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Model Paper 10 — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryModel Paper 103 Marks
Question
What transition in a hydrogen spectrum would have the same wavelength Balmer transition $n =4$ to $n =2$ of $\overline{ v }=\frac{1}{\lambda}= R _{ H } Z^2\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$ spectrum?
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Answer

For an atom, $\overline{ v }=\frac{1}{\lambda}= R _{ H } Z^2\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$
For $He ^{+}$spectrum $Z =4, n _2=4, n _1=2$
$\therefore$ For hydrogen spectrum: $\overline{ v }=\frac{3 R _{ H }}{4}$ and $Z =1$
$\therefore \overline{ v }=\frac{1}{\lambda}= R _{ H } \times 1\left(\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right)$
or $R_H\left(\frac{1}{n_1^2}-\frac{1}{m_2^2}\right)=\frac{3 R_H}{4}$ or $\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{3}{4}$
This corresponding to $n _1=1, n _2=2$ and means that the transition has taken Lyman series from $n =2$ to $n =1$.
Thus, the transition is from $n _2$ to $n _1$ in case of hydrogen spectrum.

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