Question

Uranium-233 undergoes alpha decay, and then the daughter isotope undergoes another alpha decay. Which equation correctly describes this decay series? Use the periodic table link from the tools bar to answer the question.

Answer 1

²³³₉₂U → ²²⁹₉₀Th + ⁴₂He

²²⁹₉₀Th → ²²⁵₈₈Ra + ⁴₂He

Answer 2

Explanation:

Alpha decay: In this process, alpha particles is emitted when a heavier nuclei decays into lighter nuclei. The alpha particle released has a charge of +2 units.

The general alpha decay reaction is given as:

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

$_{92}^{233}\textrm{U}\rightarrow _{90}^{229}\textrm{Th}+_2^4\alpha$

$_{90}^{229}\textrm{Th}\rightarrow _{88}^{225}\textrm{Ra}+_2^4\alpha$

When Uranium-233 undergoes alpha decay it splits into an alpha particle and thorium-229.Thorium-229 further undergoes alpha decay to produce radium-225 along with an alpha particle.