Hydrogen Atom Size and Magnetic Field Data: Understanding the Basics

In summary: Hydrogen atoms aren't spheres and therefore don't have a unique radius. Rather, the definition depends on the application. The 95% probability radius I would have to calculate myself: If r is measured in units of a (Bohr radii) then ## \int_0^{r_{95\%} }r^2 \exp(-r)dr=0.95\times \int_0^\infty r^2 \exp(-r)dr ##.
  • #1
bobie
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Can someone give me a link where I can find data on the size of an H atom (free and in main compounds like water)?
If there is none can you tell me roughly the range of sizes or at least what is the ratio between a free atom and a bound atom.
I know that for a free atom the radius is not the Bohr radius anymore , and that there is a cloud of probability, can you tell me what is the radius where probability is 95%?

If it is possible, I'd like to learn what is the official value of the magnetic field B of the ground state (1s), as on the web I found many contrasting values. I tried to calculate it myself and I got 1/8 Tesla, is that correct?
I read also that angular moment in QM is 0, is it h/2pi in classical model?
Thanks a lot for your help.
 
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Hydrogen atoms aren't spheres and therefore don't have a unique radius. Rather, the definition depends on the application. You may look for the term "covalent radius".
The 95% probability radius I would have to calculate myself:
If r is measured in units of a (Bohr radii) then
## \int_0^{r_{95\%} }r^2 \exp(-r)dr=0.95\times \int_0^\infty r^2 \exp(-r)dr ##.
Probably, Wolfram Alpha will solve that for ##r_{95\%}##.

I have never heard of the official value B field for hydrogen. Anyhow the magnetic moment vanishes both for a classical and a quantum mechnanical state with l=0, e.g. in Bohr-Sommerfeld theory.
 
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  • #3
DrDu said:
I have never heard of the official value B field for hydrogen. Anyhow the magnetic moment vanishes both for a classical and a quantum mechnanical state with l=0, e.g. in Bohr-Sommerfeld theory.
Thanks for the precious tip, I found r= 79 and covalent r 32
As to the magnetig moment I found 9.285 * 10^-24 J/T
If I divide that by μ which is (if I am not wrong q*v*r/2) h/4π, to I get the value of B?
As to angular momentum (mvr) I think it is twice as the magnetic moment h/2pi

Thanks for your help
 
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Related to Hydrogen Atom Size and Magnetic Field Data: Understanding the Basics

What is a hydrogen atom?

A hydrogen atom is the smallest and simplest type of atom, consisting of only one proton in its nucleus and one electron orbiting around it.

What is the size of a hydrogen atom?

The size of a hydrogen atom is approximately 0.1 nanometers (or 1 angstrom). This is equivalent to 1/10,000th of the width of a human hair.

How does the size of a hydrogen atom relate to its magnetic field?

The size of a hydrogen atom does not directly affect its magnetic field. However, the magnetic field of a hydrogen atom is influenced by the spin of its electron and the orientation of the electron's orbit around the nucleus.

What is the significance of understanding the basics of hydrogen atom size and magnetic field data?

Understanding the basics of hydrogen atom size and magnetic field data is crucial in many fields of science, such as chemistry, physics, and astronomy. It allows us to understand the behavior and properties of atoms, which are the building blocks of all matter.

How is hydrogen atom size and magnetic field data measured and studied?

Hydrogen atom size and magnetic field data can be measured and studied using various techniques, such as spectroscopy, electron paramagnetic resonance, and nuclear magnetic resonance. These techniques involve the use of electromagnetic radiation and magnetic fields to interact with the hydrogen atom and gather data about its properties.

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