Answered: Calculate the tunneling probability… | bartleby

Related questions

Question

Calculate the tunneling probability when the kinetic energy of the particle is 0.2 MeV , the
barrier height is 20MEV, the probability amplitude is 1.95 x105 m-1, and the width of the
barrier is 2.97×10-18
m.
(A) 0.046
(В) 0.156
(C) 0.026
(D) 0.456

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

The wave function of a particle in a one-dimensional box of width L is u(x) = A sin (7x/L). If we know the particle must be somewhere in the box, what must be the value of A?
An electron has a kinetic energy of 13.3 eV. The electron is incident upon a rectangular barrier of height 21.5 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase?
5. Consider a potential barrier represented as follows: U M 0 +a U(x) = x a Determine the transmission coefficient as a function of particle energy.
= = An electron having total energy E 4.60 eV approaches a rectangular energy barrier with U■5.10 eV and L-950 pm as shown in the figure below. Classically, the electron cannot pass through the barrier because E < U. Quantum-mechanically, however, the probability of tunneling is not zero. Energy E U 0 i (a) Calculate this probability, which is the transmission coefficient. (Use 9.11 x 10-31 kg for the mass of an electron, 1.055 x 10-34] s for h, and note that there are 1.60 x 10-19 J per eV.) (b) To what value would the width L of the potential barrier have to be increased for the chance of an incident 4.60-eV electron tunneling through the barrier to be one in one million? nm
Determine the expectation values of the position (x) (p) and the momentum 4 ħ (x)= cos cot,(p): 5V2mw 4 mah 5V 2 sin cot 2 ħ moon (x)= sin cot, (p)= COS at 52mo 2 4 h 4 moh (x)= 52mo sin cot.(p) COS 2 h s cot, (p) 5V2mco 2 moh 5V 2 sin of as a function of time for a harmonic oscillator with its initial state ())))
The wavefunction for a quantum particle tunnelling through a potential barrier of thickness L has the form ψ(x) = Ae−Cx in the classically forbidden region where A is a constant and C is given by C^2 = 2m(U − E) /h_bar^2 . (a) Show that this wavefunction is a solution to Schrodinger’s Equation. (b) Why is the probability of tunneling through the barrier proportional to e ^−2CL?
A particle with the velocity v and the probability current density J is incident from the left on a potential step of height Uo, that is, U (x) = Uo at r > 0 and U(x) = 0 at r 0.