Question

I need a solution for this problem from Thermal Physics-Blundell and Blundell

(6.5) (a) Given that the number of molecules hitting unit area of a surface per second with speeds between v and u + du and angles between θ and θ+ dθ to the normal is n f (v)dv sin θ cos θ do. 2 show that the average value of cos θ for these molecules is (b) Using the results above, show that for a gas obeying the Maxwellian distribution (i.e. f(v) oc u2e-m /2k) the average energy of all the molecules is kBT, but the average energy of those which hit the surface is 2kpT. 3

Answer 1

Solution is given in the attached image.

Final answers are boxed. Integrals involved are not solved explicitly.

< cos theta > = ? < cos theta > = integral^pi/2_0 sin theta cos^2 theta d theta times 1/2 integral V_n f(v) dv = integral^pi/2_0 sin theta cos^2 theta d theta/integral^pi/2_0 sin theta cos d theta = 1/3/1/2 = 2/3 < v^2 > =/ < v^2 > = integral^infinity_0 v^2 times v^2 e^-1/2 mv^2/k_Beta T dV/integral^infinity_0 V^2 e^-1/2 mv^2/k_Beta T dV = 3k_Beta T/m < 1/2 mv^2 > = 3/2 k_Beta T for those hitting the surface, there is an extra factor of V, < V^2 > = integral^infinity_0 v^2 V^3 e^-1/3 mv^2/k_Beta T dV/integral^infinity_0 v^3 e^-1/2 mv^2/k_Beta T dV = 4k_Beta T/M < 1/2 mv^2 > = 2k_Beta T