Question

calculate the ionisation energy of Li2+.

Answer 1

Answer: The ionization energy of $Li^{2+}$ will be 122.4 eV.

Explanation:

Ionization energy is defined as the energy that is released when a valence electron is ejected out from an isolated gaseous atom.

The formula used to calculate ionization energy follows:

$E_n=\frac{-13.6\times Z^2}{n^2}eV$

where,

Z = atomic number = 3 (for lithium)

n = Energy level

Electronic configuration for $Li^{2+}$ ion will be: $1s^1$

The transition of ionization energy is from $1\rightarrow \infty$

Putting values in above equation, we get:

$E_{1\rightarrow \infty}=E_{\infty}-E_1$

$E_{1\rightarrow \infty}=(\frac{-13.6\times 3^2}{(\infty)^2})-(\frac{-13.6\times 3^2}{1^2})\\\\E_{1\rightarrow \infty}=(0+122.4)eV=122.4eV$

Hence, the ionization energy of $Li^{2+}$ will be 122.4 eV.

Answer 2

the ionisation energy of an H atom is 13.6eV.what is the ionisation energy for li2+ ion?

where Z is atomic number and n is the shell number.

so IE li^2+=13.6×9=122.eV.