it is now almost possible to manufacture a \"box\" that traps electrons in a reg

Question

it is now almost possible to manufacture a "box" that traps electrons in a region only a few nanometers wide. Imagine that we make an essentially one-dimensional box with a length of 3.0nm . If we put 10 electrons in such a box and allow them to settle into the lowest energy states they can, consistent with the pauli exclusion principle , a) what will be the value of n for the highest energy level occupied ? b) what will be the total energy of the electrons (ignoring electrostatic repulsion) ? c) Would your answer be different if the electrons were bosons instead of fermions ?

Explanation / Answer

(a) Electron is a fermion, thus no two electrons can have same quantum numbers. We know that an electron can have two spin values, up and down. Thus, at most two electrons can be there in a given energy state.

SInce there are 10 electrons, thus for lowest energy, states upto n=5 should be filled each with two electrons. One of them being up spin and the other down. So, highest value of n occupied is 5

(b) Total energy of electrons will thus be:

(E=2*Sigma_{n=1}^{5}rac{n^2h^2}{8mL^2}=rac{h^2}{8mL^2}2*Sigma_{n=1}^{5}n^2=rac{h^2}{8mL^2}*110)

Putting the value of the constants, we get

(E=110*rac{(6.626*10^{-34})^2}{8*9.31*10^{-31}(3*10^{-9})^2=110*9.885*10^{-22}}=1.09*10^{-19}J)

(c) If the electron were a boson, then all electrons will be in the lowest energy state i.e. n=1

Thus all of them will occupy n=1 energy state. So, the highest occupied energy state will correspond to n=1

And total energy will be:-

(E=10*rac{h^2}{8mL^2}=10*9.885*10^{-22}=9.885*10^{-21} J)

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