Question:
The energy required to excite an electron from 1st to 3rd Bohr orbit in $Li^{2+}$ ion is:
The energy required to excite an electron from 1st to 3rd Bohr orbit in $Li^{2+}$ ion is:
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For hydrogen-like ions, energy levels scale with $Z^2$. Always multiply the base energy by $Z^2$.
Updated On: Jun 6, 2026
- 212.1 eV
- 136.3 eV
- 108.8 eV
- 122.4 eV
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Energy levels of hydrogen-like ions are given by $E_n = -13.6 \frac{Z^2}{n^2}$ eV.
Step 2: Meaning
For Lithium ($Li^{2+}$), the atomic number $Z = 3$. We need the energy difference between $n=1$ and $n=3$.
Step 3: Analysis
$E_1 = -13.6 \frac{3^2}{1^2} = -13.6 \times 9 = -122.4$ eV. $E_3 = -13.6 \frac{3^2}{3^2} = -13.6 \times 1 = -13.6$ eV. Energy required $\Delta E = E_3 - E_1 = -13.6 - (-122.4) = 122.4 - 13.6 = 108.8$ eV.
Step 4: Conclusion
The energy required to excite the electron is 108.8 eV.
Final Answer: (C)
Energy levels of hydrogen-like ions are given by $E_n = -13.6 \frac{Z^2}{n^2}$ eV.
Step 2: Meaning
For Lithium ($Li^{2+}$), the atomic number $Z = 3$. We need the energy difference between $n=1$ and $n=3$.
Step 3: Analysis
$E_1 = -13.6 \frac{3^2}{1^2} = -13.6 \times 9 = -122.4$ eV. $E_3 = -13.6 \frac{3^2}{3^2} = -13.6 \times 1 = -13.6$ eV. Energy required $\Delta E = E_3 - E_1 = -13.6 - (-122.4) = 122.4 - 13.6 = 108.8$ eV.
Step 4: Conclusion
The energy required to excite the electron is 108.8 eV.
Final Answer: (C)
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