Question

Calculate the energies of the first four rotational levels of 1 H127I free to rotate in three dimensions, using for its moment of inertia I=µR2 , with µ=mHmI /(mH+mI ) and R=160 pm.

Answer 1

Reduced mass μ = m_Hm_I/(m_H+m_I)

= (1 x 127)/(1 + 127) x 1.661 x 10^-27 = 1.648 x 10^-27 kg

Momemt of inertia I = μR²

= 1.648 x 10^-27 x (160 x 10^-12)² = 4.219 x 10^-47 kg m²

Rotational constant B = h²/8π²I

= (6.626 x 10^-34)²/(8 x 3.1416² x 4.219 x 10^-47) = 1.32 x 10^-22 J

Rotational energy level: E_J = BJ(J + 1)

First four rotational levels are:

E₀ = 1.32 x 10^-22 x 0 x 1 = 0 J

E₁ = 1.32 x 10^-22 x 1 x 2 = 2.64 x 10^-22 J

E₂ = 1.32 x 10^-22 x 2 x 3 = 7.92 x 10^-22 J

E₃ = 1.32 x 10^-22 x 3 x 4 = 1.58 x 10^-21 J