The angular momentum of an electron in a Bohr's orbit of He+ is 3.1652 x 10^-34 kg.m2 / sec.The wave number of the spectral line emitted when an electron falls from this level to the first excited state is:
Correct answer: B. 5R/9
- A. 3R
- B. 5R/9
- C. 3R/4
- D. 8R/9
Explanation
Recall the Rydberg Formula and Our ValuesWe know the Rydberg formula: νˉ=RZ2(1/n12 −1/n22 ) We found:Z (for He+) = 2n2 (initial level) = 3n1 (final level) = 22. Substitute the Valuesνˉ=R(22)(1/22−1/32)νˉ=4R(1/4−1/9)3. Simplify the FractionTo subtract the fractions, find a common denominator (36):νˉ=4R((9/36)−(4/36))νˉ=4R(5/36)4. Final SimplificationMultiply the 4 and the fraction:νˉ=(4×5R)/36νˉ=20R/36νˉ=5R/9
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