Calculating Energy Levels in an Infinite Potential Box: Converting to eV?

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SUMMARY

The discussion focuses on calculating energy levels for an electron in an infinite potential box, specifically converting these values to electronvolts (eV). The equation used is E = h²/8M(nx²/a²), where h is Planck's constant and M is the mass of the electron (9.11e-31 kg). The conversion from Joules to eV is emphasized, with specific constants provided: hc = 1240 eV nm and ħc = 197 eV nm. The correct combination of constants is crucial for accurate calculations, leading to the derived energy level of 0.376 eV nm².

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the infinite potential well model.
  • Familiarity with Planck's constant and its units.
  • Knowledge of energy conversion between Joules and electronvolts (eV).
  • Basic algebra for manipulating equations involving physical constants.
NEXT STEPS
  • Research the derivation of energy levels in quantum mechanics using the infinite potential well model.
  • Learn about the significance and applications of Planck's constant in quantum physics.
  • Explore the conversion methods between Joules and electronvolts in various contexts.
  • Investigate the role of fundamental constants in quantum calculations, including their units and applications.
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Students studying quantum mechanics, physicists working on particle physics, and educators teaching advanced physics concepts related to energy quantization.

leonne
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Homework Statement


well the problem is electron moving in an infinite poentail box of dimensions... calc all posible energy level in (eV)


Homework Equations


E=h2/8M(nx2/a2+...


The Attempt at a Solution


ok so no idea how they get h2/8M to = 37.6 h is planks constant the units are in m4kg s-2 but eV is in m2kg s-2 don't see how they get 37.6 for mass i used 9.11e-31
 
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Find out the energy levels in Joules and simply use the conversion between Joules and eV. Should be easily google'd/wikipedia'ed or should be in your textbook.
 
There are some useful combinations of the fundamental constants. If you know them, you can make calculations a lot faster.
\begin{align*}
h c &= 1240~\text{eV nm} \\
\hbar c &= 197~\text{eV nm} = 197~\text{MeV fm} \\
m_e c^2 &= 511000~\text{eV} = 511~\text{keV} = 0.511~\text{MeV}
\end{align*}
For your problem, you just need to introduce factors of c, the speed of light, to form the right combinations:
\frac{h^2}{8m} = \frac{(hc)^2}{8mc^2} = 0.376~\text{eV nm}^2Note that this combination has two powers of length. They are needed to cancel the units from a2 in the denominator of the formula for the energy.
 

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