Answer to Question #84465 in Molecular Physics | Thermodynamics for Ra

Question

The net energy of a system of two atoms is given by E(r)={(-A/r^2)+(B/r^8)} where r is the inter-atomic separation. If in a stable molecule the interatomic separation is 0.5 nm and the dissociation energy is 4.0eV, calculate the constants A and B.

Answer 1

The inter-atomic separation r0 in a stable molecule is determined by the position of the minimum of the net energy, and the dissociation energy Edis is the difference between the net energy at infinite inter-atomic separation (which is equal to zero) and the net energy of the stable molecule: Edis = – E(r0). To find the position r0 of the minimum of the net energy, we differentiate the expression for the net energy with respect to r and equate the result to zero: E′(r) = 2A/r3 – 8B/r9 = 0, whence we find

r_0=(4B/A)^(1⁄6) (1)

and E(r0) = – 3 A4/3/(162/3B1/3), so that

E_dis=(3A^(4⁄3))/(〖16〗^(2⁄3) B^(1⁄3) ) . (2)

We need to express A and B in terms of r0 and Edis. From equation (1), we have B=A (r_0^6)⁄4. Substituting this into (2), we obtain E_dis=3 A⁄((4r_0^2 ) ), whence A=4E_dis (r_0^2)⁄3, and B=E_dis (r_0^8)⁄3. Substituting here the values for r0 = 0.5 nm = 5 × 10–10 m and Edis = 4.0 eV, we obtain the values of the constants: A = 1.3 × 10–18 eV m2 and B = 5.2 × 10–75 eV m8.

Answer: A = 1.3 × 10–18 eV m2 , B = 5.2 × 10–75 eV m8 .