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 x 10^6 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.530 x 10^-10 m and (b) the centripetal acceleration of the electron.
a) Fc = m*v^2/ r
Fc = 9.10938188*10^-31 * (2.2*10^6)^2 / (0.53*10^-10)
Fc = 8.3188 *10^-8 N

b) ac = Fc / m =
8.31876*10^-8 / (9.10938188*10^-31) = 9.1321*10^22 m/s^2

Answer 1

Explanation:

It is given that,

Radius of the circular orbit, $r=0.53\times 10^{-10}\ m$  

Speed of the electron, $v=2.2\times 10^6\ m/s$

Mass of the electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron is centripetal force. Its formula is given by :

$F=\dfrac{mv^2}{r}$

$F=\dfrac{9.1\times 10^{-31}\times (2.2\times 10^6)^2}{0.53\times 10^{-10}}$  

$F=8.31\times 10^{-8}\ N$

(b) The centripetal acceleration of the electron is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(2.2\times 10^6)^2}{0.53\times 10^{-10}}$

$a=9.13\times 10^{22}\ m/s^2$

Hence, this is the required solution.