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As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly
ionized Li atom 12 = 3.t is

a
1.51
b
13.6
c
40.8
d
122.4

detailed solution

Correct option is D

E=−Z2×13.6eV=−9×13.6eV=−122.4eVSo  ionisation  energy=+122.4 eV

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