Question

Show that m_eff ~ m_H for HI.

Answer 1

m_eff or reduced mass is given by the formula (m₁m₂)/(m₁+m₂). In case of HI molecule, taking the mass of H to be 1.008amu and that of I to be 126.90447amu, the reduced mass is expressed in terms of amu as (1.008 x 126.90447)/(1.008 + 126.90447) = 1.000 amu which is approximately that of mass of H. In this equation, the reduced mass can also be calculated in terms of kilograms or grams by dividing the mass in atomic units by the Avagadro number. On the whole, getting N_A²/N_A in the equation, we can simplify the result in amu into grams as m_eff (in g/mol) divided by N_A (moles) which results in 1 amu as 1/(6.023x10²³) = 1.6603x10^-24g which is also approximately equal to 1.008/(6.023x10²³) = 1.6736x10^-24g.

Therefore, we can say that the reduced mass of HI is approximately equal to the mass of H atom.