A double charged lithium atom is equivalent to hydrogen whose atomic number is `3`. The wavelength of required radiation for emitted electron from first to third Bohr orbit in `Li^(++` will be (Ionization energy of hydrogen atom is `13.6 eV`)

Options

A. `18.51 Å`
B. `177.17 Å`
C. `142.25 Å`
D. `113.74 Å`

Answer

D

Explanation

`E_(n) =- 13.6 (Z^(2))/(n^(2))eV`. Required energy for said transition
`DeltaE = E_(3) - E_(1) = 13.6 Z^(2) [(1)/(1^(2)) - (1)/(3^(2))]`
`rArr DeltaE = 13.6 xx 3^(2) [(8)/(9)] = 108.8eV`
`rArr DeltaE = 108.0 xx 1.6 xx 10^(-19) J`
Now `DeltaE = (hc)/(lambda) = 108.8 xx 1.6 xx 10^(-19)`
`rArr lambda = (6.6 xx 10^(-34) xx 3xx 10^(8))/(108.8 xx 1.6 xx 10^(-19))`
`= 0.11374 xx 10^(-7)m = 113.74 Å`