Question

In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The force responsible for the electron's circular motion is the electric force of attraction between the electron and the protonIf the speed of the electron were 1.1×106m/s , what would be the corresponding orbital radius?

Answer 1

Answer:

re = 2.09 Å.

Explanation:

As the electron is moving around the stationary proton along a circular path at a constant speed, there exists a force that keeps the electron within this trajectory, which is the centripetal force.

This centripetal force, is no other than the attractive electric force between the proton and the electron, so we can write the following equality:

Fe = Fc ⇒ k*Qp*Qe / re² = me*ve²/re

⇒ re = k*Qp*Qe / m*ve², where:

k = 9*10⁹ N,  Qp =-Qe= 1.6*10⁻¹⁹ coul, me =9.1*10⁻³¹ kg, and ve= 1.1*10⁶m/s

⇒ re = 9*(1.6)²*10⁻²⁹ / 9.1*(1.1)²*10⁻¹⁹ m

⇒ re = 2.09*10⁻¹⁰ m = 2.09 Å.