Question

In the Bohr’s model of the hydrogen atom, the electron moves in a circular orbit of radius 5.041 × 10−11 m around the proton. Assume that the orbital angular momentum of the electron is equal to h. Calculate the orbital speed of the electron. Answer in units of m/s.

Answer 1

Answer:

The orbital speed of the electron is 2.296 x 10⁶ m/s

Explanation:

Given;

radius of the circular orbit, r = 5.041 × 10⁻¹¹ m

In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;

$V = \frac{nh}{2\pi mr}$

where;

h is Planck's constant

m is mass of electron

r is the radius of the circular orbit

n is the energy level of hydrogen in ground state

Substitute in these values and solve for V

$V = \frac{nh}{2\pi mr} = \frac{1*6.626*10^{-34}}{2\pi *9.11 *10^{-31}*5.041*10^{-11}}\\\\V = 2.296*10^6 \ m/s$

Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s