According to Bohr's model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionised Li atom (Z = 3) is :

40.8

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13.6

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122.4

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1.51

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Solution

The correct option is B 122.4
The doubly ionised Li atom is H-like atom
$E_{n} = \frac{- 13.6 Z^{2}}{n^{2}} e V$
The ground state energy'of doubly ionised Li atom ( $Z = 3, n = 1$ ) will be
$E_{1} = \frac{- 13.6 \times 9}{1} = - 122.4 e V$
Hence, the ionisation energy for an electron in ground state of doubly ionised Li atom 122.4 eV.

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According to Bohr's model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionised Li atom (Z = 3) is :

The correct option is B 122.4The doubly ionised Li atom is H-like atomEn=−13.6Z2n2eVThe ground state energy'of doubly ionised Li atom (Z=3,n=1) will beE1=−13.6×91=−122.4eVHence, the ionisation energy for an electron in ground state of doubly ionised Li atom 122.4 eV.