JEE Advance - Chemistry (1994 - No. 5)

According to Bohr's theory

$$mvr = {{nh} \over {2\pi }}$$

In the third orbit, n = 3

$$\therefore$$ $$mvr = 3\left( {{h \over {2\pi }}} \right)$$ ...... (i)

According to de Broglie expression

$$\lambda = {h \over {mv}}$$ ...... (ii)

Substituting this in equation (i), we get

$$\left( {{h \over \lambda }} \right)r = 3\left( {{h \over {2\pi }}} \right)$$ or $$3\lambda = 2\pi r$$

Thus, the circumference of third orbit is equal to three times the wavelength of electron i.e., the number of waves made by Bohr electron in one complete revolution is three.

In general, the number of waves made by a Bohr electron in an orbit is equal to its principal quantum number.