Question #8fb72

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SCooke
Nov 19, 2017

It loses 2 protons with the alpha particle, leaving it with an excess of 2 electrons compared to its new nucleus.

Explanation:

The common #U^(238)# decay to #Th^(234)# is by emission of an alpha particle (#alpha#).

An alpha particle is defined as a positively charged particle of a helium nuclei. An alpha particle is composed of two protons and two neutrons, so it can be represented as a Helium-4 atom. As an alpha particle breaks away from the nucleus of a radioactive atom, it has no electrons, so it has a +2 charge. Therefore, it’s a positively charged particle of a helium nuclei.
http://www.dummies.com/education/science/chemistry/the-process-of-natural-radioactive-decay/

Further in the decay sequence, #U^(234)# decay to #Th^(230)# is also by emission of an alpha particle.

http://legacy.jefferson.kctcs.edu/techcenter/Classes/Physics/AtomicNuclearandModernPhysics/AtomicNuclearandModernPhysics7.html

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