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

Question

Ionisation potential of Li and K are 5.4 and 4.3 eV respectively. Ionisation potential of Na will be: (A) 9.7 eV (B) 4.9 eV (C) 1.1 eV (D) Cannot be calculated

Tardigrade Admin · Accepted Answer

According to Dobereinerâs triads, IP of middle element (Na) is the average of IP of Li K ∴ IP of Na=(5.4 +4.3/2) =4.9 eV

Q. Ionisation potential of $L i$ and $K$ are $5.4$ and $4.3 e V$ respectively. Ionisation potential of Na will be:

$9.7 e V$

$4.9 e V$

$1.1 e V$

Cannot be calculated

According to Dobereiner’s triads, $I P$ of middle element $(N a)$ is the average of $I P$ of $L i$ & $K$
$∴ I P$ of $N a = \frac{5.4 + 4.3}{2}$
$= 4.9 e V$

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

9.7eV

4.9eV

1.1eV