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

1.51

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13.6

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50.8

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122.4

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Solution

The correct option is A 1.51
Ionization energy of electron $\propto Z^{2}$
Thus ionization energy of atom with $(Z = 3) =$ Ionization energy of H-atom $\times 3^{2} = 13.6 e V \times 3^{2} = 122.4 e V$

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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

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

The correct option is A 1.51Ionization energy of electron ∝Z2Thus ionization energy of atom with (Z=3)= Ionization energy of H-atom×32=13.6eV×32=122.4eV

The correct option is A 1.51
Ionization energy of electron $\propto Z^{2}$
Thus ionization energy of atom with $(Z = 3) =$ Ionization energy of H-atom $\times 3^{2} = 13.6 e V \times 3^{2} = 122.4 e V$