What is the Cross Section of a Hydrogen Atom in Thermal Equilibrium at 6000K?

In summary, the problem involves finding the cross section of a hydrogen atom at thermal equilibrium at 6000K for a photon inducing a transition from the ground state to the first excited state. The density of states for n=2 is four times the density of states for n=1. The lifetime of the n=2 state is 1.6 x 10^-9 s. The cross section is an area of interaction that depends on the energy of the incident photon. The number of particles in the n=1 and n=2 states can be calculated using the Maxwell-Boltzmann probability and the Boltzmann distribution.
  • #1
JohnSimpson
92
0

Homework Statement


I've been asked to find the cross section of a hydrogen atom in thermal equalibrum at 6000K for a photon which induces a transition from the ground state to the first excited state. The density of states for n=2 is 4x the density of states for n=1

i.e. g(E2) = 4*g(E1)

The lifetime of the n=2 state is 1.6 x 10^-9 s.


Homework Equations





The Attempt at a Solution



The incident photon has an energy E = hf = hc/lambda

Since the system is in thermal equlibrum, it has an average energy = kT. I am not sure what exact this is the average energy of though, the atoms? the photons? Everything?

From what I understand, the cross section is some kind of area of interaction for the process that will somehow depend on the energy of the incident photon. With this in mind, I wrote down the radius of the hydrogen atom for n =1

r1 = (epsilon)h^2 / pi * e^2 * m_e

I'm still not entirely comfortable with what the density of states represents for n=1 and n=2 respectively. I know that it sort of represents a "Price" to put an electron at that level, but I might be incorrect in saything that.

It would be very helpful if someone could clear up my misconceptions and nudge me in the right direction, thank you
 
Physics news on Phys.org
  • #2
Update: I've managed to write down the number of particles in the n=1 state at any time, which is given by

g(E1)f(E1) where f(E1) is the maxwell-boltzmann probability of finding a hydrogen atom with energy E1 corresponding to ground state, and similarly the number of n=2 hydrogen atoms is

4g(E1)f(E2)
 
  • #3
Anyone able to point me in the right direction?
 

What is a cross section calculation?

A cross section calculation is a method used in scientific research to determine the probability of a particular event occurring during an experiment. It involves calculating the area of a particle's interaction with a target material, which can help scientists understand the underlying physical processes at play.

Why is cross section calculation important in scientific research?

Cross section calculations are important because they help scientists understand the fundamental properties of matter and energy. By accurately calculating the probability of particle interactions, researchers can better understand the behavior of particles and their interactions with different materials.

What factors can affect the accuracy of a cross section calculation?

There are several factors that can affect the accuracy of a cross section calculation. These include the energy of the particles involved, the target material, and the experimental conditions. Other factors such as uncertainties in measurement and theoretical assumptions can also impact the accuracy of the calculation.

What are some techniques used for cross section calculation?

There are several techniques used for cross section calculation, including Monte Carlo simulations, perturbation theory, and numerical integration. Each technique has its own advantages and limitations, and the choice of method depends on the specific research question and experimental setup.

What are some applications of cross section calculation?

Cross section calculations are used in a wide range of scientific fields, including particle physics, nuclear physics, astrophysics, and materials science. They are particularly important in understanding the properties of atomic and subatomic particles, as well as for developing new materials and technologies.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Atomic and Condensed Matter
Replies
3
Views
1K
  • Atomic and Condensed Matter
Replies
3
Views
1K
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
14
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top