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2022-05-17

The second member of Balmer series of hydrogen atom has wavelength of $4861×{10}^{-10}$ m. Calculate the wavelength of third member of Balmer series

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To find−
Wavelength of 3rd line in Balmer series (${\lambda }_{r}$) = ?
Given −
Wavelength 2nd line in Balmer series
${\lambda }_{\beta }=4861×{10}^{-10}$ m
The second number of Balmer series i.e. ${H}_{\beta }$ line and third number is ${H}_{\gamma }$ line
For ${H}_{\beta }$ line
Therefore−
$\frac{1}{{\lambda }_{\rho }}=R\left[\frac{1}{{2}^{2}}-\frac{1}{{4}^{2}}\right]\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda }_{\rho }}=R\left[\frac{1}{4}-\frac{1}{16}\right]\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda }_{\rho }}=\frac{3R}{16}$
and
$\frac{1}{{\lambda }_{\rho }}=R\left[\frac{1}{{2}^{2}}-\frac{1}{{5}^{2}}\right]\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda }_{\rho }}=R\left[\frac{1}{4}-\frac{1}{25}\right]\phantom{\rule{0ex}{0ex}}\frac{1}{{\lambda }_{\rho }}=\frac{21R}{100}$
Dividing equations:
Therefore -
$\frac{{\lambda }_{\gamma }}{{\lambda }_{\rho }}=\frac{300}{16×21}\phantom{\rule{0ex}{0ex}}\frac{{\lambda }_{\gamma }}{{\lambda }_{\rho }}=\frac{25}{28}\phantom{\rule{0ex}{0ex}}{\lambda }_{\gamma }=\frac{25}{28}×{\lambda }_{\beta }\phantom{\rule{0ex}{0ex}}{\lambda }_{\gamma }=\frac{25}{28}×4861×{10}^{-10}\phantom{\rule{0ex}{0ex}}{\lambda }_{\gamma }=4340×{10}^{-10}m$

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