Question

If the kinetic energy of of electron moving in fourth orbit of hydrogen is €, then the total energy in first  orbit of li 2+ is

Answer 1

For hydrogen like atom.

E_n \(\propto\) -\(\frac{Ƶ^2}{n^2}\)

E_n = -k\(\frac{Ƶ^2}{n^2}\)

for hydrogen atom-

E₄ = -k\(\frac{(1)^2}{4^2}\) 

E₄ = \(\frac{-k}{16}\) 

given, E₄ = €

\(\because\) k = -€ x 16----(1)

For Li⁺²

E₁ = \(-\frac{k(3)^2}{1^2}\)

 = -k x 9

using equation (1) we get

E₁ = € x 16 x 9

E₁ = 144 €

Answer 2

Hint to find the same

We are required to find the energy required to excite doubly ionized lithium. We know that, 

En = –13.6 Z2 / n2

Hence, the excitation energy = ΔE = E3 – E1 = –13.6 × (3)2 [1/32 – 1/12]

= +13.6 × (9) [1–1/9] = 13.6 × (9) (8/9) = 108.8 eV.

Wavelength λ = hc / ΔE