An element of nuclear number A and proton number Z emits an alpha particle from its nucleus. The nucleon number and the proton number of the new element formed are respectively
(A – 2) and (Z – 2)
(A + 4) and (Z + 2)
(A– 4) and (Z+2)
(A+ 4) and (Z – 2)
(A – 4) and (Z – 2)
Correct answer is E
Let the element be S and the element formed after emission be X.
\(\implies ^{A} _{Z}S \to ^{4} _{2}He + X\)
\(\therefore X = ^{A - 4} _{Z - 2}X\).