The bond energy for the van der waals bond between two helium atoms is 7.9*10-4ev. assuming that the average kinetic energy of a helium atom is (3/2) kbt, at what temperature is the average kinetic energy equal to the bond energy between two helium atoms? use kb=8.62*10-5ev/k.

Given: van der Waal's bond energy of He = 7.9*10-4ev;

and kb=8.62*10-5ev/k

Now, according to Kinetic theory of gases we know that,

K. E = 3/2 (kb) T

where T = temperature

∴ T = (2 K. E) / (3 x kb)

Now for K. E = 7.9*10-4ev

T = (2 X 7.9*10-4) / (3 X 8.62*10-5) = 6.1 K

Thus, at temperature of 6.1 K, average kinetic energy equal to the bond energy between two helium atoms