Question

Consider the potential shown in the figure, E R V(x) where $V(x) = \frac{\hbar^2l(l+1)}{2mx^2}$. This potential represents a centrifugal barrier in 3-dimensions, however this problem is 1D. Estimate the lifetime of a particle with energy E in this potential. Take the potential at x = 0 to be +?. Express your result in terms of the dimensionless ratio u = l/kR, where l? 1 and E = ?²k²/2m is the kinetic energy of the particle, which can be taken to be free inside the well, i.e. V(x) = 0 for 0 < x < R.

          Consider the potential shown in the figure,

E
R
V(x)
where $V(x) = \frac{\hbar^2l(l+1)}{2mx^2}$. This potential represents a centrifugal barrier in 3-dimensions, however this problem is
1D.
Estimate the lifetime of a particle with energy E in this potential. Take the potential at x = 0 to be +?. Express
your result in terms of the dimensionless ratio u = l/kR, where l? 1 and E = ?²k²/2m is the kinetic energy of the
particle, which can be taken to be free inside the well, i.e. V(x) = 0 for 0 < x < R.

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Consider the potential shown in the figure,

E
R
V(x)
where V(x) = (ħ^2l(l+1))/(2mx^2). This potential represents a centrifugal barrier in 3-dimensions, however this problem is
1D.
Estimate the lifetime of a particle with energy E in this potential. Take the potential at x = 0 to be +?. Express
your result in terms of the dimensionless ratio u = l/kR, where l? 1 and E = ?²k²/2m is the kinetic energy of the
particle, which can be taken to be free inside the well, i.e. V(x) = 0 for 0 < x < R.

Added by Mark A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Dominique Jan Tan

Step 1

Since the potential is zero, the Schrödinger equation becomes: -(h^2/2m) * d^2ψ/dx^2 = E * ψ This is a simple harmonic oscillator equation, and its solutions are given by: ψ(x) = A * sin(kx) + B * cos(kx) where k = sqrt(2mE)/h and A, B are constants determined Show more…

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Consider the potential shown in the figure: (x)A R where V(x) = 1D. 2mx^2 Estimate the lifetime of a particle with energy E in this potential. Take the potential at x = 0 to be +âˆž. Express your result in terms of the dimensionless ratio u = l/kR, where l > 1 and E = h^2/2m is the kinetic energy of the particle, which can be taken to be free inside the well, i.e. V(x) = 0 for 0 < x < R.

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