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:

• A. \(\frac{3}{2}x\)
• B. \(\frac{2}{3}x\)
• C. \(\frac{9}{4}x\)
• D. \(\frac{4}{9}x\)

Hint:

For any hydrogen-like ion, the ionization energy is directly proportional to the square of the atomic number (\(Z^2\)).

Answer 1

The Correct Option is C

Step 1: Recall the energy of hydrogen-like ions.
For a single-electron species, the ionisation energy scales with the square of the atomic number, $E \propto Z^2$. The shape of the formula is the same for $He^+$ and $Li^{2+}$; only $Z$ changes.
Step 2: Find Z for the helium ion.
$He^+$ has lost one electron from helium, so its nucleus has $Z_1 = 2$.
Step 3: Find Z for the lithium ion.
$Li^{2+}$ has lost two electrons from lithium, leaving one electron around a nucleus with $Z_2 = 3$.
Step 4: Write the ratio of energies.
\[ \frac{E_2}{E_1} = \frac{Z_2^2}{Z_1^2} = \frac{3^2}{2^2} = \frac{9}{4} \]
Step 5: Insert the known value.
Since the energy for $He^+$ is $E_1 = x$, \[ E_2 = \frac{9}{4} x \]
Step 6: State the answer.
The energy needed for $Li^{2+}$ is $\tfrac{9}{4}x$, which is option 3.
\[ \boxed{\dfrac{9}{4}x} \]