Respuesta :
Answer:
(8.81 × 10⁻⁶)
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!!!
Answer:
The probability that the electron passes from one wire to the other is 5.8x10⁻⁷
Explanation:
The potential energy is:
[tex]V_{o} -E=1.5eV[/tex]
The constant k is equal:
[tex]k=\frac{\sqrt{2m(V_{o}-E) } }{h} =\frac{\sqrt{2mC^{2} (V_{o}-E)} }{hc}[/tex]
Where
m = 9.11x10⁻³¹kg
C = 3x10⁸m/s
[tex]k=\frac{\sqrt{2*9.11x10^{-31}kg(3x10^{8})^{2}\frac{1eV}{1.6x10^{-19}J } *1.5eV } }{1.054x10^{-34}Js\frac{1eV}{1.6x10^{-19}J }3x10^{8}m/s } =6.27x10^{9} m^{-1}[/tex]
The probability is:
[tex]P=2e^{-2kL}[/tex]
Where
L = 1.2 nm = 1.2x10⁻⁹m
[tex]P=2e^{-2*6.27x10^{9}*1.2x10^{-9} } =5.8x10^{-7}[/tex]
