The first member of Balmer series of hydrogen atom has a wavelength of 6561 Å. Find wavelength of the second member of the Balmer series (in nm)

quorums15lep
2022-05-10
Answered

The first member of Balmer series of hydrogen atom has a wavelength of 6561 Å. Find wavelength of the second member of the Balmer series (in nm)

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Duncan Cox

Answered 2022-05-11
Author has **18** answers

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å.

We have to find,

Wavelength of the second member of the Balmer series (in nm)

We know,

$\mathrm{\u25b3}E=\frac{hc}{\lambda}=\mathrm{\u25b3}{E}_{0}[\frac{1}{4}-\frac{1}{9}]$

So,

$\frac{hc}{\lambda}=\mathrm{\u25b3}{E}_{0}[\frac{1}{4}-\frac{1}{16}]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\chi}{\lambda}=\frac{5\times 16}{9\times 4\times 3}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}^{\prime}=\frac{5\times 4\times 656.1}{9\times 3}\phantom{\rule{0ex}{0ex}}=486\text{}nm$

We have to find,

Wavelength of the second member of the Balmer series (in nm)

We know,

$\mathrm{\u25b3}E=\frac{hc}{\lambda}=\mathrm{\u25b3}{E}_{0}[\frac{1}{4}-\frac{1}{9}]$

So,

$\frac{hc}{\lambda}=\mathrm{\u25b3}{E}_{0}[\frac{1}{4}-\frac{1}{16}]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\chi}{\lambda}=\frac{5\times 16}{9\times 4\times 3}\phantom{\rule{0ex}{0ex}}\Rightarrow {\lambda}^{\prime}=\frac{5\times 4\times 656.1}{9\times 3}\phantom{\rule{0ex}{0ex}}=486\text{}nm$

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