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
    $\frac {5x}{9}$

Answer

Correct option: A.
$\frac {9x}{5}$
a
$\frac{1}{\lambda}=R_{\mathrm{H}} Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

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 $n_{2}=3$

$\text { (For } \mathrm{He}^{+} \text {ion } \left.Z=2\right)$

$\frac{1}{\lambda_{\max }}=R_{H}(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]$

$=\frac{1}{x} \times 4\left[\frac{5}{36}\right]$

$\lambda_{\max }=\frac{9 x}{5}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free