Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^{-2CL}T≃e ​−

Question

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^{-2CL}T≃e ​−2CL ​​ , where L is the width of the barrier and C is a term that includes the particle energy and barrier height. If the tunneling coefficient is found to be T = 0.050T=0.050 for a given value of LL, for what new value of L\text{‘}L’ is the tunneling coefficient T\text{‘} = 0.025T’=0.025 ? (All other parameters remain unchanged.) Express L\text{‘}L’ in terms of the original LL.

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Nho 4 years 2021-09-02T14:17:05+00:00 1 Answers 19 views 0

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  1. haidang
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    2021-09-02T14:18:55+00:00
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    Answer:

    L’ = 1.231L

    Explanation:

    The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

    T \approx e^{-2CL}

    L: width of the barrier

    C: constant that includes particle energy and barrier height

    You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let’s say, L’, the value of the transmission coefficient is T’=0.025.

    To find the new value of the L’ you can write down both situation for T and T’, as in the following:

    0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)

    Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

    ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)

    Next, you divide the equation (3) into (4), and finally, you solve for L’:

    \frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L

    hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L’ is 1.231 L

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