(You may have done these you\'ll use the temperatures to determine the heat capa

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(You may have done these you'll use the temperatures to determine the heat capacity) temperature calculations before, but it is useful to practice with these concepts, and this time The interatomic spring stiffness for copper is determined from Young's modulus measurements to be 28 N/m. The m one mole of copper is 0.064 kg. If we model a block of copper as a collection of atomic "osclilators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscilators is 4 times the calculated value given above. ass of Use these precise values for the constants: = 1.0546 x 10.34 J . s (Planck's constant divided by 2) Avogadro's number = 6.0221 x 1023 molecules/mole ka = 1.3807 x 10-23 J/K (the Boltzmann constant) In order for WebAssign to be able to give you sensible feedback, you must give 6 figures in your answers forE and S. Otherwise there is a build-up of round-off errors when you take differences of E and S, which could result in correct answers being marked wrong one quantum = [3.6599 joules Here Is a table containing the number of ways to arrange a given number of quanta of energy in a particular block of copper Complete the remainder of the table, including calculating the temperature of the block. The energy E is measured from the ground state. Be sure to give the temperature to the nearest 0.1 degree.

Explanation / Answer

The eigenvalues or energies of a quantum harmonic oscillator (QHO) is given by:

E(n) = (n+½), where n = 0, 1, 2 ... corresponding to discrete levels of quantum states that the QHO can occupy, and

= sqrt(k/m)

k is the bond force constant, and m is the reduced mass

Here we are given that the mass of one mole of copper is:

Mcopper= 0.064 kg

Therefore the mass of one copper atom is given by:

m = Mcopper/Avogadro's number = 0.064/6.022×10^23
= 1.06x10^-25 kg

So = sqrt(k/m) = sqrt[4(28)/(1.06x10^-25)]
= 3.2505 x 10^13 rad/sec

Also, = 1.0546 x 10^-34 J sec

Then the ground state of the copper quantum harmonic oscillator is:

E(0) = (1.0546 x 10^-34)(3.2505 x 10^13) / 2

= 17.14 x 10^-22 J =0.0107625 eV

b) One Quantum/ delta Temp/ 100 atoms = Heat Capacity

Heat Capacity=1.565e-23 J/K/atom

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