Question #207304

The Potential Step (E > U0):

(a) Express the reflection and transmission probabilities in terms of E and U0. What is the probability

that the incident particle will be reflected for the case U0 = 0.7 E?

(b) Obtain the probability density and the real and imaginary parts of the wavefunction for t = π/2 ω .

Also, plot them.


1
Expert's answer
2021-06-16T14:22:12-0400

Part a

The physics of a potential step for the case E>U0E>U_0

U(x)=[+U00 x>0x<0U(x)=[_{+U_0}^0 \space _{x>0}^{x<0}

u(x)=eikx+Reikxu(x)=e^{ikx}+Re^{-ikx}

k=2mE2k=\sqrt{\frac{2mE}{\hbar^2}}

u(x)=Teikxu'(x)=Te^{-ik'x}

k=2mEU02k'=\sqrt{\frac{2mE-U_0}{\hbar^2}}

Continuity wave function =x=0=x=0 implies

1+R=T1+R=T

[ikeikxikReikx]x=0=[ikTeikx]x=0[ike^{ikx}-ikRe^{-ikx}]_{x=0}=[ik'Te^{ik'x}]_{x=0}

k(1R)=k(1+R)k(1-R)=k'(1+R)

(k+k)R=(kk)(k+k')R=(k-k')

R=kkk+kR= \frac{k-k'}{k+k'}

T=1R=2kk+kT=1-R= \frac{2k}{k+k'}

u(x)=[eikxkkk+keikxu(x)=[^{e^{ikx}} {\frac{k-k'}{k+k'}}e^{-ikx} x<0x<0

2kk+keikx\frac{2k}{k+k'}e^{ik'x} x>0x>0

Preflection=R2=(kkk+k)2P_{reflection}=|R|^2=(\frac{k-k'}{k+k'})^2

Ptransmission=1Preflection=4kk(k+k)2P_{transmission}=1-P_{reflection}=\frac{4kk'}{(k+k')^2}

The transmission probability goes to 1 k=kk=k' (since there is no step)

The transmission probability goes for 0 k=0k'=0 (since the kinetic energy is zero)


Part b

P(x,x+Δx)Ψ(x,t)2δx.P(x,x+Δx)≈|Ψ(x,t)|^2δx.

P(x,x+Δx)=xx+ΔxΨ(x,t)2dxP(x,x+Δx)=∫^{x+Δx}_x|Ψ(x,t)|^2dx

P(,+)=Ψ(x,t)2dx=1.P(−∞,+∞)=∫^∞_{−∞}|Ψ(x,t)|^2dx=1.

P(,+)=C2dx=1.P(∞,+∞)=∫^∞_{−∞}|C|2dx=1.

P(x=0,L)=0LC2dx=1.P(x=0,L)=∫^L_0|C|^2dx=1.

C=1LC=\sqrt\frac{1}{L} where L=π2L= \frac{\pi}{2}

P(x=0,L/2)=0L/21π22P(x=0,L/2)=∫^{L/2}_0 |\sqrt{\frac{1}{\frac{\pi}{2}}}|^2

=(10.5π)0.5π2=0.25=(\frac{1}{0.5 \pi}) \frac{0.5 \pi}{2} = 0.25


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS