Energy of Particle in Circular Orbit: Find n & \lambda

In summary: T=(1/2)mv^2Now, what is this is polar coordinates? Well, v^2=\dot{r^2}+r^2 \dot{\theta^2}And, of course, E = T + V(r)...-DanBTW, you can get rid of the \dot{\theta} by using angular momentum conservation.
  • #1
Reshma
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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?
 
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  • #2
Reshma said:
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?

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
 
  • #3
topsquark said:
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
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:
  • #4
Reshma said:
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?

[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.
 

What is the energy of a particle in a circular orbit?

The energy of a particle in a circular orbit is given by the equation E=-13.6eV/n^2, where n is the principal quantum number.

How do you find the principal quantum number (n) for a particle in a circular orbit?

The principal quantum number (n) can be found by using the Rydberg formula, which is given by n2 = R(1/λ), where R is the Rydberg constant and λ is the wavelength of the electromagnetic radiation emitted or absorbed by the particle.

What is the significance of the principal quantum number (n) in a circular orbit?

The principal quantum number (n) represents the energy level of the particle in the orbit. As n increases, the energy of the particle also increases, and the orbit becomes larger and less stable.

How does the energy of a particle in a circular orbit change with the change in principal quantum number (n)?

The energy of a particle in a circular orbit is inversely proportional to the square of the principal quantum number (n). This means that as n increases, the energy decreases, and as n decreases, the energy increases.

What is the relationship between the energy of a particle in a circular orbit and its wavelength (λ)?

The energy of a particle in a circular orbit is directly proportional to the wavelength (λ) of the electromagnetic radiation emitted or absorbed by the particle. This means that as the energy increases, the wavelength also increases, and vice versa.

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