Question

Two nanowires are separated by 1.2 nm as measured by STM. Inside the wires the potential energy is zero, but between the wires the potential energy is greater than the electron's energy by only 1.5 eV. Estimate the probability that the electron passes from one wire to the other.

Answer 1

The probability that the electron passes from one wire to the other is 5.8x10⁻⁷

Step-by-step explanation:

The potential energy is:

`V_(o) -E=1.5eV`

The constant k is equal:

`k=\frac{\sqrt{2m(V_(o)-E) } }{h} =\frac{\sqrt{2mC^(2) (V_(o)-E)} }{hc}`

Where

m = 9.11x10⁻³¹kg

C = 3x10⁸m/s

`k=\frac{\sqrt{2*9.11x10^(-31)kg(3x10^(8))^(2)(1eV)/(1.6x10^(-19)J ) *1.5eV } }{1.054x10^(-34)Js(1eV)/(1.6x10^(-19)J )3x10^(8)m/s } =6.27x10^(9) m^(-1)`

The probability is:

`P=2e^(-2kL)`

Where

L = 1.2 nm = 1.2x10⁻⁹m

`P=2e^{-2*6.27x10^(9)*1.2x10^(-9) } =5.8x10^(-7)`

Answer 2

(8.81 × 10⁻⁶)

Step-by-step explanation:

Probability of tunneling is T˜e^(-2CL)

L = 1.2nm = (1.2 × 10⁻⁹) m

C=√[2m(U-E)] ÷ (h/2π)

m = mass of electron = (9.11 × 10⁻³¹) kg

(U-E) =0.9eV = (0.9 × 1.6 × 10⁻¹⁹) J

= (1.44 × 10⁻¹⁹) J

h = Planck's constant = (6.63 × 10⁻³⁴) Js

(h/2π) = (1.0552 × 10⁻³⁴) Js

Therefore,

C =√[(2)(9.11×10⁻³¹)(1.44×10⁻¹⁹)] ÷ (1.0552 × 10⁻³⁴)]

= (4.85 × 10⁹) m⁻¹

Then T˜e^[-(2)(4.85*10⁹)(1.2×10⁻⁹)

= 0.0000088067 = (8.81 × 10⁻⁶)

Hope this Helps!!!