Energy of a particle

  1. Feb 23, 2006 #1
    A particle of mass 'm' is moving in a circular orbit under the influence of the potential [itex]V(x) = \frac{ar^4}{4}[/itex] where 'a' is a constant. Given that the allowed orbits are those whose circumference is [itex]n\lambda[/itex], where 'n' is an integer and [itex]\lambda[/itex] is the de-Broglie wavelength of the particle. Obtain the energy of the particle as a function of 'n' and [itex]\lambda[/itex].

    So,
    [tex]2\pi r_n = n\lambda[/tex]
    I don't understand how the potential of the orbit comes into the picture here. Isn't PE = 2KE for Bohr orbits? How is the energy calculated?
     
  2. jcsd
  3. Feb 23, 2006 #2
    Total energy is always potential plus kinetic. From general central motion techniques, how do you find the total energy of an orbit when the central potential goes as r^4?

    -Dan
     
  4. Feb 24, 2006 #3
    I don't have a clue. The force can be shown as [itex]\vec F = -\vec \nabla V[/itex]. The work done will be: [itex]W = \int \vec F \cdot dr[/itex]. Will this equal the energy?
     
    Last edited: Feb 24, 2006
  5. Feb 24, 2006 #4
    [tex]T=(1/2)mv^2[/tex]
    Now, what is this is polar coordinates? Well, [tex]v^2=\dot{r^2}+r^2 \dot{\theta^2}[/tex]

    And, of course, E = T + V(r)...

    -Dan

    BTW, you can get rid of the [tex]\dot{\theta}[/tex] by using angular momentum conservation.
     
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