Question

Sketch the nuclear potential for protons and neutrons as a function of radial distance. Use your sketch to explain why it is possible to have more than one for a heavy element and why a decay often prevails over proton emission. proton-em itting isotope

Answer 1

We know, force F = -nabla U, where U = scalar potential F = 0 and hence, partial differential U/partial differential r = 0 gives us the point of equilibrium. The point at which the equilibrium is attained (approximately -50 MeV) is shown in the figure. An alpha particle is a Helium atom^3H e_2, i.e., two nucleons, So, from the figure, we can conclude that the alpha particle has a very high binding energy and hence very little energy is needed to emit an alpha particle. This energy is mostly supplied from the energy within the nucleus itself. Other decays like those of single proton emission require additional energy and also less probable to come to pass. The alpha decays cause the number of nucleons to diminish by 2 We can have more than one alpha decay resulting in more than one isotope of heavy element.

We know,

force $F=-\nabla U$, where U = scalar potential.

F = 0 and hence, $\frac{\partial U}{\partial r}=0$ gives us the point of equilibrium. The point at which the equilibrium is attained (approximately -50 MeV) is shown in the figure. An alpha particle is a Helium atom $^3He_2$, i.e., two nucleons. So, from the figure, we can conclude that the alpha particle has a very high binding energy and hence very little energy is needed to emit an alpha particle. This energy is mostly supplied from the energy within the nucleus itself. Other decays like those of single proton emission require additional energy and also less probable to come to pass.

The alpha decays cause the number of nucleons to diminish by 2. We can have more than one alpha decay resulting in more than one isotope of heavy element.