The de-Broglie wavelength of an electron in 4th orbit is (where, r=radius of 1st orbit)

Options

A. `2pir`
B. `4pir`
C. `8pir`
D. `16pir`

Answer

C

Explanation

de-Broglie wavelength, `lamda=(h)/(P)` . . (i)
According to Bohr's quantisation condition
`mvr_(n)=(nh)/(2pi) implies pr_(n)=(nh)/(2pi)` . . (ii)
From eqs. (i) and (ii), we get
`lamda=(hxx2pir_(n))/(nh)=(2pir_(n))/(n)`
For fourth orbit `(n=4),lamda=(2pir_(4))/(4)` . . . (iii)
Moreover, `rpropn^(2)`
`implies (r_(1))/(r_(4))=((1)^(2))/((4)^(2)) implies r_(4)=16r_(1)`
Substituting the value of `r_(4)` in Eq, (iii), we get
`lamda=(2pi[16r_(1)])/(4)=8pir_(1)=8pir`