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

1.51

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13.6

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40.8

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122.4

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Solution

The correct option is D
122.4

For hydrogen and hydrogen like atoms $E_{n} = - 13.6 \frac{(Z^{2})}{n^{2}} e V$ therefore, ground state energy of doubly ionized lithium atom (Z = 3, n = 1) will be $E_{1} = (- 13.6) \frac{(3)^{2}}{(1)^{2}} = 122.4 e V$

$∴$ Ionization energy of an electron in ground state of doubly ionized atom will be 122.4 eV

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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 (Z = 3) is

The correct option is D 122.4 For hydrogen and hydrogen like atoms En=−13.6(Z2)n2eV therefore, ground state energy of doubly ionized lithium atom (Z = 3, n = 1) will be E1=(−13.6)(3)2(1)2=122.4eV ∴ Ionization energy of an electron in ground state of doubly ionized atom will be 122.4 eV