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    Born approximation problem

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    Demonstrate, for a particle scattering from a finite-range, spherically symmetric potential V(r), which is weak enough so that the Born approximation for symmetric potentials is valid, that the total cross section, at very low energies, is a linear function of the energy σ(E) = σ0(1+αE), where σ0 is related to the volume integral of the potential... Please see attached for full question.

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    SOLUTION This solution is FREE courtesy of BrainMass!

    The differential cross section is given by:

    dsigma/dOmega = m^2/((2 pi)^2 h-bar^4)|T(k',k)|^2 (1)

    k' is the wave vector after scattering, k before.

    In the Born approximation T(k',k) is given by:

    T(k',k)= Integral[d^3r Exp[i (k-k') dot r] V(r)] (2)

    The potential is of short range, so only r within some small region can contribute to the integral. When the energy goes to zero, the k vectors go to zero to, so the term (k-k') dot r in the exponential can be regarded as a small parameter which we are allowed to expand in. It is convenient to put d = k - k'. We will need the magnitude of d, so let's calculate this first.

    In elastic collisions |k| = |k'|. It thus follows that:

    d^2 = (k - k')^2 = k^2 + k'^2 - 2 k dot k' =
    2k^2 - 2 k^2 cos(theta_s) (3)

    Here I used the notation vector^2 = inner product of the vector with itself, and theta_s is the scattering angle, which in this case is the angle between k and k'. Using the trigonometric formula:

    cos(2 x) = 1-2 sin^2(x)

    you find that:

    |d| = 2 |k| sin(theta_s/2) (4)

    Now let's expand the exponential in (2) to second order:

    Exp[i d dot r] = 1 + i d dot r -1/2 (d dot r)^2+... (5)

    The term i d dot r vanishes when integrated over r. Using (4) the third term can be written as:

    -1/2 (|d||r| cos(theta))^2= -2 k^2 sin^2(theta_s) r^2 cos^2(theta) (6)

    Here theta is the angle between r and d.

    Let's now perform the integration in (2) by substituting the expansion (5) in (2) and using (6). The first term is just the integral of V(r). The second term vanishes, and the third term is:

    - 2 k^2 sin^2(theta_s)Integral[d^3r r^2 cos^2(theta) V(r)]

    Simplify the integral by evaluating the theta integral. In spherical coordinates d^3r = r^2 d|r| dtheta dphi sin(theta). Since V only depends on |r| you can evaluate the theta integral. If you rewrite the result in terms of

    I_2 = Integral[4 pi r^4 V(r) d|r|] = Integral[d^3r r^2 V(r)]

    you find:

    T(k',k)= I_0 - 2/3 k^2 Sin^2(theta_s/2)I_2 (7)

    According to (1) we need the square of this and then integrate that over the scattering direction. In the square keep only the leading and nest to leading terms (remember that k^2 times I_2 is a small parameter). So:

    T^2 = (I_0)^2 - 4/3 I_0 I_2 k^2 Sin^2(theta_s/2) (8)

    (I've suppressed the arguments k and k' of T)

    The cross section is proportional to the integral of T^2 over all angles and this is:

    Integral[dtheta_s dphi sin(theta_s)T^2] =

    4 pi (I_0)^2 [1-2/3 k^2 I_2/I_0] (9)

    This is the desired low energy expansion because the energy is proportional to k^2. You need to plug in the factor m^2/((2 pi)^2 h-bar^4) from (1) to get the final result.

    To find the cross section of an electron in case the range of the potential is rb = 1 Bohr radius and V = 1 eV inside this range you can take I_0 to be 4/3 pi r_b^3 times 1 eV. Then, in the low energy limit (ignoring the second term in (9)) you should find that the cross section is about 2 times 10^(-19) cm^2.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:08 pm ad1c9bdddf>