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 `H^(+)` ion

Answer

2

Explanation

`lambda=(h)/(mv)`Now`" "v alpha ((z)/(n))" "so" "lambdaalpha((n)/(z))`
for second orbit of `L^(2+)" "lambda alpha((2)/(3))" "`for`" "n=4` of `C^(5+) `ion, `lambda alpha((4)/(6))=(2)/(3)` Hence the result.