Modern Physics - Length of infinite well that has an electron

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SUMMARY

The discussion centers on calculating the size of an infinite potential well for an electron with a lowest energy level of 1.0 eV, using the formula E = n²h²/(8mL²). The variables include energy (E = 1.602e-19 J), Planck's constant (h = 6.626e-34 J*s), and electron mass (m = 9.109e-31 kg). A critical error identified in the calculations was the omission of a square on Planck's constant, leading to an incorrect size of L = 7.534e8 m, which is excessively large.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically particle in a box model.
  • Familiarity with energy quantization and its mathematical representation.
  • Knowledge of fundamental constants such as Planck's constant and electron mass.
  • Ability to manipulate equations involving physical constants and units.
NEXT STEPS
  • Review the derivation of energy levels for a particle in an infinite potential well.
  • Learn about the implications of quantum confinement on electron behavior.
  • Explore the effects of varying dimensions on energy levels in quantum systems.
  • Investigate common mistakes in quantum mechanics calculations and how to avoid them.
USEFUL FOR

This discussion is beneficial for students and educators in physics, particularly those studying quantum mechanics, as well as anyone involved in solving problems related to particle confinement and energy quantization.

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Homework Statement


** My book doesn't have any solutions in the back , and I trying to find out if I am doing the problems correctly. My book is Modern Physics for Scientists and Engineers by John C. Morrison. If you know anywhere I can find the Answers. I would greatly appreciate it!

The lowest energy level of an electron confined to a one-dimensional region is 1.0 eV. (a) By describing the electron as a particle in an infinite well, find the size of the region.


Homework Equations



Energy levels of particle in an infinite well:
E=\frac{n^{2}*h^{2}}{8mL^{2}}

The Attempt at a Solution


Formula:
E=\frac{n^{2}*h^{2}}{8mL^{2}}

Variables:
E= 1.0 eV = 1.602e-19 J
n =1
h = 6.626e-34 J*s
m= 9.109e-31 kg
L = ?

1.602*10^{-19}J=\frac{(1)^{2}*(6.626*10^{-34}J*s)}{8*(9.109*10^{-31}kg)*(L)^{2}}

L = 7.534*10^{8}m
 
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You're missing a square on your h, so you're getting an answer which is far too large...
 

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