The ratio of the radius of second orbit of `Li^(2+)` to that of thired orbit of `Be^(3+)` is

Options

A. `(9)/(8)`
B. `(8)/(9)`
C. `(27)/(16)`
D. `(16)/(27)`

Answer

D

Explanation

Radius of an electron `(r_(n))` in any orbit can be calculated as follows :
` r_(n) = (0.52 xx10^(10) n^(2))/(Z) m `
where , n = number of orbit, Z = atomic number
For ` Li^(2+) - n = 2, Z = 3 `
For ` Be^(3+) - n = 3 , Z = 4 `
Therefore , `(""^(r) (Li^(2+)))/(""^(r)(Be^(3+)) )= [(n^(2))/(Z)]_((Li^(2+)) )xx [(Z)/(n^(2))]_((Be^(3+)))= [(2^(2))/(3)]xx [(4)/(3^(2))] = (16)/(27)`
Hence, option (d) is the correct anser .