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What is the distance of closest approach of an alpha particle to a silver nucleus is the kinetic energy of the alpha particle is 0.40 MeV.

Answer»


Solution :The APPROACH distance of an alpha-particle to the nucleus will be a minimum in the CASE of a central collision, when the ENTIRE kinetic energy of the alpha-particle is transformed into potential energy: `K=U=(Z_(1)Z_(2)e^(2))/(4piepsi_(0)r),"where "Z_(1)andZ_(2)` are the respective ATOMIC numbers.


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