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.

Answer 1

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

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

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

=>        $a = 1.9322 *10^{-9} \ m$