The wavelength of $\mathrm{H}_{\alpha}$ line of Balmer series is $X$ $Å$. What is the wavelength of…

The wavelength of $\mathrm{H}_{\alpha}$ line of Balmer series is $X$ $Å$. What is the wavelength of $\mathrm{H}_{\beta}$ line of $\mathrm{Balmer}$ series.
  1. $\mathrm{X} \frac{108}{80} Å$
  2. $\mathrm{X} \frac{80}{108} Å$
  3. $\frac{1}{\mathrm{X}} \frac{80}{108} Å$
  4. $\frac{1}{\mathrm{X}} \frac{108}{80} Å$

Solution

$\mathrm{H}_{\alpha}$ line of Balmer series means first line of Balmer series. $n_{1}=2, n_{2}=3$
$\bar{v}=\frac{1}{\lambda_{\alpha}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}ight)=\frac{5 \mathrm{R}}{36}$
$\therefore \lambda_{\alpha}=\frac{36}{5 \mathrm{R}}=\mathrm{X}$
$\mathrm{H}_{\beta}$ line of Balmer series means, second line of Balmer series,
$\mathrm{n}_{1}=2, \mathrm{n}_{2}=4$ $\bar{v}=\frac{1}{\lambda_{\beta}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}ight)=\frac{3 \mathrm{R}}{16}$
$\therefore \lambda_{\beta}=\frac{16}{3 \mathrm{R}}=\mathrm{X}$
when $\frac{36}{3 R}=X$
Then $\frac{16}{3 R}=\frac{X \times 5 R \times 16}{36 \times 3 R}=\frac{80 X}{108} Å$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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