MCQ
If the shortest wavelength of $H-$ atom in Lyman series is $x$ , then longest wavelength in Balmer series of $He^+$ is
- ✓$\frac {9x}{5}$
- B$\frac {36x}{5}$
- C$\frac {x}{4}$
- D$\frac {5x}{9}$
For Lyman series,
$n_{1}=1$ and $\lambda$ is shortest if $n_{2}=\infty$
$\therefore R_{\mathrm{H}}=\frac{1}{\mathrm{x}},$ for $\mathrm{H}-$ atom $(Z=1)$
For Balmer series,
$n_{1}=2$ and $\lambda$ is longest if $n_{2}=3$
$\text { (For } \mathrm{He}^{+} \text {ion } \left.Z=2\right)$
$\frac{1}{\lambda_{\max }}=R_{H}(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]$
$=\frac{1}{x} \times 4\left[\frac{5}{36}\right]$
$\lambda_{\max }=\frac{9 x}{5}$
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