2. Cubic potential (25 points total) Valley Velley I A cubic potential of the fo

Question

2. Cubic potential (25 points total) Valley Velley I A cubic potential of the form V(Q)m 2 x (Fig 1) can be used to model decay of a metastable state, e.g. a hydrogen atom trapped by a weekly bounding complex. Consider a particle of mass m 2000 me (me1 a.uthe mass of an electron) in a cubic potential with the parameters Ha a) 0.01 and b 0.2981 Bohr Figure 1. Schematics of a cubic potential. (5 points) Compute the coordinate Xn where the potential has a local minimum and the potential energy at that point. (5 points) Compute the coordinate Xmax where the potential has a local maximum and the potential energy at that point. (5 points) Compute the energy barrier, AE, of this potential. By assuming the potential can be approximated by the harmonic potential in the valley I, compute the number of quantum states this valley can accommodate (5 points) The particle is initially placed at the coordinate xo =-0.2 0.1. compute the classical energy of this point. To which quantum state is the energy of this state is closest? (5 points) Assuming the potential in the valley I can be approximated by the harmonic potential, compute ZPE-this is the minimal energy a system can have in this valley

Explanation / Answer

Answer:

for the given cubic potential

V(q) = 0.5mw^2*x^2 - (1/3)bx^3

m = 2000 me

me = 1 au = 9.1*10^-31 kg

w = 0.01 (Ha/Bohr^2. me)^1/2

b = 0.2981 Ha/Bohr^3

a. from condiciotn for maxima minima

dV/dx = 0

hence

m*w^2*x - bx^2 = 0

=> m*w^2*x = bx^2

x = 0, mw^2/b

now, V(0) = 0

V(mw^2/b) = 0.5*m*w^2*m^2*w^4/b^2 - (1/3)b*m^3w^6/b^3

V(mw^2/b) = m^3*w^6/6b^2

hence local minima is at x = 0

potential energy at this point, V = 0

b. local maxima is at x = mw^2/b = 2000 me * 0.01 ^ 2 (Ha / bohr^2 * me) *bohr^3/0.2981 Ha

x = 2000 * 0.01 ^ 2 *bohr^2/0.2981 = 0.6709158 bohr

the value of PE at this point is

V = 2000^3 *me^3 *0.01^6 * Ha^3 *Bohr^6/ Bohr^6 * me^3 * 6*0.2981^2 Ha^2

V = 2000^3 *0.01^6 * Ha / 6*0.2981^2 = 0.015004267 Ha

c. dE = V - 0 = 0.015004267 Ha

now, 1 Ha = 27.211396 eV

hence

dE = 0.4082 eV

for harmonic osscilator

En = nh/2*pi

hence

n*6.67*10^-34/2*pi*1.6*10^-19 = dE

hence

n = 615374681755027.08269 energy elvels

d. at xo = -0.2 bohr

V = 0.004 eV

this is closest to mth energy state

m*h/2*pi = V

hence

m = 7226975632721 energy level

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