# 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

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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

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For hydrogen like atom.

En $\propto$ -$\frac{Ƶ^2}{n^2}$

En = -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 €

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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