Quantum Numbers
Energy difference (ΔE) between n=2 and n=4 levels of hydrogen atom is approximately (in eV):
Select the correct option:
Solution
2.55 eV
- Bohr's Energy Formula: For Hydrogen, En=−n213.6 eV.
- Identify States:
- n1=2⟹E2=−413.6=−3.40 eV.
- n2=4⟹E4=−1613.6=−0.85 eV.
- Calculate Difference:
- ΔE=∣E4−E2∣
- ΔE=∣−0.85−(−3.40)∣=2.55 eV.
- Conclusion: The energy required for this transition (or emitted during decay) is 2.55 eV.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- quantum numbers
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2.55 eV
- Bohr's Energy Formula: For Hydrogen, En=−n213.6 eV.
- Identify States:
- n1=2⟹E2=−413.6=−3.40 eV.
- n2=4⟹E4=−1613.6=−0.85 eV.
- Calculate Difference:
- ΔE=∣E4−E2∣
- ΔE=∣−0.85−(−3.40)∣=2.55 eV.
- Conclusion: The energy required for this transition (or emitted during decay) is 2.55 eV.
This medium difficulty chemistry question is from the chapter atomic structure, covering the topic of quantum numbers. It appeared in the 2025 exam.
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