Question a0dfc

Feb 4, 2017

You can do it like this:

Explanation:

If we consider the electron as a standing wave then an integral number of wavelengths must fit into one circular orbit:

$\therefore$$\textsf{n \lambda = 2 \pi r \text{ } \textcolor{red}{\left(1\right)}}$

From de Broglie we get:

$\textsf{\lambda = \frac{h}{m v} \text{ } \textcolor{red}{\left(2\right)}}$

Substituting for $\textsf{\lambda}$ from sf(color(red)((1)) into $\textcolor{red}{\left(2\right)} \Rightarrow$

$\textsf{\frac{n h}{m v} = 2 \pi r}$

The angular momentum is given by $\textsf{m v r}$.

$\therefore$sf(mvr=(nh)/(2pi)#

This shows the angular momentum to be quantised, where n is the principal quantum number.