By using Rydberg's equation,I find out wavelengths of different spectral lines of Lyman series of the hydrogen atom. Wavelength of photon emitted when electron in the hydrogen atom jumps from n=2 to n=1 is 4/3R, where R is Rydberg's constant. Similarly, For n=3 to n=1, it is 9/8R, for n=4 to n=1 it is 16/15R and so on. These wavelengths are nearly equal to R(Rydberg's constant). Why these wavelengths are so same?

  • $\begingroup$ What do you mean by "why are they so similar"? The ratio between 9/8 and 16/15 is 1.2, and we can very easily detect a 20% change in wavelength. That's huge by atomic physics standards - we can easily measure frequencies down to a part per million or better. $\endgroup$ – probably_someone May 29 '17 at 8:34

The energy variation of an Hydrogen atom is given by: $$E(n)=-\frac{E_o}{n^2}$$

Let's consider $E(k)-E(1)$, and call it $\Delta E$: $$\Delta E=E_o(1-\frac{1}{k^2})$$

$\frac{\Delta E}{E_o}$ gets closer and closer to one as you increase $k$.

What this basically means is that because of the $n^{-2}$ variation, the change in energy between successive energy levels becomes lesser and lesser as you increase $n$, and hence there is no significant change in the wavelength emitted for these successive levels.

(Italicized "significant" because these small differences are important on the atomic level, but might not be so to a layperson.)


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