Atomic Energy Levels Homework: De Broglie Wavelength & Ionisation of Fe

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SUMMARY

The discussion focuses on calculating the De Broglie wavelength for electrons with an energy of 50 eV and determining the ionization energy of iron (Fe) in its ground state. The formula used for the De Broglie wavelength is λ = h / (m√(2Em)), where h is Planck's constant, m is the mass of the electron, and E is the energy. The calculated wavelength of 1.74 x 10^-10 m was identified as incorrect, and clarification was provided regarding the positive nature of ionization energy, emphasizing that it should not be negative.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically De Broglie wavelength.
  • Familiarity with the energy levels of electrons in atoms, particularly iron (Fe).
  • Knowledge of basic physics equations involving Planck's constant and electron mass.
  • Ability to interpret energy diagrams and ionization energies.
NEXT STEPS
  • Research the calculation of De Broglie wavelengths for various energies.
  • Study the ionization energies of transition metals, focusing on iron (Fe).
  • Learn about the implications of negative versus positive energy in quantum mechanics.
  • Explore the relationship between energy levels and electron transitions in atoms.
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in atomic structure and electron behavior in elements like iron.

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Homework Statement


Couple of questions

State the De Broglie Wavelength for electrons with an energy of 50eV

and

the following is a picture is a diagram of the energy levels in Fe: http://www.freeimagehosting.net/image.php?eeddf8b465.jpg
What is the energy required to ionise Fe in in the ground state

Homework Equations


*lambda* = h/p = h/mv = h/sqr(2Em)

The Attempt at a Solution


for the first question i got 1.74x10^-10m but that is not the solution

and for the next question

is it -11.6 or +11.6
 
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you mean

[tex]\lambda = \frac{h}{m \sqrt{2Em}}[/tex]

did you remember to multiply by the mass again in the denominator?

We'd need to see the diagram to help you with the next part of the question.
 
If it's "energy required", it's positive; in this case it makes no sense to talk about adding a negative amount of energy.
 

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