Question

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 occurring electrons is removed. The radius of the orbit is 4.68 × 10-11 m. Determine the magnitude of the electron's centripetal acceleration.

Answer 1

The magnitude of the electron's centripetal acceleration can be defined from the Newton's second law:

$$a = \dfrac{F}{m_e}$$

where $F$ is the force acting on the electron from the protons and $m_e = 9.1\times 10^{-31} kg$ is the mass of electron.

The force $F$ can be found using the Coulomb's law:

$$F = k\dfrac{e\cdot 2p}{r^2}$$

where $e =p= 1.6\times 10^{-19}C$ is the charge of the electron and proton, $k = 9\times 10^{9} N\cdot m^2/C^2$ is the Coulomb's constant and $r = 4.68\times 10^{-11}m$ is the radius of the orbit.

Combining it together, obtain:

$$a = k\dfrac{2pe}{m_er^2} = 9\times 10^9\dfrac{2\cdot 1.6^2\times 10^{-38}}{9.1\times 10^{-31}\cdot 4.68^2\cdot 10^{-22}} \approx 2.2\times 10^{23} m/s^2$$

Answer. 2.2*10^23.