TS EAMCET · Physics · Dual Nature of Matter
The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to (use \(\mathrm{R}_{\mathrm{H}}=1.097 \times 10^7 \mathrm{~m}^{-1}\) )
- A \(3646 Å\)
- B \(912 Å\)
- C \(364.6 Å\)
- D \(91.2 Å\)
Answer & Solution
Correct Answer
(A) \(3646 Å\)
Step-by-step Solution
Detailed explanation
For shortest wavelength \(\mathrm{n}_{\mathrm{f}}=2\) and \(\mathrm{n}_{\mathrm{i}}=\infty\)…
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