Calculate the tunneling probability when the kinetic energy of the particle is 0.2 MeV, the barrier height is 20 MeV, the probability amplitude is 1.95×10¹5 m²¹, and the width of the barrier is 2.97x10-¹8 m. (A) 0.046 (B) 0.156 (C) 0.026 (D) 0.456

Suppose we had a classical particle in a frictionless box, bouncing back and forth at constant speed. The probability density of the position of the particle in soma box of length L is given by: 0 ans-fawr (7) p(x)= 0 x L a. Sketch the probability density as a function of position b. What must A be in order for p(x) to be normalized? Remember that you are welcome to use resources to solve integrals such as Wolfram Alpha, a table of integrals etc.

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.

24. Consider a modified box potential with V(x) = V₁x, Vi(ar), x a Use the orthogonal trial function = c₁f₁+c₂f₂ with f₁ = √√sin (H) and f2 = √√ √√sin sin (2) to determine the upper bound to ground state energy.