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

Generally the barrier width is
`a = 1.9322 *10^(-9) \ m`

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{ (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{ (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 * (E)/(V_o ) * [1 - (E)/(V_o) ] * e^(-2 * k * a)`

So

`1.0 *10^(-5) = 16 * (0.04)/(0.4 ) * [1 - (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`