Consider two straight wires lying on the x-axis, separated by a gap of 4 nm. The potential energy in the gap is about 3 eV higher than the energy of a ondu tion ele tron in either wire. What is the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire?

Answer: 1.3 * 10^-31

Explanation:

the required probability is P = e^ (-2αL)

Firstly, evaluate (-2αL)

α = 1/hc √2mc^2 (U - E)

h = modified planck's constant

where,

(-2αL)

= - (2L) / (h/2π) * √2mc^2 (U - E)

= - (2L) / (hc^2/π) * √2mc^2 (U - E)

(hc^2/2pi) = 197*eV. nm (standard constant)

2*L = 8 nm

mc^2 = 0.511*10^6 eV

Where m = mass electron

C = speed of light

(-2αL) = [-8nm / (197 eV. nm) ] * (1.022 * 10^6 eV**3 eV) ^0.5

(-2αL) = - 71.1

Probability = e^ (-2αL) = e^-71.1 = 1.3 * 10^-31

Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 * 10^-31.