Question

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rotates in a circular orbit about this nucleus. In the n = 1 orbit the electron is 5.29 10-11 m from the nucleus and it rotates with an angular speed of 4.12 1016 rad/s. Determine the electron's centripetal acceleration in m/s2.

Answer 1

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

$a_c = \omega^2 r$

Here,

$\omega =$Angular velocity

r = Radius

Our values are given as,

$\omega = 4.12*10^{16}rad/s$

$r = 5.29*10^{-11}$

Replacing,

$a_c = (4.12*10^{16})^2( 5.29*10^{-11})$

$a_c = 8.979*10^{22}m/s^2$

Therefore the electron's centripetal acceleration is $8.979*10^{22}m/s^2$