in alpha decay, alpha particles are ejected from the nucleus. which equation represents alpha…

in alpha decay, alpha particles are ejected from the nucleus. which equation represents alpha decay?\n$_{95}^{241}am\\rightarrow_{93}^{237}np + _{2}^{4}he$\n$_{9}^{18}f\\rightarrow_{8}^{18}o+_{1}^{0}e$\n$_{6}^{14}c\\rightarrow_{7}^{14}n+_{ - 1}^{0}e$\n$_{66}^{152}dy\\rightarrow_{66}^{152}dy + y$

in alpha decay, alpha particles are ejected from the nucleus. which equation represents alpha decay?\n$_{95}^{241}am\\rightarrow_{93}^{237}np + _{2}^{4}he$\n$_{9}^{18}f\\rightarrow_{8}^{18}o+_{1}^{0}e$\n$_{6}^{14}c\\rightarrow_{7}^{14}n+_{ - 1}^{0}e$\n$_{66}^{152}dy\\rightarrow_{66}^{152}dy + y$

Answer

Brief Explanations:

In alpha - decay, an alpha particle (${2}^{4}He$) is emitted from the nucleus. The atomic number of the parent nucleus decreases by 2 and the mass number decreases by 4. In the equation ${95}^{241}Am\rightarrow_{93}^{237}Np + {2}^{4}He$, an alpha particle (${2}^{4}He$) is emitted, with the atomic number of Americium ($Am$) decreasing from 95 to 93 (forming Neptunium ($Np$)) and the mass number decreasing from 241 to 237. The other equations do not involve the emission of an alpha - particle.

Answer:

${95}^{241}Am\rightarrow{93}^{237}Np + _{2}^{4}He$