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Step-by-Step Solution
Step 1: Calculate Energy Released by Hydrogen Transition (n = 2 to n = 1)
When a hydrogen atom transitions from the second energy level (n = 2) to the first level (n = 1), the energy of the emitted photon $E_1$ is given by the energy difference between these levels in hydrogen:
$ E_1 = 13.6 \times \Bigl(\frac{1}{1^2} - \frac{1}{2^2}\Bigr)\,\text{eV} $
Simplifying:
$ E_1 = 13.6 \times \Bigl(1 - \frac{1}{4}\Bigr)\,\text{eV} = 13.6 \times \frac{3}{4}\,\text{eV} = 10.2\,\text{eV} $
Step 2: Write Expression for Ionization Energy of Doubly Ionized Lithium (Z = 3)
A doubly ionized lithium ion (Li2+) has atomic number $Z = 3$. The energy required to remove (ionize) an electron from the nth level of this ion is:
$ E_2 = \frac{13.6 \times Z^2}{n^2} = \frac{13.6 \times 3^2}{n^2}\,\text{eV} = \frac{13.6 \times 9}{n^2}\,\text{eV} = \frac{122.4}{n^2}\,\text{eV} $
Step 3: Condition for Ionization by the Photon
The photon emitted from the hydrogen atom must have at least this much energy to fully remove the electron from the Li2+ ion:
$ E_1 \ge E_2 $
Substituting the expressions:
$ 10.2 \,\text{eV} \ge \frac{122.4}{n^2}\,\text{eV} $
Step 4: Solve for the Quantum Number n
Rewriting the inequality:
$ 10.2 \ge \frac{122.4}{n^2} $
Multiply both sides by $n^2$:
$ 10.2 \,n^2 \ge 122.4 $
$ n^2 \ge \frac{122.4}{10.2} = 12 $
Therefore:
$ n \ge \sqrt{12} \approx 3.46 $
Since n must be an integer (principal quantum number):
$ n \ge 4 $
Step 5: Conclusion
The least possible excited state that satisfies $n \ge 4$ is n = 4. Hence, the answer is:
4
