Question

In the Bohr model of the hydrogen atom (see Section 38.5), in the lowest energy state the electron orbits the proton at a speed of 2.2 * 106 m/s in a circular orbit of radius 5.3 * 10-11 m. (a) What is the orbital period of the electron?

Answer 1

Answer:

Orbital period of electron is 1.5×10⁻¹⁶s

Explanation:

Given data

Proton speed v=2.2×10⁶m/s

Radius r=5.3×10⁻¹¹m

To find

Orbital period of electron T

Solution

We need to find the orbital period of electron the period is given by:

$T=\frac{2\pi r}{v}$

Substitute the given values

$T=\frac{2\pi (5.3*10^{-11}m)}{2.2*10^{6}m/s }\\ T=1.5*10^{-16}s$

Orbital period of electron is 1.5×10⁻¹⁶s