron, and
r is the radius of the orbit.
To find v, we need to use the equation that relates centripetal force, electric force, and the orbital velocity:
F_c = F_e
mv² / r = k * (e * Ze) / r²
where:
F_c is the centripetal force,
F_e is the electric force,
m is the mass of the electron,
v is the orbital velocity of the electron,
k is the Coulomb's constant (8.99 x 10^9 N m²/C²),
e is the charge of an electron (-1.6 x 10^-19 C),
Z is the atomic number of the nucleus. In this case, Z = 2 since there are two protons.
Rearranging the equation to solve for v, we have:
v = √(k * (e * Ze) / (m * r))
Now, we can substitute the given values to calculate v. The mass of the electron (m) is 9.11 x 10^-31 kg, the charge of an electron (e) is -1.6 x 10^-19 C, and the radius of the orbit (r) is 2.65 x 10^-11 m.
Calculating v using the given values, we have:
v = √((8.99 x 10^9 N m²/C²) * (-1.6 x 10^-19 C) * 2 / (9.11 x 10^-31 kg * 2.65 x 10^-11 m))
v ≈ 2.19 x 10^6 m/s
Now that we have the velocity (v), we can substitute it into the formula for centripetal acceleration:
a = v² / r
a = (2.19 x 10^6 m/s)² / (2.65 x 10^-11 m)
a ≈ 1.80 x 10^22 m/s²
Therefore, the magnitude of the electron's centripetal acceleration is approximately 1.80 x 10^22 m/s².