The ratio of the de Broglie wavelengths of the electron in the 2nd excited state of hydrogen atom and the 1st excited state of Li?' ion is:

Options

A. `3/2`
B. `9/2`
C. `7/2`
D. `1/2`

Answer

B

Explanation

`E prop - Z^2/n^2` ,
In the second excited state `(n=3)` of H-atom,
`E_H = (13.6)/((3)^2) =-1.51eV` In the Ist excited state `(n=2) of Li^(2+)` ion
`E_(Li)= -(13.6xx(3)^2)/((2)^2) = -30.6 eV`
`:.` Ratio of de Broglie wavelengths is
`lambda_H/lambda_(Li)=(h//sqrt(2mE_H))/(h//sqrt(2mE_(Li))) = sqrt(E_(Li)/E_H) = sqrt(30.6/1.51) =4.5 =9/2`