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Dimensionless Radial Equation Hydrogen Atom

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that in terms of the dimensionless variable ##\xi## the radial equation becomes ##\frac{\mathrm{d}^{2} u}{\mathrm{d} \xi^{2}}=(\frac{l(l+1)}{\xi^{2}}-\frac{2}{\xi}-K)u##

    2. Relevant equations
    ##u(r)\equiv rR(r)##
    ##\xi \equiv \sqrt{2\mu U_{0}}\frac{r}{\hbar}## dimensionless variable
    ##K\equiv \frac{E}{U_{0}}##
    Radial equation: ##-\frac{\hbar^{2}}{2\mu}\frac{1}{r^{2}}\frac{\mathrm{d} }{\mathrm{d} r}(r^{2}\frac{\mathrm{d} }{\mathrm{d} r}R)+\frac{\hbar^{2}l(l+1)}{2\mu r^{2}}R+U(r)R=ER##
    and ##U(r)## is the coulomb potential
    3. The attempt at a solution
    The radial equation becomes:
    ##-\frac{U_{0}}{\xi^{2}}\frac{\mathrm{d} }{\mathrm{d} r}(r^{2}\frac{\mathrm{d} }{\mathrm{d} r}\frac{u}{r})+\frac{U_{0}l(l+1)}{\xi^{2}}\frac{u}{r}+U(r)\frac{u}{r}=E\frac{u}{r}##
    and after taking the derivatives and some canceling:
    ##-\frac{r}{\xi^{2}}\frac{\mathrm{d}^{2}}{\mathrm{d}r^{2}}u+\frac{l(l+1)}{\xi^{2}}\frac{u}{r}+\frac{U(r)}{U_{0}}\frac{u}{r}=K\frac{u}{r}##
    and so
    ##\frac{r^{2}}{\xi^{2}}\frac{\mathrm{d}^{2} }{\mathrm{d} r^{2}}u=(-K+\frac{l(l+1)}{\xi^{2}}-\frac{e^{2}}{4\pi \varepsilon _{0}rU_{0}})u##

    so two of the terms on the right are fine. I then substituted the ground state of hydrogen in for ##U_{0}## in the last term, and the only way that comes out correctly is if ##\mu## (the effective mass) ##=16m##. Is this the case? I don't really understand how ##\mu## is defined. and I don't understand how to integrate with respect to ##\xi## in the first term. Thank you so much in advance!

    edit: actually, not sure where I got the extra factor of 4 in there but ##\mu=m## what is the point of defining ##\mu##???
     
  2. jcsd
  3. Nov 7, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    μ is the reduced mass:
    $$
    \mu = \frac{m_1 m_2}{m_1+m_2}
    $$
    It appears when one separates the center-of-mass motion (which depends on the total mass) from the relative motion (which depends on the reduced mass) for a system of two particles.

    For a hydrogen atom, since the mass of the proton is 3 orders of magnitude bigger, one gets
    $$
    \mu = \frac{m_e m_p}{m_e+m_p} \approx \frac{m_e m_p}{m_p} = m_e
    $$
     
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