Under certain circumstances, a nucleus can decay by emitting a particle heavier than an particle. Such decays are very rare. Consider the decays

We have to calculate the -values for these decays and determine that both are energetically possible. (b) The Coulomb barrier height for particles in this decay is 30 MeV. We have to find the barrier height for decay.

Solution:             Click For PDF Version

(a)

We will determine first whether the following reactions are energetically allowed:

We will use the data on values of the atomic masses given in the statement of the problem.

The for the reaction is given by

It is positive. Therefore, this reaction is allowed.

The for the reaction

is given by

It is positive. Therefore, this reaction is allowed.

(b)

The atomic number of is 88. It is given that the Coulomb barrier height for particles in this decay is 30 MeV. Let the radius of nucleus be m. The Coulomb barrier height for an particle when it tunnels out of nucleus is the potential energy between charge 86 e and 2 e separated by r m. That is

We will use the above result for finding the barrier height for decay. As the charge in a nucleus is 6 e, the height of the barrier for tunnelling of from nucleus will be