The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively

# The minimum and maximum values of wavelength in the Lyman series of a H atom are, respectively

1. A

364.3nm and 653.4 nm

2. B

91.2 nm and 121.5 nm

3. C

41.2 nm and 102.6 nm

4. D

9.12 nm and 121.5 nm

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### Solution:

For hydrogen like species , use the following formula to calculate the wavelength Of radiation of spectral series lines.

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right]$

Case 1: Minimum wavelength in the Lyman series:

Then, $\frac{1}{\mathrm{\lambda }}=109677\left[\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right]$

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[1-\frac{1}{4}\right]=\frac{3\mathrm{R}}{4}$

$⇒\mathrm{\lambda }=\frac{4}{3\mathrm{R}}=\frac{4}{3×109677{\mathrm{cm}}^{-1}}$

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_{1}^{2}}-\frac{1}{{\mathrm{n}}_{2}^{2}}\right]$

Case 2: Maximum wavelength in the Lyman series:

Then, wavelength of line is

$\frac{1}{\mathrm{\lambda }}=109677\left[\frac{1}{{\left(1\right)}^{2}}-\frac{1}{{\left(\mathrm{\alpha }\right)}^{2}}\right]$

$\frac{1}{\mathrm{\lambda }}=109677\left[1-0\right]=109677{\mathrm{cm}}^{-1}$

$⇒\mathrm{\lambda }=\frac{1}{109677{\mathrm{cm}}^{-1}}=9.117×{10}^{-6}\mathrm{cm}$

$\therefore \mathrm{\lambda }=91.2\mathrm{nm}$

Note: