Atomic StructureEasy
Question
The ionization energy of He+ ion is x eV. The ionization energy of Be3+ ion should be
Options
A.4x eV
B.2x eV
C.$\frac{x}{4}$eV
D.$\frac{x}{2}$eV
Solution
$I.E. \propto z^{2}$
$\therefore\frac{(I.E.)_{Be^{3 +}}}{(I.E.)_{He^{+}}} = \left( \frac{z_{Be^{2 +}}}{z_{He^{+}}} \right) \Rightarrow \frac{(I.E.)_{Be^{3 +}}}{x} = \left( \frac{9}{2} \right)^{2} $$$\therefore(I.E.)_{Be^{3 +}} = 4x\text{eV}$$
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