Ionisation energy of Li (Lithium) atom in gro | vedclass

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(D) The total energy required to remove all three electrons is the sum of the energy required to remove each electron sequentially.$1$. Energy to remo

Vedclass · Accepted Answer

$156.6$

Vedclass · Answer

(D) The total energy required to remove all three electrons is the sum of the energy required to remove each electron sequentially.$1$. Energy to remove the $1^{st}$ electron (Ionisation energy of $Li$): $E_1 = 5.4 \ eV$.$2$. After removing the $1^{st}$ electron,we are left with $Li^+$. The binding energy of an electron in $Li^+$ is $75.6 \ eV$. Since there are two electrons remaining in the $Li^+$ ion,the energy required to remove both is $2 	imes 75.6 \ eV = 151.2 \ eV$.$3$. Total energy required = $E_1 + E_2 + E_3 = 5.4 \ eV + 151.2 \ eV = 156.6 \ eV$.

Ionisation energy of $Li$ (Lithium) atom in ground state is $5.4 \ eV$. Binding energy of an electron in $Li^+$ ion in ground state is $75.6 \ eV$. Energy required to remove all three electrons of Lithium $(Li)$ atom is ........... $eV$.

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