Question

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.20 3 106 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.529 3 10210 m and (b) the centripetal acceleration of the electron.

Answer 1

Part a)

`F_c = 8.3 * 10^(-8) N`

Part b)

`a_c = 9.15 * 10^(22) m/s^2`

Step-by-step explanation:

Part a)

While moving in circular path we know that the acceleration of particle is known as centripetal acceleration

so here we will have

`a_c = (v^2)/(R)`

now the net force on the moving electron is given as

`F_c = m(v^2)/(R)`

now plug in all values in it

`F_c = (9.1* 10^(-31))((2.20 * 10^6)^2)/(0.529 * 10^(-10))`

now we have

`F_c = 8.3 * 10^(-8) N`

Part b)

Centripetal acceleration is given as

`a_c = (F_c)/(m)`

`a_c = 9.15 * 10^(22) m/s^2`