D

Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 * 10^{-10} m.

8 viewed last edited 1 month ago
Anonym0us
0

[Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]

Qalaxia Master Bot
0

I found an answer from www.khanacademy.org

De Broglie wavelength (video) | Khan Academy


Bohr model energy levels (derivation using physics) ... Since visible light has a wavelength of about 500 nanometers, this means that visible light ... Since electrons have a rest mass, unlike photons, they have a de Broglie wavelength ... For example in the double slit experiment.... electron waves diffract and then ' condense' ...


For more information, see De Broglie wavelength (video) | Khan Academy

Qalaxia Knowlege Bot
0

I found an answer from stanford.edu

PHYSICS 430 Lecture Notes on Quantum Mechanics


The wave equation for De Broglie waves, and the Born interpretation. Why an electron is not a wave. 4. The Quantum State. How does the electron get from A to ...


For more information, see PHYSICS 430 Lecture Notes on Quantum Mechanics

Qalaxia QA Bot
0

I found an answer from physics.stackexchange.com

Kinetic energy of electron in metals - Physics Stack Exchange


Sep 2, 2016 ... At room temperature the electrons in metals are actually a degenerate Fermi gas and can be treated as if near absolute zero. Quoting wikipedia ...


For more information, see Kinetic energy of electron in metals - Physics Stack Exchange

Veda
0

Given that

de Broglie wavelength of an electron \lambda = ?

Temperature T = 27\degree C

Mean separation between two electrons r = 2 * 10^{-10}


Step 1: Ge an expression for the de Broglie wavelength of an electron

             Gaseous particle kinetic energy at a certain temperature  K.E = \frac{3}{2} kT ............................(1)

                    Where,  k - Boltzmann constant. = 1.38 * 10^{-23} J/K


              de Broglie wavelength \lambda = \frac{h}{p}

                   Where, Plank constant h = 6.63 * 10^{-34} and p - momentum

                                   \lambda = \frac{h}{\sqrt{2mK.E}}                                \because K.E = \frac{p^2}{2m}

                                   \lambda = \frac{h}{3mkT}                                      \because \text{ equation (1)}


Step 2: Finding the de Broglie wavelength of an electron

              Mass of an electron m = 9.11 * 10^{-31} kg                

                                                 \lambda = \frac{6.63 * 10^{-34}}{3 * 9.11 * 10^{-31} * 1.38 * 10^{-23} * 300 }

                                                     = 6.2 * 10^{- 9} m


                         de Broglie wavelength of an electron \lambda=6.2*10^{-9} m

               Therefore, Mean separation between two electrons < de Broglie wavelength of an electron (r < \lambda)