Question
A positronium atom is a bound state of an electron (e–) and its antiparticle, the positron (e+) revolving round their centre of mass. In which part of the em spectrum does the system radiate when it de-excites from its first excited state to the ground state?
Solution
1. The motion of nucleus is ignored in an ordinary atom as the first approximation because the atom is too heavy.
2. In a positronium atom, a positron replaces proton of hydrogen atom. As electron and positron masses are equal, the motion of the positron cannot be ignored.
3. The motion of electron and positron about their centre of mass is considered.
4. A detailed analysis will show that formulae of Bohr model apply to positronium atom provided that we replace mass of electron (me ) by what is known as reduced mass of the electron. The reduced mass is me/2 for positronium atom.
5. In the transition from n = 2 to n = 1, the wavelength of radiation emitted is double than that of the corresponding radiation emitted for a similar transition in hydrogen atom, which has a wavelength of 1217 Å and, is equal to 2 × 1217 = 2434 Å.
6.This radiation lies in the ultra-violet part of the electromagnetic spectrum.
2. In a positronium atom, a positron replaces proton of hydrogen atom. As electron and positron masses are equal, the motion of the positron cannot be ignored.
3. The motion of electron and positron about their centre of mass is considered.
4. A detailed analysis will show that formulae of Bohr model apply to positronium atom provided that we replace mass of electron (me ) by what is known as reduced mass of the electron. The reduced mass is me/2 for positronium atom.
5. In the transition from n = 2 to n = 1, the wavelength of radiation emitted is double than that of the corresponding radiation emitted for a similar transition in hydrogen atom, which has a wavelength of 1217 Å and, is equal to 2 × 1217 = 2434 Å.
6.This radiation lies in the ultra-violet part of the electromagnetic spectrum.
