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Ionisation Energy

Ionisation po...

Question

Ionisation potential (IP) of the elements/ion with atomic number Z is given by ${(I P)}_{Z} = 13.6 (\frac{Z^{x}}{n^{2}}) e V$ . The value of x is 2. If true enter 1, else enter 0.

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Solution

For He, $Z^{}$ =1.70
$∴ {(I P)}_{H e} = 13.6 \times {(1.70)}^{2} = 39.30 e V$ (for comparison only)
$∴ {(I P)}_{H e} = 13.6 \times {(1.70)}^{2} = 39.30 e V F o r {H e}^{+}, Z^{} = Z = 2 {I P}_{H e +} = 13.6 \times {(2)}^{2} = 54.40 e V T h u s, {I P}_{H e}^{+} > {(I P)}_{H e}$

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