Question

Ratio of ionisation energies of li+2 and be+3

Answer 1

Answer:

Use formula::: 13.6×Z^2 ev....by solving u can easily get...

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Answer 2

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.

Hope it helps you,

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