A certain semiconductor device requires a tunneling probability of T = 10-5 for an electron tunneling through a rectangular barrier with a b

Question

A certain semiconductor device requires a tunneling probability of T = 10-5 for an electron tunneling through a rectangular barrier with a barrier height of Vo = 0.4 eV, the electron energy is 0.04 eV Determine the maximum barrier width.

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Khải Quang 1 month 2021-07-31T19:38:29+00:00 1 Answers 2 views 0

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    2021-07-31T19:39:31+00:00

    Answer:

    Generally the barrier width is a = 1.9322 *10^{-9} \ m

    Step-by-step explanation:

    From the question we are told that

         The tunneling probability required is  T  = 1 * 10^{-5}

          The barrier height is  V_o  = 0.4 eV

           The electron energy is  E = 0.08eV

    Generally the wave number is mathematically represented as

          k  =  \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }

    Here m is the mass of the electron with the value  m  =  9.11 *10^{-31} \  kg

             h  is is know as h-bar and the value is  \= h = 1.054*10^{-34} \  J \cdot s

    So

              k  =  \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }

    =>      k = 3.073582 *10^{9}  \ m^{-1}

    Generally the tunneling probability is mathematically represented as

              T  = 16 * \frac{E}{V_o }  * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}

    So

            1.0 *10^{-5} = 16 * \frac{0.04}{0.4 }  * [1 - \frac{0.04}{0.4} ] * e^{-2 * 3.0736 *10^{9} * a}

    =>    6.944*10^{-6}= e^{-2 * 3.0736 *10^{9} * a}

    Taking natural log of both sides

              ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}

    =>        -11.8776  = -2 * 3.0736 *10^{9} * a}

    =>        a = 1.9322 *10^{-9} \ m

           

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