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Answer 1 · 2017-02-04T18:41:35

You can do it like this:

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

#:.##sf(nlambda=2pir" "color(red)((1)))#

From de Broglie we get:

#sf(lambda=h/(mv)" "color(red)((2)))#

Substituting for #sf(lambda)# from #sf(color(red)((1))# into #color(red)((2))rArr#

#sf((nh)/(mv)=2pir)#

The angular momentum is given by #sf(mvr)#.

#:.##sf(mvr=(nh)/(2pi)#

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