(a)
Show that the total number of energy states is
(a)
Answer to Problem 66P
It is showed that the total number of energy states is
Explanation of Solution
Given:
The density of the electron states in a metal is
Formula used:
The number of energy states is given by,
Calculation:
The number of energy states in a metal is calculated as:
Conclusion:
Therefore, it is showed that the total number of energy states is
(b)
The fraction of the
(b)
Answer to Problem 66P
The fraction of the conduction electrons that are within
Explanation of Solution
Formula used:
The number of energy states is given by,
Calculation:
The fraction of number of states that is within
Conclusion:
Therefore, the fraction of the conduction electrons that are within
(c)
The fraction of the conduction electrons of copper that are within
(c)
Answer to Problem 66P
The fraction of the conduction electrons of copper that are within
Explanation of Solution
Formula used:
The expression forthe fraction of the conduction electrons that are within
Calculation:
The value of Fermi energy of copper is
Conclusion:
Therefore, the fraction of the conduction electrons of copper that are within
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Chapter 38 Solutions
Physics for Scientists and Engineers, Vol. 3
- If the energy gap for an insulating material is 4.5 eV, what is the probability that an electron will be promoted to the conduction band when the temperature is 100 °C? You may assume that the Fermi energy is in the middle of the energy gap.arrow_forwardConsider a n-type Si crystal at room temperature (300K) doped with 6 x1016 cm-3 arsenic impurity atoms and with certain number of shallowholes. Find out the equilibrium electron concentration, hole concentrationand Fermi level EF with respect to Ei, and the conduction band edge EC.For Si at 300K, the value of ni is 1.45 x 1010 cm-3 and k = 1.38 x 10-23 J/K,1eV = 1.60 x 10-19J. The band gap energy, Eg, of Si is 1.2eV.Solution:n @ Nd = 6 x 1016 cm-3.In equilibrium condition, hole concentration = 3.5 x 103 cm-3.EF – EI = 0.396eVEC – EF = 0.164eV.arrow_forwardCalculate No(E), the density of occupied states, for copper atT = 1000 K for an energy E of (a) 4.00 eV, (b) 6.75 eV, (c) 7.00 eV,(d) 7.25 eV, and (e) 9.00 eV..The Fermi energy for copper is 7.00 eVarrow_forward
- Consider a copper wire that is carrying, say, a few amperes of current. Is the drift speed vd of the conduction electrons that form that current about equal to, much greater than, or much less than the Fermi speed vF for copper (the speed associated with the Fermi energy for copper)?arrow_forwardWhat is the probability that a state 0.0620 eV above the Fermi energy will be occupied at (a) T= 0 K and (b) T =320 K?arrow_forwardThe Fermi energy of aluminum is 11.6 eV; its density and molar mass are 2.70 g/cm3 and 27.0 g/mol, respectively. From these data, determine the number of conduction electrons per atom.arrow_forward
- For a solid metal having a Fermi energy of 8.500 eV, what is the probability, at room temperature, that a state having an energy of 8.520 eV is occupied by an electron?arrow_forwardFor a certain semiconductor, the Fermi energy is in the middle of its band gap. If the temperature of the semiconductor is 285 K, find the probability that a state at the bottom of the conduction band is occupied if the band gap is 1.5 eV.arrow_forwardWhat mass of phosphorus is needed to dope 1.0 g of silicon so that the number density of conduction electrons in the silicon is increased by a multiply factor of 106 from the 10^16 m-3 in pure silicon.arrow_forward
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning