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26 December, 20:39

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.

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  1. 26 December, 22:48
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    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
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