MCQ
The ionization potentials of $Li$ and $K$ are $5.4$ and $4.3\, eV$ respectively. The ionization potential of $Na$ will be .............. $\mathrm{eV}$
- A$9.7$
- B$1.1$
- ✓$4.9$
- Dcannot be calculated
✓
Answer
Correct option: C.
$4.9$
c
$Li, Na, K$ are elements of same group ie $-IA$
$Li, Na, K$ are elements of same group ie $-IA$
In a group from top to bottom $I.P.$ value decreases.
So, the $I.P$ value of $Na$ will be lessthan Lithium and greater than Potassium
$I.P.$ value $=Li\,>Na\,>K$
So, it will be between $5.4$ and $4.3$ i.e $4.9 \,eV$.
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