On the Rydberg Constant and the Emission Lines

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Gabrielmonteiro
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During the study of Rutherford's atomic model and hydrogen lines, I had doubts about the Rydberg constant, about the variation of its value and its physical meaning.
With regard to Rutherford's atomic model, and Rydberg's discovery in general for the hydrogen distribution lines, what does Rydberg's constant physically mean? Its unit is m ^ -1, as if it were a rate, but it was not clear to me its physical meaning.

And why does it grow with atomic mass? From the hydrogen series equations, we come to the conclusion that the wave number is proportional to the Rydberg constant, therefore, considering the relationship c = yf, would it be correct to say that elements with greater atomic mass have higher emission frequencies?

The references to this question were taken from the book Quantum Physics - Eisberg & Resnick, 26th Ed. Chapter 4.
 

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PeroK
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Summary:: During the study of Rutherford's atomic model and hydrogen lines, I had doubts about the Rydberg constant, about the variation of its value and its physical meaning.

With regard to Rutherford's atomic model, and Rydberg's discovery in general for the hydrogen distribution lines, what does Rydberg's constant physically mean? Its unit is m ^ -1, as if it were a rate, but it was not clear to me its physical meaning.

And why does it grow with atomic mass? From the hydrogen series equations, we come to the conclusion that the wave number is proportional to the Rydberg constant, therefore, considering the relationship c = yf, would it be correct to say that elements with greater atomic mass have higher emission frequencies?

The references to this question were taken from the book Quantum Physics - Eisberg & Resnick, 26th Ed. Chapter 4.
The physical meaning is probably best understood by its relationship to the wavelengths of the emission spectrum.

For hydrogenic atoms, it takes more energy to free an electron from a greater number of nuclear protons. The greater the energy of the photon aborbed or emitted corrtesponds to a shorter wavelength and, hence, a greater Rydberg constant for that system.
 
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  • #3
Gabrielmonteiro
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The physical meaning is probably best understood by its relationship to the wavelengths of the emission spectrum.

For hydrogenic atoms, it takes more energy to free an electron from a greater number of nuclear protons. The greater the energy of the photon aborbed or emitted corrtesponds to a shorter wavelength and, hence, a greater Rydberg constant for that system.
The question of physical meaning was not very clear to me. I believe it is a conceptual issue. I will look for some material on, if I have any indication, I will be grateful. And with respect to the variation of the Rydberg constant for larger atomic masses, it really makes sense, I hadn't looked at it that way. Thank you for the explanation
 
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PeroK
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The question of physical meaning was not very clear to me. I believe it is a conceptual issue.
If you invert the Rydberg constant you get a characteristic (wave)length. I wouldn't worry too much how to interpret ##R## physically, as opposed to ##1/R## being interpreted as a length. You may not find a deeper meaning at all.
 
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vanhees71
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The Rydberg constant is the ionization energy of atomic hydrogen. In the most simple approximation (in both the old Bohr-Sommerfeld model and modern quantum theory) for the energy levels of the hydrogen atom is
$$E_n=-\frac{1 \; \text{Ry}}{n^2}, \quad n \in \mathbb{N}=\{1,2,\ldots \}.$$
It's entirely given by natural constants (expressed in SI units),
$$1 \; \text{Ry}=\frac{\mu e^4}{32 \pi \epsilon_0^2 \hbar^2} \simeq 13.6 \; \text{eV}$$
with
$$\mu=\frac{m_e m_p}{m_e+m_p}.$$
 
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  • #7
Gabrielmonteiro
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The Rydberg constant is the ionization energy of atomic hydrogen. In the most simple approximation (in both the old Bohr-Sommerfeld model and modern quantum theory) for the energy levels of the hydrogen atom is
$$E_n=-\frac{1 \; \text{Ry}}{n^2}, \quad n \in \mathbb{N}=\{1,2,\ldots \}.$$
It's entirely given by natural constants (expressed in SI units),
$$1 \; \text{Ry}=\frac{\mu e^4}{32 \pi \epsilon_0^2 \hbar^2} \simeq 13.6 \; \text{eV}$$
with
$$\mu=\frac{m_e m_p}{m_e+m_p}.$$
Would this correction for infinite nuclear mass need a second correction in the event of a relativistic collision? In this situation, would we take the mass rest convention?
 

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