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

1.	Calculate the de Broglie wavelength of the electron in third orbit of hydrogen atom
Answer» Solution :`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`

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