The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one node. Upon absorbing light , the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom.
Energy of the state `S_(1)` in units of the hydrogen atom ground state energy is

Options

A. `0.75`
B. `1.50`
C. `2.25`
D. `4.50`

Answer

C

Explanation

`S_(1)` state sphere symmeterical means `l = 0`
Number oif nodel point `= 1, (n - l - l )= 1 (n - 0 - l)= 1,:. N = 2`
Hence `2s` orbit corresponding to `S_(1)` state
`S_(2)`state has only on radial and E = ground state of H atom
`E_(H_2) n = - (R )/(l^(2))`(for ground state )
`E is S_(2)` state`Li^(2+) = (R )/(12) ("but this is equal" - (Z^(2))/(n^(2)) xx R)- R = - R,(Z^(2))/(n_(2)), "since" Z = 3 for Li^(2+) "ion" :. n = 3`
Since it has only one radial node
`(n - l - 1, rArr (3 - l - 1):. l = 1`
Therefore `S_(2)` state corresponding in `3p(n = 3, l = 1)`
`E of S_(1)`state = `(RZ^(2))/(n^(2)) = R xx (3^(2))/(22) = (9)/(4)R = 2.25 R = 2.25 xx E_(H_2) n ` is ground state
Since `S_(2)` state corresponding to `l = 1 (2p)`
Angular momentum `= sqrt(2(l = 1)) (h)/(2pi) = sqrt(2) (h)/(2pi)`