The ionisation energy of He^(+) " is " 8.72 xx 10^(-18)J " atom"^(-1). Calculate the energy of the first stationary state of Li^(2+)

The ionisation energy of He^(+) " is " 8.72 xx 10^(-18)J " atom"^(-1).

1.	The ionisation energy of He^(+) " is " 8.72 xx 10^(-18)J " atom"^(-1). Calculate the energy of the first stationary state of Li^(2+)
Answer» Solution :`E_(n) = - (2pi^(2) mZ^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))` (K = constant) I.E. of `He^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))= 4K` Hence, `4K = 8.72 xx 10^(-18)J "atom"^(-1)` (GIVEN) or `K = 2.18 xx 10^(-18) J " atom"^(-1)` For `Li^(2+), Z =3` and for 1st stationary STATE, n = 1 `:. E_(1) = -K (Z^(2))/(n^(2)) = - 2.18 xx 10^(-18) xx (3^(2))/(1^(2)) = - 19.62 xx 10^(-18) J "atom"^(-1)`