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Question

# Calculate the height of the potential barrier for a head on collisionof two deuterons. (Hint: The height of the potential barrier is givenby the Coulomb repulsion between the two deuterons when theyjust touch each other. Assume that they can be taken as hardspheres of radius 2.0 fm.)

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Solution

## Given, the radius of deuteron nucleus is 2 fm. Let the distance between the centres of two deuterons be d. Formula for the distance dbetween the two deuterons is, d=radius of 1 deuteron +radius of 2 deuteron Substitute the values in the above equation. d=2× 10 −15 +2× 10 −15 =4× 10 −15  m Charge on a deuteron nucleus is e=1.6× 10 −19 C Potential energy of the two-deuteron system is given by, V= e 2 4π ε ° d Here, ε 0 is the permittivity of free space.. Substituting the values in the above equation, we get: V= e 2 4π ε ° d = ( 1.6× 10 −19 ) 2 4×π×8.85× 10 −12 ×4× 10 −15 =360 keV Thus, the height of the potential barrier of the two-deuteron system is 360 keV.

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