- #1

- 1,026

- 0

## Homework Statement

In 1989, a new type of exotic atom called protonium was produced. Its

structure is hydrogen-like, with the electron replaced by an antiproton. As

the antiparticle of the proton, an antiproton has the same mass but

opposite charge. In this problem, you should consider only the

electromagnetic interaction between the antiproton and the proton, by

analogy with ordinary hydrogen. The mass of the proton is 1836 times

larger than the mass of an electron.

2)

What is the energy in eV of a photon emitted when protonium

undergoes a transition from the state with principal quantum

number n = 4 to the state with n = 2?

## Homework Equations

[tex] E_n = \frac{E_r}{n} [/tex]

[tex] E_r = -\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2 [/tex]

## The Attempt at a Solution

Okay, so I have done this question, but my answer seems really large...

using:

[tex] \Delta E_{4 > 2} = \frac{E_r}{4^2} - \frac{E_r}{2^2} [/tex]

using the constants:

m = 1.67 e-27

hbar = 1.054e-34

e = -1.6e-19

epsilon-naught = 8.85e-12

I have gotten the Rydberg energy, [tex] E_r = 1.73 * 10^{13} [/tex]

I have thus got a value for the energy shift of -3.24 e12 Joules, which is [tex]2 *10^{31}[/tex] eV. This seems to big. does this answer seem reasonable...?