Bohr's Model
Hardchemistry
The wavelength of the first line in the Balmer series of hydrogen spectrum is 656 nm. The wavelength of the second line of this series is approximately:
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- bohr's model
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
486 nm
The wavelength of spectral lines is given by the Rydberg Formula: λ1=RH(n121−n221) For the Balmer Series, n1=2.
- First Line (Hα line): Transition from n2=3→n1=2. 6561=RH(221−321)=RH(41−91)=RH(365)
- Second Line (Hβ line): Transition from n2=4→n1=2. λ21=RH(221−421)=RH(41−161)=RH(163) Dividing Eq. 1 by Eq. 2: 656λ2=3/165/36=365×316=2720 λ2=656×2720≈485.9≈486 nm
This hard difficulty chemistry question is from the chapter atomic structure, covering the topic of bohr's model. It appeared in the 2025 exam.
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