How Small Can a Quantum Well Be for an Electron's Total Energy to Be Zero?

[D__grant]

1. Homework Statement

-This is a problem on my practice final that I haven't been able to solve. Hoping someone out there can take a crack & clarify it for me.

Quantum wells are devices which can be used to trap electrons in semiconductors. If the electron is in the well it has a lower energy than if it is outside, so it tends to stay in the well. Suppose we have a quantum well which has a width of DeltaX and a depth of 1.0 eV , i.e. if the electron is in the well it has a potential energy of -1.0 eV and if it is outside it has a potential energy of 0 eV. Use the uncertainty principle to find the value of DeltaX for which total energy kinetic & potential of an electron in the well is zero.
Note: This is the smallest size well we can have because if deltaX is any smaller, the total energy of the electron in the well will be bigger than zero, and escape.



2. Homework Equations
1. E=KE+PE
2. Vo= -1 eV
3. Total Energy > 1/2m x (h/2piDeltaX)^2 - Vo

3. The Attempt at a Solution

I set the Total Energy=0 and attempted to solve for deltaX. My first solution was the the order of 10^-11 but I doubt I answered it correctly. Also, the mass of an electron was not given on the exam so I'm wondering if there's a different path to take. Thank you

 

[BruceW]

I think you have the right idea. If the mass of the electron was not given, then what constants were given?

 

[D__grant]

Charge of an electron in joules? I noticed somewhere in my notes he used the mass of an electron in eV.
--

I am just completely unsure as to whether this is "the way to do it."

Thank you though, Bruce

 

[BruceW]

Yeah, I'm pretty sure that's the right way. This is one of those examples that are used over and over again to get students used to making "order-of-magnitude" calculations.

 

[Nickie]

for any help.



Based on the given information, we can use the uncertainty principle to solve for the minimum width (DeltaX) of the quantum well in which the total energy of the electron is zero. The uncertainty principle states that the product of the uncertainty in position (DeltaX) and the uncertainty in momentum (DeltaP) must be greater than or equal to h/2pi, where h is the Planck's constant. In this case, we can assume that the uncertainty in momentum is equal to the uncertainty in the kinetic energy (KE) of the electron.

Using the given equation E=KE+PE, where PE is the potential energy of the electron, we can rewrite it as KE=E-PE. Since we want the total energy to be zero, we can substitute E=0 and PE=-1 eV (given in the problem) to get KE=1 eV. From the uncertainty principle, we know that DeltaX * DeltaP >= h/2pi. We can rewrite DeltaP as m * DeltaV (where m is the mass of the electron and DeltaV is the uncertainty in velocity). We also know that KE=1/2 * m * DeltaV^2. Substituting this into the uncertainty principle equation, we get:

DeltaX * m * DeltaV^2 >= h/2pi

Solving for DeltaX, we get:

DeltaX >= (h/2pi) / (m * DeltaV^2)

Since we want the minimum value of DeltaX, we can assume that DeltaV is equal to the maximum velocity of the electron, which is the velocity when it is at the edge of the well. This velocity can be calculated using the classical equation KE=1/2 * m * v^2, where v is the maximum velocity. Substituting this into the above equation, we get:

DeltaX >= (h/2pi) / (m * (2 * KE/m)^2)

Simplifying this, we get:

DeltaX >= (h^2/16pi^2) * (1/KE)

Substituting the given value of KE=1 eV, we get:

DeltaX >= (6.626 x 10^-34 J*s)^2 / (16 * (3.14)^2 * 1 eV)

DeltaX >= 1.37 x 10^-10 m

Therefore, the minimum width of the