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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Chemistry
Chapter
Structure of Atom

BITSAT 2016 · Chemistry · Structure of Atom

The first emission line in the atomic spectrum of hydrogen in the...

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at

Options

  1. A9R400cm−1\frac{9 R}{400} \mathrm{ cm}^{-1}
  2. B7R144cm−1\frac{7 \mathrm{R}}{144} \mathrm{ cm}^{-1}
  3. C3R4cm−1\frac{3 \mathrm{R}}{4} \mathrm{ cm}^{-1}
  4. D5R36cm−1\frac{5 \mathrm{R}}{36} \mathrm{ cm}^{-1}

Correct answer

D. 5R36cm−1\frac{5 \mathrm{R}}{36} \mathrm{ cm}^{-1}

Step-by-step solution

For Balmer n1=2\mathrm{n}_{1}=2 and n2=3\mathrm{n}_{2}=3; v=R(122−132)=5R36cm−1v=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36} \mathrm{ cm}^{-1}
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