Question

Which answer choice represents a balanced alpha emission nuclear equation?
Superscript 52 subscript 26 upper F e right arrow superscript 52 subscript 25 upper M n plus superscript o subscript plus 1 e.
Superscript 235 subscript 92 upper U right arrow superscript 239 subscript 94 upper P u s plus superscript 4 subscript 2 upper H e.
Superscript 189 subscript 83 upper Bi right arrow superscript 185 subscript 81 upper T l plus superscript 4 subscript 2 H e.
Superscript 14 subscript 6 upper C right arrow superscript 14 subscript 7 upper N plus superscript 0 subscript negative 1 e.

Answer 1

The third choice: 189/83 Bi → 185/81 Tl + 4/2 He.

Check: mass numbers 189 = 185 + 4, atomic numbers 83 = 81 + 2. That matches alpha emission (alpha = 4/2 He). The others are positron decay (1), an unbalanced equation (2), and beta‑minus decay (4).