Question

Go back to the derivation of Bohrs formula for quantized orbits. In reality, the electron is not merely moving around the proton, rather the two particles are moving around the center of mass and we can reduce the problem to that of a single particle (of reduced mass) moving around this center of mass. This is what you did in Mechanics course to solve two body central force problem. Explain why did then we put mass of electron in all our formulas instead of reduced mass. Positronium is an atom made up of a positive electron (a positron, the anti-particle of electron) and a usual negative electron moving around each other. For this atom, find the Bohrs radius and first few frequencies of “Balmer” series, i.e. when this atom makes transitions to n = 2 orbit from n = 3, 4, 5 orbits. Compare the frequencies with Hydrogen atom.

Answer 1

The reduced mass is the ”effective” inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem.

In case of Hydrogen atom, electron and proton are moving around the center of mass. As we know that the mass of proton is very large as compare to the mass of electron, therefore center of mass shifted nearer to the proton. Infact proton is at the position of the center of mass and may be considered as stationary, while electron moving around it. Reduced mass for such a system can be calculated by using the equation,

Reduced mass = μ = (m_em_p)/(m_e + m_p),

where m_p >> m_e, ⇒ m_e + m_p ≈ m_p. Thus reduced mass will become,

μ = (m_em_p)/(m_e + m_p),

μ ≈ (m_em_p)/m_p

µ ≈ m_e.

From above we can see that the reduced mass is approximately equal to the mass of electron, that is why in our formulas we put mass of electron instead of reduced mass. Now let’s consider the positronium. Positronium is made up of a positron (anti particle of electron) and an electron. Since electron and positron both have the same masses the center of mass will be at the center of the line joining the two particles. In this case the reduced mass is given by,

Substitute values of constants, we get,

r = (2 x (1.05 x 10^-34J.s)²/(9 x 10⁹Nm²/C² x 9.1 x 10^-31kg x (1.6 x 10^-19C)²

= 1.05 x 10^-10m

= 1.05Å

Numerical value for Bohr’s radius for hydrogen atom is 0.529 ˚A. Therefore Bohr’s radius for positronium is increased by 2 times. As we know that Rydberg constant in case of hydrogen atom is given by,

Since Rydberg constant for positronium is reduced to half, formula for Balmer series for positronium is given by,

For Hydrogen atom frequencies are calculated as below. 

f₁ = 2f'₁ = 2 × 2.28 × 10¹⁴ Hz = 4.57 × 10¹⁴ Hz 

f₂ = 2f'₂ = 2 × 3.08 × 10¹⁴ Hz = 6.17 × 10¹⁴ Hz 

f₃ = 2f'₃ = 2 × 3.46 × 10¹⁴ Hz = 6.92 × 10¹⁴ Hz.