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Interpretation:
The expression for the transmission probability has to be derived. The expression for the condition when transmission probability reduces to T≈16ε(1−ε)e−κW when κW is much higher than 1 has to be shown.
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Answer to Problem 7D.12P
The expression for the transmission probability is shown below.
T=((eκW−e−κW)216ε(1−ε)+1)−1
When κW≫1, then the expression for the transmission probability is shown below.
T=16ε(1−ε)e2κW
Explanation of Solution
The transmission probability is given by the expression shown below.
T=|A'|2|A|2 (1)
Where,
- A and A' are the coefficients.
At the edge of barrier (x=0), the relationship between the coefficients A, B, C, and D is given by the expression as shown below.
A+B=C+D (2)
At the edge of barrier (x=W), the relationship between the coefficients A', C, and D is given by the expression as shown below.
CeκW+De−κW=A'eikW (3)
Where,
- W is the potential energy width.
- κ is a constant.
At the edge of barrier (x=0), the slope is given by the expression as shown below.
ikA−ikB=κC−κD (4)
Where,
- k is a constant.
At the edge of barrier (x=0), the slope is given by the expression as shown below. κCeκW−κDe−κW=ikA'eikW (5)
Rearrange the equation (2) for the value of B.
B=C+D−A (6)
Rearrange the equation (4) for the value of B.
B=ikA−κC+κDik (7)
Substitute the value of equation (6) in equation (7).
C+D−A=ikA−κC+κDikC+D−A=A−κCik+κDik (8)
Rearrange the equation (8) for the value of C.
C=2Aik+D(κ−ik)κ+ik (9)
Rearrange the equation (3) for the value of A'.
A'=CeκW+De−κWeikWA'=(CeκW+De−κW)e−ikW (10)
Rearrange the equation (5) for the value of A'.
A'=κe−ikW(CeκW−De−κW)ik (11)
Substitute the value of equation (10) in equation (11).
(CeκW+De−κW)e−ikW=κe−ikW(CeκW−De−κW)ik (12)
Rearrange the equation (12) for the value of C.
C=(κik+1)De−2κWκik−1C=(κ+ik)De−2κWκ−ik (13)
Substitute the value of equation (9) in the equation (13).
(κ+ik)De−2κWκ−ik=2Aik+D(κ−ik)κ+ik (14)
Rearrange the equation (14) for the value of D.
D=2Aik(κ−ik)(κ+ik)2e−2κW−(κ−ik)2 (15)
Substitute the value of D from equation (15) to the equation (13).
C=(κ+ik)(2Aik(κ−ik)(κ+ik)2e−2κW−(κ−ik)2)e−2κWκ−ikC=2Aik(κ+ik)e−2κW(κ+ik)2e−2κW−(κ−ik)2 (16)
Substitute the values of D from equation (15) and C from equation (16) to the equation (10).
A'=((2Aik(κ+ik)e−2κW(κ+ik)2e−2κW−(κ−ik)2)eκW+(2Aik(κ−ik)(κ+ik)2e−2κW−(κ−ik)2)e−κW)e−ikWA'=2Aik(κ+ik)2e−2κW−(κ−ik)2((κ+ik)e−κW+(κ−ik)e−κW)e−ikWA'=4Aiκke−κWe−ikW(κ+ik)2e−2κW−(κ−ik)2A'=4Aiκke−ikW(κ+ik)2e−κW−(κ−ik)2eκW
Substitute the value of A' from above equation in the equation (1).
T=|4Aiκke−ikW(κ+ik)2e−κW−(κ−ik)2eκW|2|A|2=(4iκke−ikW(κ+ik)2e−κW−(κ−ik)2eκW)(−4iκkeikW(κ+ik)2e−κW−(κ−ik)2eκW)=−16i2κ2k2((κ+ik)2e−κW−(κ−ik)2eκW)2
Substitute the value i2=−1 on the above equation.
T=−16(−1)κ2k2((κ+ik)2e−κW−(κ−ik)2eκW)2=16κ2k2((κ+ik)2e−κW−(κ−ik)2eκW)2
The above equation is further solved to get the expression shown below.
T=16κ2k2(κ2+k2)2(eκW−e−κW)2+16κ2k2=((κ2+k2)2(eκW−e−κW)2+16κ2k216κ2k2)−1=((κ2+k2)2(eκW−e−κW)216κ2k2+1)−1 (17)
The constants κ and k can be written in the terms of energy as shown below.
κ=[2m(V0−E)]1/2ℏk=[2mE]1/2ℏ
Where,
- E is the total energy.
- V is the potential energy.
- m is the mass.
- ℏ is a constant.
Substitute the value of κ and k in the equation (17).
T=((([2m(V0−E)]1/2ℏ)2+([2mE]1/2ℏ)2)2(eκW−e−κW)216([2m(V0−E)]1/2ℏ)2([2mE]1/2ℏ)2+1)−1
The above equation is further simplified to get an expression shown below.
T=((V02)(eκW−e−κW)216E(V0−E)+1)−1T=((eκW−e−κW)216EV0(1−EV0)+1)−1 (20)
The ratio of total energy and potential energy is given by the expression shown below.
ε=EV0
Substitute the value of EV0 in the equation (20).
T=((eκW−e−κW)216ε(1−ε)+1)−1
When κW≫1, then the eκW term is very negligible and the e−κW term is very large in comparison to 1. Therefore, the above expression can be simplified as shown below.
T=((−e−κW)216ε(1−ε))−1=(e−2κW16ε(1−ε))−1=16ε(1−ε)e−2κW=16ε(1−ε)e2κW
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