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

Options

A. `2pi r `
B. `4 pi r `
C. `6 pi r `
D. `16 pi r `

Answer

C

Explanation

de-Broglie wavelength
`lambda = (h)/(P) " " …. (i)`
According to Bohr's quantisation condition
`mv r_n = (nh)/(2pi) implies pr_(n) = (nh)/(2pi) " " … (ii)`
From Eqs. (i) and (ii) , we get
`lambda = (h xx 2pi r_(n))/(nh) implies lambda = (2 pi r_(n))/(n)`
For fourth orbit (n = 4)
`lambda = (2pi r_(4))/(4) " " ... (iii)`
Moreover , `r prop n^(2)`
`implies (r_(1))/(r_(4)) = ((1)^(2))/((4)^(2)) implies rho_(4) = 16rho_(1)`
Substituting the value of `r_(4)` in Eq. (iii) , we get
`lambda = (2pi [16 r_(1)])/(4) = 8 pi r_(1) = 8 pi r`