The radius of first Bohr orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearly

Options

A. `x/3`
B. 9x
C. `2pi x`
D. `6pi x`

Answer

D

Explanation

`r_(n) = a_(0) (n^(2))/(Z)`. But `mvr_(n) = (nh)/(2pi)`
`:. mv ((a_(0) n^(2))/(Z)) = (nh)/(2pi)`
For 3rd orbit, n = 3 and for hydrogen, Z = 1
`a_(0) = x` (Given, Bohr radius of H-atom)
Hence, `mv(x xx 3^(2))/(1) = (3h)/(2pi) or mv = (h)/(6pi x)`
`lamda = (h)/(mv) = (h)/(h//6pi x_ 6pi x`