1861. Запишите реакцию радиоактивного распада плутония, в результате которого $${}^{239}_{94}Pu$$ превращается в уран $${}^{235}_{92}U$$.

The problem asks for the nuclear reaction where Plutonium-239 ($${}^{239}_{94}Pu$$) decays into Uranium-235 ($${}^{235}_{92}U$$). To solve this, we need to determine what particle is emitted during this decay. Nuclear reactions must conserve both mass number (the superscript) and atomic number (the subscript). Let the unknown emitted particle be represented by $${}^{A}_{Z}X$$, where A is the mass number and Z is the atomic number. The reaction can be written as: $${}^{239}_{94}Pu \rightarrow {}^{235}_{92}U + {}^{A}_{Z}X$$ To find A, we balance the mass numbers: $$239 = 235 + A$$ $$A = 239 - 235$$ $$A = 4$$ To find Z, we balance the atomic numbers: $$94 = 92 + Z$$ $$Z = 94 - 92$$ $$Z = 2$$ The particle $${}^{A}_{Z}X$$ has a mass number of 4 and an atomic number of 2. This corresponds to an alpha particle, which is a helium nucleus ($${}^{4}_{2}He$$). Therefore, the nuclear reaction is: $${}^{239}_{94}Pu \rightarrow {}^{235}_{92}U + {}^{4}_{2}He$$

Реакция:

• $${}^{239}_{94}Pu \rightarrow {}^{235}_{92}U + {}^{4}_{2}He$$

Ответ: $${}^{239}_{94}Pu \rightarrow {}^{235}_{92}U + {}^{4}_{2}He$$