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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Chemistry
Chapter
Structure of Atom

BITSAT 2014 · Chemistry · Structure of Atom

The wave number of the limiting line in Lyman series of hydrogen is...

The wave number of the limiting line in Lyman series of hydrogen is 109678cm−1109678 \mathrm{ cm}^{-1}. The wave number of the limiting line in Balmer series of He+\mathrm{He}^{+} would be :

Options

  1. A54839cm−154839 \mathrm{ cm}^{-1}
  2. B109678cm−1109678 \mathrm{ cm}^{-1}
  3. C219356cm−1219356 \mathrm{ cm}^{-1}
  4. D438712cm−1438712 \mathrm{ cm}^{-1}

Correct answer

B. 109678cm−1109678 \mathrm{ cm}^{-1}

Step-by-step solution

RH=109678cm−1\mathrm{R}_{\mathrm{H}}=109678 \mathrm{ cm}^{-1} Wave number of the limiting line in Balmer series of He+\mathrm{He}^{+}=RH⋅Z2[1n12−1n22]=109678\ttimes(2)2[1[2)2−1∞]=109678cm−1\begin{array}{l}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^{2} \left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]\\ =109678 \t times(2)^{2} \left[\frac{1}{[2)^{2}}-\frac{1}{\infty}\right]\\ =109678 \mathrm{ cm}^{-1} \end{array}$$
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