Answer :
Answer:
Generally the barrier width is [tex]a = 1.9322 *10^{-9} \ m[/tex]
Step-by-step explanation:
From the question we are told that
The tunneling probability required is [tex]T = 1 * 10^{-5}[/tex]
The barrier height is [tex]V_o = 0.4 eV[/tex]
The electron energy is [tex]E = 0.08eV[/tex]
Generally the wave number is mathematically represented as
[tex]k = \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }[/tex]
Here m is the mass of the electron with the value [tex]m = 9.11 *10^{-31} \ kg[/tex]
h is is know as h-bar and the value is [tex]\= h = 1.054*10^{-34} \ J \cdot s[/tex]
So
[tex]k = \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }[/tex]
=> [tex]k = 3.073582 *10^{9} \ m^{-1}[/tex]
Generally the tunneling probability is mathematically represented as
[tex]T = 16 * \frac{E}{V_o } * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}[/tex]
So
[tex]1.0 *10^{-5} = 16 * \frac{0.04}{0.4 } * [1 - \frac{0.04}{0.4} ] * e^{-2 * 3.0736 *10^{9} * a}[/tex]
=> [tex]6.944*10^{-6}= e^{-2 * 3.0736 *10^{9} * a}[/tex]
Taking natural log of both sides
[tex]ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}[/tex]
=> [tex]-11.8776 = -2 * 3.0736 *10^{9} * a}[/tex]
=> [tex]a = 1.9322 *10^{-9} \ m[/tex]