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We have the value for expression of line spectrum of hydrogen.

$\Delta E = 13.6 \cdot Z^2 \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Amount of energy required by electron to change its orbit.

Now for wave number $\pu{1.09677e7} \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Does this mean the wavelength of light that it can absorb.

If yes , then why do we compare it with this .How is this compared with it.enter image description here

Since this shows the light emitted by spectrum and only certain colours show up.

My main question is that does any of these formula have any relation with the fact that you know , we see certain colours emitted when seen on a black screen and not all the colours. Form the picture that I have shown.

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    $\begingroup$ The differences in formulas must be in parentheses. $\endgroup$
    – Poutnik
    Dec 19 '20 at 7:44
  • $\begingroup$ Wavenumber is number of waves that fits the chosen length unit. Conventionally, it is usually 1 cm, so the wavenumber unit is 1/cm. $\endgroup$
    – Poutnik
    Dec 19 '20 at 7:53
  • $\begingroup$ My main question is that does any of these formula have any relation with the fact that you know , we see certain colours emitted when seen on a black screen and not all the colours@Poutnik $\endgroup$
    – srijan Sri
    Dec 19 '20 at 8:22
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Both ( equivalent ) formulas, after recalculation to wavelengths from energy differences or wavenumbers, predict particular wavelengths to be either emitted by hot and excited hydrogen atoms, either absorbed by hydrogen atoms being excited.

Hot atoms emit just particular wavelengths, not continuous spectrum like condensed matter. So their emission spectrum is linewise on black background.

OTOH, relatively cold atoms absorb at the same wavelengths they emit when excited. So their absorption spectrum is linewise as well, with the same lines as the emmision spectrum, but this time dark lines on bright background.

Both phenomena are the opposite direction of the common process, how atoms deal with discrete energy levels of their electrons. Emission of photons to reach lower energy, or gaining energy by absorption of photons. These photons belong to light of the very particular wavelength.


Comment feedback:

Incandescent bulb filaments have as condensed matter continuous spectrum. If their light passes through gas containing free hydrogen atoms ( i.e. not only molecular hydrogen ), particular wavelengths are absorbed, leaving dark lines in the spectrum.

Excited hydrogen atoms then either reemit at the same wavelength in any direction, or convert it during collisions in kinetic energy.

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  • $\begingroup$ Yes.You understood my question.Like for example we a bulb.Now the filament of becomes hot when switch is on.So when that light is passes through prism.Only certain colours appear on screen.Whereas , if we pass light through hydrogen gas , it first absorbs and then emits different wavelength of light. $\endgroup$
    – srijan Sri
    Dec 19 '20 at 8:43
  • $\begingroup$ So,What is the relation of the formula with this? $\endgroup$
    – srijan Sri
    Dec 19 '20 at 8:44
  • $\begingroup$ Does the formula tell us that there will be some black spaces in the formula ? Is the wavelength of light that it absorb same as wavelength of light it emits ? $\endgroup$
    – srijan Sri
    Dec 19 '20 at 8:45
  • $\begingroup$ I hope you get my question. $\endgroup$
    – srijan Sri
    Dec 19 '20 at 8:45
  • $\begingroup$ Read what I have written. Then read again. Then again, until you get it. $\endgroup$
    – Poutnik
    Dec 19 '20 at 8:45

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