What Are the Characteristics of Protonium?

[stunner5000pt]

Protonium consists of Proton and an antiproton in a bound state

first of all waht is a bound state - is it a proton orbiting an antiproton or vice versa??

a) what is teh Binding energy of protonium?
do i use this formula??

$$E = \frac{ \mu e^4}{8 \epsilon_{0}^2 h^2}$$

b) How far apart are the antiproton and proton in this ground state??

$$r = \frac{\epsilon_{0} h^2}{\mu \pi Z e^2} n^2$$

where $$\mu = \frac{m_{p}^2}{2m_{p}} = \frac{m_{p}}{2}$$

c) What is teh angular momentum of the proton or antiproton in the 3rd excited state?

so the 3rd excited state is n = 4

$$L = \mu v r = n \frac{h}{2 \pi}$$

and i find v using the formula given??

andi right so far if you could correct me on any of these please please tell me!

also i coul also use your help for this thread too, similar topic question!

https://www.physicsforums.com/showthread.php?t=52658

 

[anathema]

Hello,

Thank you for your questions. I can provide you with some answers and clarifications.

Firstly, a bound state refers to a system of particles that are held together by a force, such as electromagnetic or nuclear forces. In the case of protonium, it is a bound state of a proton and an antiproton, where they are held together by the strong nuclear force.

a) The binding energy of protonium can be calculated using the formula you provided, where \mu is the reduced mass of the system, e is the elementary charge, \epsilon_{0} is the permittivity of free space, and h is Planck's constant. This formula gives the energy required to break the bond between the proton and antiproton.

b) The distance between the proton and antiproton in the ground state can be calculated using the formula you provided, where r is the distance, \epsilon_{0} is the permittivity of free space, h is Planck's constant, \mu is the reduced mass, Z is the atomic number (in this case, Z = 1 for both proton and antiproton), e is the elementary charge, and n is the principal quantum number. For the ground state (n = 1), the distance will be the smallest, and it will increase as n increases.

c) The angular momentum of the proton or antiproton in the 3rd excited state can be calculated using the formula you provided, where L is the angular momentum, \mu is the reduced mass, v is the velocity, r is the distance, and n is the principal quantum number. For the 3rd excited state (n = 4), the angular momentum will be larger than in the ground state (n = 1).

I have checked the thread you provided, and it seems like you have the right understanding of the topic. However, I would recommend double-checking the formulas and units to ensure accuracy.

I hope this helps. Let me know if you have any further questions. Good luck with your studies!

 

[wais]

a) The binding energy of protonium can be calculated using the formula you provided. However, it is important to note that this formula is for the binding energy of a single electron to a nucleus, so it may not accurately represent the binding energy of a proton-antiproton pair. A more accurate calculation would involve solving the Schrodinger equation for the protonium system.

b) The formula you provided for the distance between the antiproton and proton in the ground state is correct. However, the value of \mu you have used is incorrect. It should be \mu = \frac{m_{p}m_{\bar{p}}}{m_{p}+m_{\bar{p}}} where m_{p} and m_{\bar{p}} are the masses of the proton and antiproton, respectively.

c) To find the angular momentum in the 3rd excited state, you can use the formula L = \mu v r = n \frac{h}{2 \pi}. However, you will need to use the correct value of \mu as mentioned above, and also substitute the value of n = 4.

I am not able to access the thread you mentioned, so I am unable to provide any help for it. However, I suggest seeking guidance from your instructor or a physics tutor for any further assistance with this topic.