A Uranium-(238,92) nucleus decays into a Thorium-234 nucleus by the emission of an alpha-particle. Given Thorium has a chemical symbol Th build a nuclear equation.

A nuclear equation expresses all of the elements before (LHS of = sign) and after (RHS of = sign) the reaction in terms of nucleon (N) and proton number (A), both quantities are conserved during the reaction. Uranium has the symbol U and both N=238 and A=92 are provided. Thorium has the symbol Th, only N=234 is provided not A. Alpha particle has the symbol α, neither N nor A are provided.Before the reaction you just have a Uranium-238 nucleus, expressed as U(238,92), 1st number is nucleon number and 2nd number is proton number. After the reaction you have thorium-234 nucleus, expressed as Th(234,y) and an alpha particle.Given an alpha particle is a nucleus of two neutrons and two protons, it has a nucleon number and proton number of 4 and 2 respectively, it can expressed with an alpha symbol as α(4,2).Note that the Th proton number is currently expressed as an unknown y.Hence the nuclear decay equation becomes .....U(238,92)→ T(234,y)+α(4,2)Solve for y be recalling that the proton number is conserved during the reaction, hence you can form the proton number equation 92=y+2, where y=90. So the final nuclear equation is......U(238,92)→ T(234,90)+α(4,2) Final Answer