I was solving the following problem the following way:(adsbygoogle = window.adsbygoogle || []).push({});

In a solid the atoms are regularly arranged in space. The potential seen by an electron is thus periodic and the energy levels are arranged in bands in which the energy of a state k is given by:

e_{k}= E_{at}+ 2t cos(ka)

The relevant band structure of the metal Silver (Ag) can be roughly modelled with a band of that form with t= 4 eV and a= 0.409 nm. Assume Ag has one conduction electron per atom.

a) Calculate the maximum velocity of an electron.

e_{k}= E_{at}+ 2t cos(ka)

with:

a= 0.409* 10^{-9}m

t= 4 eV= 4* 1.6*10^{-19}= 6.4*10^{-19}J

Therefore:

e_{k}= E_{at}+ 1.28*10^{-18}cos(0.409* 10^{-9}k)

v_{g}= 1/h_{with line}* (d e_{k}/dk)

d e_{k}/dk= -1.28*10^{-18}*0.409* 10^{-9}sin(0.409* 10^{-9}k)

v_{g}=-4.96*10^{6}sin(0.409* 10^{-9}k)

The velocity is maximal when the sine is maximal, therefore when the sine=1. In that case the velocity is -4.96*10^{6}m/s.

b) Calculate the drift velocity (average velocity) of electrons in a silver wire of 1mm^{2}cross-sectional area through which a current of 1A is flowing, knowing v_{drift}= I/nAq and n= 5.85*10^{28}electrons per m^{3}.

v_{drift}= I/nAq

v_{drift}= 1/5.85*10^{28}* 1*10^{-6}* 1.6*10^{-19}= -1.069*10^{-4}m/s

Now my question is: why are the velocities calculated in a and b so different from each other?!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: QM- Drift & Maximum velocity of an electron

**Physics Forums | Science Articles, Homework Help, Discussion**