Question

What is the maximum wavelength line in the Lyman series for $H{e}^{+}$ ion?

• A. $3R$
• B. $\frac{1}{3R}$
• C. $\frac{4}{4R}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{3R}$

For $H{e}^{+},Z=2$
Wave number, $\stackrel{-}{v}=\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

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,
$\frac{1}{{\lambda }_{max}}=4R[\frac{1}{1}-\frac{1}{4}]$

$\Rightarrow {\lambda }_{max}=\frac{1}{3R}$