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