# Estimating Diameter of Hydrogen Atom from Lyman Alpha Radiation

• ultimateguy
In summary, using the simplified model of an infinite 1D potential well, an estimate of the diameter of the hydrogen atom is derived by relating the energy levels and frequency of light emitted from an electron transition. The length of the potential well is used as an approximation for the diameter of the atom, with the correct expression for length being L=\sqrt{\frac{3h\lambda}{8mc}}. This results in a diameter of 3.325x10^-10m, using the mass of an electron.
ultimateguy

## Homework Statement

The Lyman alpha radiation for hydrogen (nf=2, ni=1) has a wavelength of 1216 angstroms. Using the simplified model of an infinite 1D potential well, derive an estimate of the diameter of the hydrogen atom. How does this value compare with twice the Bohr radius 2r=1.06anstroms?

## Homework Equations

I imagine maybe De Broglie's relation, or the energy of a potential well.

## The Attempt at a Solution

I have no idea how I can relate a potential well to the radiation given off by hydrogen. I realize that I'm supposed to work this out myself, but my exam is tomorrow morning and would appreciate if I could get the solution, or at least a hefty hint.

Hi ultimateguy, you know the energy levels of an infinite well, right? What frequency of light would be emitted if an electron made a transition from n=2 to n=1 in an infinite well of size L?

The energy levels for an infinite well are $$E_{n}=\frac{n^2h^2}{8mL^2}$$ and the frequency is $$\omega=\frac{E}{\hbar}$$.

EDIT: Is the length of the potential well the same as the diameter of the atom?

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ultimateguy said:
The energy levels for an infinite well are $$E_{n}=\frac{n^2h^2}{8mL^2}$$ and the frequency is $$\omega=\frac{E}{\hbar}$$.

EDIT: Is the length of the potential well the same as the diameter of the atom?

First, the frequency for n=2 to n=1 is $$\omega = (E_2 - E_1)/\hbar$$, not just E.

Second, they clearly want you to approximate the atom with the potential well, so what do you think will correspond to the atom diameter?

I think that the length of the well approximates the atom. For $$E_1$$ the wavelength $$\lambda=2L$$ and for $$E_2$$ that $$\lambda=L$$. Although this is the wavelength of the wavefunction and not the photon.

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If the length of the well approximates the diameter of the atom then you should figure out what the length of the well is.

I tried doing that already. I had $$E_2-E_1=\frac{3h^2}{8mL^2}$$. I then substituted into $$\omega=\frac{E_2-E_1}{\hbar}$$ to get $$\omega=\frac{3h\pi}{4mL^2}$$ and solving for L I get $$L=\sqrt{\frac{3h\lambda}{4m}}$$. I got 7.77x10^-21m using this.

You did your algebra wrong. For one thing you forget a factor of c, the speed of light.

Don't know where c comes in.

What are the units of frequency? How did you relate angular frequency to wavelength? What are the units of wavelength? Notice anything? Your current expression does not have the right units, so there is no way you could get some number of meters from it.

$$\omega=\frac{2c\pi}{\lambda}$$. Damn it I'm an idiot.

My equation for length is $$L=\sqrt{\frac{3h\lambda}{8mc}}$$. If I use the proton mass for m I get 7.76x10^-12m. This isn't right, is it?

Nope. Why are you using the proton mass?

I figured it was a hydrogen atom, has one proton. Now I see that it is an electron mass I need to use, and I get 3.325x10^-10m. I guess the electron mass is used because that is what's in a potential well. That must be right :P.

## 1. What is Lyman Alpha radiation?

Lyman Alpha radiation is a type of electromagnetic radiation that is emitted when an electron in a hydrogen atom transitions from a higher energy state to the ground state.

## 2. How can we estimate the diameter of a hydrogen atom from Lyman Alpha radiation?

The diameter of a hydrogen atom can be estimated by measuring the wavelength of the Lyman Alpha radiation and using the Rydberg formula, which relates the wavelength to the energy levels of the atom. The diameter can then be calculated using the known value of the Bohr radius.

## 3. Why is Lyman Alpha radiation used for estimating the diameter of a hydrogen atom?

Lyman Alpha radiation is used because it has a very specific wavelength of 121.6 nanometers and is easily detectable. This allows for accurate measurements and calculations to be made.

## 4. What are the potential sources of error when estimating the diameter of a hydrogen atom from Lyman Alpha radiation?

Potential sources of error include instrumental limitations, such as the precision of the measuring equipment, and environmental factors, such as interference from other sources of radiation. Additionally, the Rydberg formula assumes the hydrogen atom is in a vacuum, which may not be the case in a laboratory setting.

## 5. How is this method of estimating the diameter of a hydrogen atom useful in scientific research?

Knowing the diameter of a hydrogen atom is useful in various fields of science, including chemistry and physics. It can help in understanding the behavior and interactions of atoms, as well as in developing new technologies and materials.

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