Question

Why does 238U not spontaneously decay by emitting a proton?

A. 238U --> 237Pa + 1H would not conserve the number of protons. There would have to be a beta decay.
B. 238U --> 237Pa + 1H would not conserve the number of neutrons. There would have to be a beta decay.
C. No energy would be released in that reaction. Energy would have to be added to make this reaction possible.
D. 238U does spontaneously decay by emitting a proton.
E. No nucleus can decay by spontaneously emitting a proton.

Answer 1

238U does not spontaneously decay by emitting a proton due to energy considerations; alpha decay to 234Th is more energetically favorable and leads to a more stable nuclide.

Step-by-step explanation:

The reason 238U does not spontaneously decay by emitting a proton is related to nuclear stability and energy considerations. In nuclear reactions, energy is released when a nucleus decays into a more stable configuration. For 238U, the emission of an alpha particle (2 protons and 2 neutrons), which results in the formation of 234Th, is a more favorable process in comparison to the emission of a single proton.

Proton emission from a nucleus is less common and often seen in nuclei with an extreme excess of protons, making this type of decay energetically favorable for such isotopes. This is not the case for 238U, which has a neutron to proton ratio that is conducive to alpha decay. Therefore, the most likely decay process for 238U is alpha decay, which is more energetically favorable and contributes to a lower energy state of the resulting nuclide.