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

Answer:

Part a)

$F_c = 8.3 \times 10^{-8} N$

Part b)

$a_c = 9.15 \times 10^{22} m/s^2$

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

$a_c = \frac{v^2}{R}$

now the net force on the moving electron is given as

$F_c = m\frac{v^2}{R}$

now plug in all values in it

$F_c = (9.1\times 10^{-31})\frac{(2.20 \times 10^6)^2}{0.529 \times 10^{-10}}$

now we have

$F_c = 8.3 \times 10^{-8} N$

Part b)

Centripetal acceleration is given as

$a_c = \frac{F_c}{m}$

$a_c = \frac{(8.3 \times 10^{-8} N){9.1 \times 10^{-31}}$

$a_c = 9.15 \times 10^{22} m/s^2$