u* of finding the electron per unit volume is zero outside 0 a, and ys is determined by the Schrödinger equation in 0 a, and y is determined by the Schrödinger equation in 0 < x < a %3D whe 0. Therefore, in the region 0 < x < a d²y 2me Eys = 0 %3D dx [3.22] Thic is a second-order linear differential equation. As a general solution, we an take ys(x) = A exp(jkx) + B exp(-jkx) [3 231
u* of finding the electron per unit volume is zero outside 0 a, and ys is determined by the Schrödinger equation in 0 a, and y is determined by the Schrödinger equation in 0 < x < a %3D whe 0. Therefore, in the region 0 < x < a d²y 2me Eys = 0 %3D dx [3.22] Thic is a second-order linear differential equation. As a general solution, we an take ys(x) = A exp(jkx) + B exp(-jkx) [3 231
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By using "Solution of Schrodinger's Wave Equation for a Particle trapped in a Well with Width = a", and with infinitely high walls. Derive the equations for, Wavefunction and Energy E.
Please take a look at the pictures that is the point you need to begin
![3.3
nsider une inside that region and infinite outside, as shown in Figure 3.16. The
of ond x > a, and y is determined by the Schrödinger equation in 0 < x < a
We= 0. Therefore, in the region 0 < x < a
%3D
d²y 2me
.2
%D
[3.22]
Thic is a second-order linear differential equation. As a general solution, we
can take
(x)sh
y (x) = A exp(jkx) + B exp(-jkx)
where k is some constant (to be determined) and substitute this in Equation 3.22 to
find k. We first note that y(0) = 0; therefore, B = -A so that
[3.23]
y (x) = A[exp((jkx) – exp(-jkx)] = 2Aj sin kx
[3.24]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe94f166b-cfef-43be-b644-d0de530f1378%2Fb64bdfef-2d3f-4e10-ada5-81336014e96f%2Fj0n9r5a.jpeg&w=3840&q=75)
Transcribed Image Text:3.3
nsider une inside that region and infinite outside, as shown in Figure 3.16. The
of ond x > a, and y is determined by the Schrödinger equation in 0 < x < a
We= 0. Therefore, in the region 0 < x < a
%3D
d²y 2me
.2
%D
[3.22]
Thic is a second-order linear differential equation. As a general solution, we
can take
(x)sh
y (x) = A exp(jkx) + B exp(-jkx)
where k is some constant (to be determined) and substitute this in Equation 3.22 to
find k. We first note that y(0) = 0; therefore, B = -A so that
[3.23]
y (x) = A[exp((jkx) – exp(-jkx)] = 2Aj sin kx
[3.24]
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