Calculate the de Broglie wavelength of the electron in third orbit of hydrogen atom

Explanation

`r_(n) = r_(0) xx n^(2)`
`:. r_(3) = 3^(2) xx 0.529 Å = 4.761 Å`
By Bohr's postulate of angular momentum
`mv r = (n h)/(2pi)`
`:. mv = (3h)/(2pi r)`
`lamda= (h)/(mv) = (h)/(3h//2pir) = (2pi r)/(3) = (2(3.143) (4.761 Å))/(3) = 9.975 Å`