Question

The ratio of ionisation energies of li2+ and be3+ is

Answer 1

For li2+ => Ionisation energy of any Hydrogen like elements is given by:-

E = -2.18 × 10⁻¹⁸ (Z²/n²)

For lithium Z = 3 and n = 1

then E = -19.62 × 10⁻¹⁸ Joules.

For the Be3+ => ion the Ionisation energy is 217.6 eV. Again, this is 16 times larger

than the ionization energy for the hydrogen atom.

Now you can find ratio by comparing the value that I have solved.

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Answer 2

Answer:

The ionization energy of $Li^{2+}$ and  $Be^{3+}$ ions  are 122.4eV and 217.6eV respectively.

Explanation:

We know,

The ionization energy for Hydrogen atom $= 13.6(\frac{Z^{2}}{n^{2}} )eV$

where Z is the atomic number

and,  n is principle quantum number which has value equal to 1 for  hydrogen or every hydrogen like ion.

$Li^{2+}$ and $Be^{3+}$ are hydrogen like ions as they have one electron.

The ionization energy for $Li^{2+}$ ion   $= 13.6(\frac{3^{2}}{1^{2}} )eV=122.4eV$

The ionization energy for  $Be^{3+}$ ion $= 13.6(\frac{4^{2}}{1^{2}} )eV=217.6eV$

Therefore, I.E. of $Li^{2+}$ and $Be^{3+}$ is equal to 122.4eV and 217.6eV  respectively.