Question:
If the energy required to remove an electron from the ground state of \(He^+\) is \(x\) J, the energy (in J) required to remove an electron from the ground state of \(Li^{2+}\) is:
If the energy required to remove an electron from the ground state of \(He^+\) is \(x\) J, the energy (in J) required to remove an electron from the ground state of \(Li^{2+}\) is:
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For any hydrogen-like ion, the ionization energy is directly proportional to the square of the atomic number (\(Z^2\)).
Updated On: Jun 9, 2026
- \(\frac{3}{2}x\)
- \(\frac{2}{3}x\)
- \(\frac{9}{4}x\)
- \(\frac{4}{9}x\)
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The Correct Option is C
Solution and Explanation
Concept:
The energy required to remove an electron from the ground state (ionisation energy) of a hydrogen-like species is given by the formula:
\[
E_n \propto Z^2
\]
where \(Z\) is the atomic number of the species[cite: 1796].
Step 1: Identify the atomic numbers (\(Z\)).
For \(He^+\) (Helium ion): \[ Z_1 = 2 \] For \(Li^{2+}\) (Lithium ion): \[ Z_2 = 3 \]
Step 2: Set up the ratio of energies.
Let \(E_1\) be the energy for \(He^+\) (\(= x\)) and \(E_2\) be the energy for \(Li^{2+}\). \[ \frac{E_2}{E_1} = \frac{(Z_2)^2}{(Z_1)^2} \]
Step 3: Calculate \(E_2\).
\[ \frac{E_2}{x} = \frac{(3)^2}{(2)^2} \] \[ \frac{E_2}{x} = \frac{9}{4} \] \[ E_2 = \frac{9}{4}x \] \[ \boxed{\frac{9}{4}x} \]
Step 1: Identify the atomic numbers (\(Z\)).
For \(He^+\) (Helium ion): \[ Z_1 = 2 \] For \(Li^{2+}\) (Lithium ion): \[ Z_2 = 3 \]
Step 2: Set up the ratio of energies.
Let \(E_1\) be the energy for \(He^+\) (\(= x\)) and \(E_2\) be the energy for \(Li^{2+}\). \[ \frac{E_2}{E_1} = \frac{(Z_2)^2}{(Z_1)^2} \]
Step 3: Calculate \(E_2\).
\[ \frac{E_2}{x} = \frac{(3)^2}{(2)^2} \] \[ \frac{E_2}{x} = \frac{9}{4} \] \[ E_2 = \frac{9}{4}x \] \[ \boxed{\frac{9}{4}x} \]
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