Photon Emission
Hardphysics
Photon frequency for transition n=3 to n=2 in H: ν =? (E_n = -13.6/n² eV; h = 4.14×10^{-15} eV·s)
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Solution
Incorrect! Answer:
4.57×1014Hz
- Energy Transition: When an electron jumps from n=3 to n=2, it emits a photon with energy ΔE=E3−E2.
- Calculate Energies:
- E3=−13.6/32=−13.6/9≈−1.51 eV
- E2=−13.6/22=−13.6/4=−3.40 eV
- Energy Difference:
- ΔE=(−1.51)−(−3.40)=1.89 eV (or more precisely 13.6×[1/4−1/9]=13.6×5/36≈1.889 eV).
- Frequency Formula: E=hν⟹ν=E/h.
- Substitution:
- ν=1.889/(4.14×10−15)≈0.456×1015 Hz=4.56×1014 Hz.
- Result: Rounded to two decimal places, the frequency is 4.57×1014 Hz.
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About This Question
- Subject
- physics
- Chapter
- atoms and nuclei
- Topic
- photon emission
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
4.57×1014Hz
- Energy Transition: When an electron jumps from n=3 to n=2, it emits a photon with energy ΔE=E3−E2.
- Calculate Energies:
- E3=−13.6/32=−13.6/9≈−1.51 eV
- E2=−13.6/22=−13.6/4=−3.40 eV
- Energy Difference:
- ΔE=(−1.51)−(−3.40)=1.89 eV (or more precisely 13.6×[1/4−1/9]=13.6×5/36≈1.889 eV).
- Frequency Formula: E=hν⟹ν=E/h.
- Substitution:
- ν=1.889/(4.14×10−15)≈0.456×1015 Hz=4.56×1014 Hz.
- Result: Rounded to two decimal places, the frequency is 4.57×1014 Hz.
This hard difficulty physics question is from the chapter atoms and nuclei, covering the topic of photon emission. It appeared in the 2025 exam.
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