Calculating Minimum Energy of Photons in Ionization and Relativistic Corrections

[BRN]

Hi at all! I need one more help from you.

1. Homework Statement
123.4 eV photons ionize further a rarefied gas of $B^{2+}$ ions. A small fraction of electrons emitted in this process is immediately captured by $B^{3+}$ ion, going to occupy the states 2p, 3p, and 3d. Calculate the minimum energy of the photons emitted in the process of capturing, in the not relativistic approximation. How does the result taking into account the relativistic corrections order α²? Do you remember the relativistic correction (multiplicative) to the eigenenergy of motion in the Coulomb potential: $[1+\frac{(Z\alpha)^2}{n}(\frac{1}{j+1/2}-\frac{3}{4n})]$.

The Attempt at a Solution

I admit that I have no idea on how to solve this exercise. The minimum energy of emitted photons is

$E=h\nu=h\frac{c}{\lambda}$

How can I know the wavelength knowing the states that are occupied?
Or maybe I have to totally change approach with this exercise ...

 

[DrClaude]

How can I know the wavelength knowing the states that are occupied?

Think about the process: an electron is being captured by an ion. What is the smallest change in energy possible of the electron + ion system?

 

[BRN]

But the states 2p 3p 3d give me useful information?

 

[DrClaude]

But the states 2p 3p 3d give me useful information?

What do you think? Does this knowledge change the answer, in other words, if the final state could be 4d, would that change the minimum energy of the emitted photon?

Again, think about the process and the change in energy of the system. If a photon is emitted, its energy must come from the ion + electron system.

 

[BRN]

After capture process, I have this configuration: 1s² 2p 3p 3d

Then, the system energy is:

$E_{tot}=-\frac{\mu}{m_e}\frac{E_{Ha}}{2}(\frac{Z^2}{1^2}+\frac{Z^2}{2^2}+\frac{Z^2}{3^2})$

with:

$E_{Ha}=\frac{m_e e^4}{\hbar^2}$ Hartree energy;

$\mu$= reduced mass.

Is correct?

Don't kill me please!

 

[DrClaude]

Is correct?

No.

Don't kill me please!

Too late. You already sold us your soul when you registered at PF. Didn't you read the rules?

Seriously, I think you need to consider that only one electron is captured by a given atom. Also, you need only to consider the difference in energy for the before and after situation.

 

[BRN]

Sorry for delay,

If I consider one electron, I have:

before 1s² 2s² 2p

$E_{tot}=-\frac{\mu}{m_e}\frac{E_{Ha}}{2}(\frac{Z^2}{1^2}+\frac{Z^2}{2^2})=-425.1547 [eV]$

after 1s² 3d
$E_{tot}=-\frac{\mu}{m_e}\frac{E_{Ha}}{2}(\frac{Z^2}{1^2}+\frac{Z^2}{3^2})=-377.9153 [eV]$

with:

$M=A*1.6605*10^{-27}[kg]=1.7933*10^{-26}[kg] \Rightarrow \mu=\frac{Mm_e}{M+m_e}=9.1088*10^{-31}[kg]$

then:

$\Delta E= 47.23 [eV]$

but solutions are: 14.5545 eV; with relativistic corrections: 14.5400 eV

 

[DrClaude]

Consider a single electron going from ionized to captured by the atom.

 

[BRN]

There is something that is not clear...

But, how I conseder Z?

Before $B^{2+}=1s^22s \rightarrow Z=2$
After $B^{3+}=1s^2 2p$ or $1s^2 3p$ or $1s^2 3d \rightarrow Z=3$

 

[BRN]

No one else can help me?