MCQ
What is the maximum wavelength line in the Lyman series of $He^+$ ion ?
- A$3R$
- ✓$1/3R$
- C$4/4R$
- DNone of these
Wave number, $\bar{v}=\frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
For Lyman series, $n _1=1$ and for maximum wavelength, $n _2$ has to be the smallest possible energy level.
$\therefore n _2=2$
Therefore,
$\frac{1}{\lambda_{\max }}=4 R \left[\frac{1}{1}-\frac{1}{4}\right]$
$\Rightarrow \lambda_{\max }=\frac{1}{3 R }$
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${\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\longrightarrow} {\text { (Major product) }{ }^{\prime} \mathrm{A}^{\prime}}$
Product $A$ is :
$C{H_3}C \equiv \,CC{H_3}\,\xrightarrow[{{\text{heat}}}]{{NaN{H_2}/{\text{inert solvent}}}}P$