1. Tardigrade
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  4. Ionisation potential of Li and K are 5.4 and 4.3 eV respectively. Ionisation potential of Na will be:

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Q. Ionisation potential of $Li$ and $K$ are $5.4$ and $4.3\, eV$ respectively. Ionisation potential of Na will be:

Classification of Elements and Periodicity in Properties

Solution:

According to Dobereiner’s triads, $IP$ of middle element $(Na)$ is the average of $IP$ of $Li$ & $K$
$\therefore IP$ of $Na=\frac{5.4 +4.3}{2}$
$=4.9\,eV$