A spectral line is measured to have a wavelength of 1000nm. Is this within the Balmer series?

Joshua Foley
2022-07-14
Answered

A spectral line is measured to have a wavelength of 1000nm. Is this within the Balmer series?

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vrtuljakwb

Answered 2022-07-15
Author has **13** answers

For Balmer series, shortest wavelength is when transition takes from n = 3 to n = 2.

Wavelength for transition is given by, $\frac{1}{\lambda}=(1.097\times {10}^{7})(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})$

Substituting values,

$\frac{1}{\lambda}=(1.097\times {10}^{7})(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}\lambda =6.56\times {10}^{-7}m=656nm$

Since 1000 nm > 656 nm, hence, it belongs to Balmer series.

Wavelength for transition is given by, $\frac{1}{\lambda}=(1.097\times {10}^{7})(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})$

Substituting values,

$\frac{1}{\lambda}=(1.097\times {10}^{7})(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}\lambda =6.56\times {10}^{-7}m=656nm$

Since 1000 nm > 656 nm, hence, it belongs to Balmer series.

asked 2022-05-08

From my quick investigation, the spectrum is based on the Rydberg formula, and with a small change, would lead to

$\frac{1}{{\lambda}_{\mu}}=\frac{{m}_{\mu}}{{m}_{e}}\left(R(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})\right)$

where ${m}_{\mu}$ is the mass of a muon.

So, taking hydrogen as an example, we would observe similar bands, shifted into the x-ray/gamma range.

Is this correct?

$\frac{1}{{\lambda}_{\mu}}=\frac{{m}_{\mu}}{{m}_{e}}\left(R(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})\right)$

where ${m}_{\mu}$ is the mass of a muon.

So, taking hydrogen as an example, we would observe similar bands, shifted into the x-ray/gamma range.

Is this correct?

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From the figure find the shortest wavelength present in the Balmer series of hydrogen, If $R=1.097\times {10}^{7}\text{}{m}^{-1}$

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