Question

Gaseous HBr has an absorption band centered at about 2645/cm consisting of a series of line approximately equally spaced with an interval of 16.9/cm. For gaseous DBr estimate the frequency in wave number of the band center and the interval between lines.

Answer 18 828.56 /cm

Answer 1

The rotation constant Bv is given by, Bv = h/(8*pi²*c*mu*d²), mu - reduced mass of the molecule. The difference between HBr and DBr is in the reduced mass and all other parameters are same for both the molecules and frequency of DBr and HBr are related by,

nu(HBr)/nu(DBr) = {mu(DBr)/mu(HBr)}^1/2 and the spacing is related by S(HBr)/S(DBr) = mu(DBr)/mu(HBr)

Therefore, the frequency of DBr absorption, nu(DBr) = nu(HBr)/(1.951/0.988)^1/2 = 2645*0.7116 = 1882/cm and the spacing is 16.9/(1.951/0.988) = 8.558/cm.