p42_033 - . 013)(4 . 7 eV) 3(8 . 62 10 5 eV / K) = 4 . 7 10...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
33. The fraction f of electrons with energies greater than the Fermi energy is (approximately) given in Problem 42-32: f = 3 kT/ 2 E F where T is the temperature on the Kelvin scale, k is the Boltzmann constant, and E F is the Fermi energy. We solve for T : T = 2 fE F 3 k = 2(0
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 013)(4 . 7 eV) 3(8 . 62 10 5 eV / K) = 4 . 7 10 2 K . It should be noted that the numerical approach, discussed briey in part (c) of problem 32, would lead to a value closer to T = 6 . 5 10 2 K....
View Full Document

Ask a homework question - tutors are online