The Fermi energy for copper is EF 7.00 ev: For copper at T = 1030 K, (a) find the energy of the energy level whose probability of being occupled by an electron Is 0...

The Fermi energy for copper is 7.00 eV. For copper at $1000 \mathrm{K},(\mathrm{a})$ find the energy of the energy level whose probability of
being occupied by an electron is $0.900 .$ For this energy, evaluate $(\mathrm{b})$
the density of states $N(E)$ and $(\mathrm{c})$ the density of occupied states
$N_{\mathrm{o}}(E) .$

For this problem a density of folk abide state and nod. Um, he is given by density off estates times the probability of United States with energy e This is given by C times E one or one over two are times so one or two years in the exponents so e energy minus e f the f the Fermi Energy divided by Katie plus plus one minus one in power where C is ah eight Swerved off to buy mass Ah, three over two divided by itch three though substituting your values for ah electron mass in the point constant we obtained the constant c two B 1.62 times 10 to the power 56 kg, uh three or two for Joel tree or 2nd 3 in this skin beard in a electron world six point 81 times 10 to the power 27 For me to cube Electron world three minus three or two For about a it energy E is equal to fall Electron, Walt. Then we can right the occupy density off the states, uh, will be substituting your values. So this will be 6.821 times 10 to the power to any seven per meter. So that is minus three times electron world minus three over two times a four electron world one or two this divided by, um, exponential energy four electron world minus seven. Electron world divided by a bolt man Constant. The times for the temperature 2000 Calvin plus one Here. Um so this will view us. This is will give us They're not the occupied state number off. Provide state. They're not to be 1.36 times 10 to the power 28 farm Nature cube elect for electron worlds pulled apart to be for energy for energy of e is equal to are 6.75 electron wall. We use the same procedure that we did for about A and substitute the value for the tea to be a 1000 Calvin. We attend the occupy density off the states and not to be 1.68 times 10 to the power 28 per meter cube electro but electron worlds Similarly, for a party for energy equal to seven electron world, the value for and lord is mine for 01 times 10 to the power 27 per meter queue for electron world body At energy, he's equal to 7.25 across the world. The value off a nod we'll turn here is 9.56 times 10 to the power 26. Well part you need to cube electron world butter lip from you Then for a part e at energy nine Electron world So energy we have here is a nine electron worlds. We obtained the value and Lord and no number off occupied state to be 1.71 times 10 to the power 18 for the energy nine electron world So this is the power 18 for me to pube for electron won't

In part of the problem we're plugging, um, energy is above the firm energy by a positive 0.12 Your troubles. After changing it into jewels and plugging an expression for F, we're fine. Af you close, um, 8.6 times 10 to the negative three. This is a reasonable result. Reasonable result. Because that's this low temperature of 293 K. It is, um, there is a very small probability for the state to be occupied because the status is very far above two. For me, enter, you report Be of the problem. We plugging immodesty f equals negative their appointed 12 electoral waltz under also converting into standard units plugging a temperature 293 kelvin with on F equals zero point 9914 So part see, the probability that the stadium's that import beast unoccupied is simply one linus the previous F through a 0.9 and one for and this is exactly the expression with way found in port, eh? In the 8.6 times, 10 to 73. So there is kind of a symmetry for this, uh, this rip distribution f

But a healthy given problem. The density off the states for the Fermi Energy and Off E F um from the given expression is given by four Divided by itch square times, M Cube Root three three by square. And, um, since the desire to medical answer, use this electron world's women by denominator in new monitor by C Square. And we obtain an expression for density of states for for me energy to be a four m c square divided by ich sie whole square times it see paltry by square. And, um, we can extract this and out So the expression would come, Let's say everything together, this all together, which I would like to dance she her times Ah, and one work third. So we opened anyone worth that? Um, we just need to find the values for, um for this value, we can find it by substituting the value for rest mass energy off electron, which is a four times 511 times gentle power. Three electron world. We can do this by looking the table from the book divided by Etc Square. Let's see we have is a 12 40 electron World Nana Meter on Do we take the square off? This times went applying a constant which is three, uh, by the square here, times off course and one or three. Then we attend the the density off the states for the Fermi Energy to be 4.11 Nanami True minus two minus electron world times Uh, one over three which is the equal into the given value in the problem photo. But me, since there are 10 to the power 27 Cuban enemy Georgina cubic meter. Um then the result for the given problem can return His hand isn't called eight point 49 times 10 to the power 28 Ah, poor a meter cube is 84.9 man a meter cube the cube root door for this end one over three, approximately equal to 4.4 for Nana meter. Hence the expression in about a leads us to and off He off f is equal to 4.11 nano meter are square Our daughter Electron world times four point For Nana Meter man and nature, this equals d'oh! This equals 2 18 men of major uh, Electron World Spur, which is a call to 1.8 times gentle power. 28 for a meter cube. But electron won't. Then if we multiply this by 10 to the power 10 to the power 27 meter cube, part Nano Meter Cube. We see this compare Ready? Well, with the they carved that we obtained that we have in a figure 41.6 generated 87 electron world.

So here in this problem there two parts. In the a part, we have to calculate the average energy of a conduction. If the former energy the copper is 7.5 electron volt at 300 Calvin, that means if we're given us, it is seven point 05 Electron world with the average energy is given by E average. It is equals to three by five times E If so, three by five e if we're given US 7.5 electron walled. So from here, average energy comes out to be as 4.23 electron world, so this is required answer for the a part Now. Next, it will be solved in the be part where we have to calculate at what temperature the average transformational energy off a molecule will be equal to the energy Calculated in the part A. That means 4.23 electron world. It should be equal to the average translation energy, which is given by three by two times k b times t. So from here, temperature T comes out to be asked to buy three times four point 23 Electron world divide by K B. We have 1.38 times 10 to the power minus 23. Jules spoke Kelvin on. We're going to change the unit here. That is one electron world we know. It is equal toe 1.6 times 10 to the power, minus 19 jewels. So from here, temperature T comes out to be as nearly 3.27 times 10 to the power for Cohen. So this is the answer for the be part. So I hope you have understood the problem. Thank you.