Question

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary proton.What is the speed of the electron in its orbit?

Answer 1

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by  2.068 x 10^6 m / s.