Question

How do you find the ionization energy? I am drawing a blank, and I need to calculate it for beryllium.

Answer #1

ionisation eneregy is the energy required to remove the last electron from it's orbit.

so for a atom if electron is removed from n^{th} state
than ionisation energy is given by (- Energy of electron in
n^{th} state)

for berylium (Be) electronic config. is 1s^{2}
2s^{2} , so last electron is in 2^{nd} state.

so ionisation energy for berylium is = (-E_{2})

this can be caluclated using Bohr's theory. acc. to bohr energy
of an electron in n^{th} state = -13.6 Z^{2} /
n^{2} ev

so for berylium Z = 4 and n = 2 so

ionisation energy = - E_{2} = -(-13.6 * 4*4 / 2*2)

= 54.4 electron volt

= 54.4 * 1.6 * 10^{-19} joule

**ionisation energy = 87.4 * 10 ^{-19} joule = 54.4
electron volt**

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