de Broglie wavelength of electron in second orbit of `Li^(2+)` ion will be equal to de Broglie's wavelength of electron in :

Options

A. `n=3` of H-atom
B. `n=4` of `C^(5+)` ion
C. `n=6` of `Be^(3+)` ion
D. `n=3` of `He^(+)` ion

Answer

B

Explanation

`lambda prop n/Z :. N_(1)/Z_(1)=n_(2)/Z_(2) "or" 2/3=4/6 (n=4 "of" C^(5+) "ion")`