Question

What is the total energy of electron in an orbit n for Li2+ if radius of orbit n is 1.587 Å in Li+2 ion?
–122.4 eV



–1.51 eV



–13.6 eV



–30.6 eV

Answer 1

we have to find the total energy of electron in an orbit n for Li²⁺ ion if radius of orbit n is 1.587 A° in Li²⁺ ion.

solution : using Bohr's atomic radius formula,

r = 0.529 × n²/Z A°

here, atomic no = Z = 3 , radius of orbit, r = 1.587 A°

so, 1.587 = 0.529 × n²/3

⇒3 × 3 = n²

⇒n = 3

now, total energy of electron in nth orbit (i.e., 3rd) orbit of Lithium ion (Li²⁺) , T.E = -13.6 × Z²/n² eV

here, Z = 3 , n = 3

so total energy , T.E = -13.6 × (3)²/(3)² = -13.6 eV

Therefore the total energy of electron in 3rd orbit of Lithium ion is -13.6 eV.

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Answer 2

Explanation:

$\huge \tt - 13.6 \: \: ev$