Suppose there exists an alpha particle in the nucleus. Within a radius of 2 femtometer, the dominating force is the nuclear force but beyond this radius, the Coulomb force becomes the effective dominating force due to it's longer range force. Outside this radius is a daughter nucleus.So let's suppose the alpha particle moves beyond the effective range of the nuclear force,the electric potential energy due to the force between the alpha particle and the daughter nucleus is given by as U B = 1 4 π ε 0 ( 2 ( Z − 2 ) e 2 R ) .Why is there the term 2 ( Z − 2 ) ?

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Suppose there exists an alpha particle in the nucleus. Within a radius of 2 femtometer, the dominating force is the nuclear force but beyond this radius, the Coulomb force becomes the effective dominating force due to it&#039;s longer range force. Outside this radius is a daughter nucleus.So let&#039;s suppose the alpha particle moves beyond the effective range of the nuclear force,the electric potential energy due to the force between the alpha particle and the daughter nucleus is given by as      U          B        =      1          4      π              ε                  0                      (            2      (      Z      −      2      )              e                  2                      R    )  .Why is there the term  2  (  Z  −  2  )     ?

percolarse2rzd · Accepted Answer

The electric potential between two point charges depends on the product of the charge of each.The charge in the nucleus is due to protons, and the count of protons is given as   Z for a total charge of   Z  e.The alpha decay removes two protons (and two uncharged neutrons). The charge of the alpha particle is   2  e, leaving a charge of   Z  e  −  2  e in the nucleus after decay. Multiplying the two and rearranging, that becomes   2  (  Z  −  2  )      e    2  .

Suppose there exists an alpha particle in the nucleus. Within a radius of 2 femtometer, the dominati

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