Behaviour of electron in hydrogen atom when hit by photon

serverxeon
Messages
100
Reaction score
0
1. Given that the ground energy level of a hydrogen atom is -13.6eV, the 1st excited state -3.4eV. The difference is 10.2eV. Imagine a photon of energy 11.0eV hits the electron at ground state. What will be the final result?

Electron doesn't gets excited, photo bounces off with 11.0eV?
or
Electron goes to 1st excited state, and photon gets red-shifted by 10.2eV.


3. Somehow I think is the latter. There is partial energy loss by the photon, and red-shift happens.
Am I right?
 
Physics news on Phys.org
hint: absorption spectrum
 
so,

the answer is the former?

as given a spectrum of EM radiation, the atom will only absorb specific frequency which correspond to the energy difference between the energy levels.

Hence, only the specific frequencies absorbed are missing from the EM spectrum observed.

correct?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top