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JEE Mains · Chemistry · STD 11 - 2. structure of atom
The wave number of the first emission line in the Balmer series of \(H-\) Spectrum is ( \(R =\) Rydberg constant)
- A \(\frac {5}{36}\,\,R\)
- B \(\frac {9}{400}\,\,R\)
- C \(\frac {7}{6}\,\,R\)
- D \(\frac {3}{4}\,\,R\)
Answer & Solution
Correct Answer
(A) \(\frac {5}{36}\,\,R\)
Step-by-step Solution
Detailed explanation
\(\bar v = R{Z^2}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\) \( = R\left( {\frac{1}{4} - \frac{1}{9}} \right)\) \( = \frac{{5R}}{{36}}\)
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