contestada

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?

Respuesta :

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.

Otras preguntas