Question

Question 5 The rotational energy levels of a diatomic molecule are given by E,= BJ(J+1) with B the rotational constant equal to 8.02 cm Each level is (2) +1)-times degenerate. (wavenumber units) in the present case (a) Calculate the energy (in wavenumber units) and the statistical weight (degeneracy) of the levels with J =0,1,2. Sketch your results on an energy level diagram. (4 marks) (b) The characteristic rotational temperature is defined as where k, is the Boltzmann constant. Calculate the value of , in kelvin. (2 marks) (c) (0) Use the Maxwell-Boltzmann distribution to calculate the ratios T = 20K where N, N, and N, are the populations of the levels with / 0,7 = 1 and J=2, respectively. (3 marks) (1) Find an approximation for the rotational partition function at T = 20 K if it is given by the following expansion and hence, calculate the partial pressures P. (due to molecules in N) and P. (due to molecules in N) of a gas composed of a molecules per unit volume (number density) if the total pressure of the gas, considered to be ideal, ispank, 1 Pa. (8 marks) –9x5)- GS- (d) Calculate an expression for the value (1 ) of the most populated rotational level of population N, ). Estimate at T = 20 K and T = 2000 K. Hint: use the Maxwell- Boltzmann distribution to express the ratic in terms of the rotational partition function (N is the total population) and then solve the equation O for J. (8 marks)

Answer 1

- 1(3+ degeneracy and = (25+1) 7 for J=1, E = B*1*(1 +1) = 16.04 cm strength = - 8.02x2 (2x1+1) = 3 statistical weight = (2x0+1) = - For T-2 2 - - 6x2 (2+1) = 6x8.02 = 48-12 cm Bx2(2 +1) = 6 statistical weight = (2x2 + 1) = 5 Rolational temperature or= o ho B= 8.02 cm = 802 m B = 802the youle = 802 x 6.626x1034 x3x108 = 1.5942 x 10-22 s. 15942 x1622 1:38x10 23 = 11055k - co s According to Maxwell Accordion to Relative population Boltzman distribution, N =é Eslket - BJ(J+1) he KT N 8.02x1x(1+1)x6.626x10s x3x1010 - 1.35x10 23 x 20 - 0.315 8.02 x 2 x (2+1) X6.626x103Y Xaxido 1038x10 23 x 20 = 0.03125