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Sever21 [200]
3 years ago
14

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

ier with a barrier height of Vo = 0.4 eV, the electron energy is 0.04 eV Determine the maximum barrier width.
Mathematics
1 answer:
Goryan [66]3 years ago
4 0

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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