Question

which nuclear reaction modeled in the equations below would release an alpha decay particle?
$^{14}_{6}c\
ightarrow^{14}_{7}n + ___$
$^{42}_{19}k\
ightarrow^{42}_{19}k + ___$
$^{226}_{88}ra\
ightarrow^{222}_{86}rn + ___$
$^{2}_{1}h+^{2}_{1}h\
ightarrow^{3}_{2}he + ___$

Explanation:

Step1: Recall alpha - decay definition

Alpha - decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle ($^{4}_{2}He$). In alpha - decay, the mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2.

Step2: Analyze each option

For $^{14}_{6}C
ightarrow^{14}_{7}N+...$, the mass number remains the same (14), so it's not alpha - decay.
For $^{42}_{19}K
ightarrow^{42}_{19}K+...$, the mass number and atomic number of the parent and daughter nuclei are the same, so it's not alpha - decay.
For $^{226}_{88}Ra
ightarrow^{222}_{86}Rn+...$, the mass number of the parent nucleus ($^{226}_{88}Ra$) is 226 and of the daughter nucleus ($^{222}_{86}Rn$) is 222 (a decrease of 4), and the atomic number of the parent is 88 and of the daughter is 86 (a decrease of 2). This is an alpha - decay, and the emitted particle is $^{4}_{2}He$.
For $^{2}_{1}H+^{2}_{1}H
ightarrow^{3}_{2}He+...$, this is a fusion reaction, not alpha - decay.

Answer:

$^{226}_{88}Ra
ightarrow^{222}_{86}Rn+^{4}_{2}He$