1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Energy of a particle
  1. Particle Energy (Replies: 1)

Loading...