Question

1. $_{91}^{231}pa\

ightarrow _{2}^{4}he + _{89}^{227}\\\\2. $_{95}^{241}am\
ightarrow \\\\3. $_{94}^{292}pu\
ightarrow \\\\4. \\\\ \
ightarrow _{2}^{4}he + _{86}^{222}rn$

Explanation:

Step1: Apply conservation of mass - number and atomic - number in nuclear reactions.

In a nuclear reaction, the sum of mass - numbers on the left - hand side equals the sum of mass - numbers on the right - hand side, and the sum of atomic - numbers on the left - hand side equals the sum of atomic - numbers on the right - hand side.

For the first reaction:

The mass - number of $^{231}_{91}Pa$ is 231 and the atomic - number is 91. The mass - number of $^{4}_{2}He$ is 4 and the atomic - number is 2. Let the unknown nuclide be $^{A}_{Z}X$.
Mass - number: $231=4 + A$, so $A = 231 - 4=227$.
Atomic - number: $91=2 + Z$, so $Z = 91 - 2 = 89$. The element with atomic - number 89 is Actinium (Ac). The complete reaction is $^{231}_{91}Pa
ightarrow^{4}_{2}He+^{227}_{89}Ac$.

For the second reaction:

Let the product of the alpha - decay of $^{241}_{95}Am$ be $^{A}_{Z}X$.
Mass - number: $241=4 + A$, so $A = 241 - 4 = 237$.
Atomic - number: $95=2+Z$, so $Z = 95 - 2=93$. The element with atomic - number 93 is Neptunium (Np). The reaction is $^{241}_{95}Am
ightarrow^{4}_{2}He+^{237}_{93}Np$.

For the third reaction:

Let the product of the alpha - decay of $^{292}_{94}Pu$ be $^{A}_{Z}X$.
Mass - number: $292=4 + A$, so $A = 292 - 4=288$.
Atomic - number: $94=2 + Z$, so $Z = 94 - 2 = 92$. The element with atomic - number 92 is Uranium (U). The reaction is $^{292}_{94}Pu
ightarrow^{4}_{2}He+^{288}_{92}U$.

For the fourth reaction:

Let the parent nuclide be $^{A}_{Z}X$.
Mass - number: $A=4 + 222=226$.
Atomic - number: $Z=2 + 86 = 88$. The element with atomic - number 88 is Radium (Ra). The reaction is $^{226}_{88}Ra
ightarrow^{4}_{2}He+^{222}_{86}Rn$.

Answer:

1. $^{231}_{91}Pa

ightarrow^{4}_{2}He+^{227}_{89}Ac$

1. $^{241}_{95}Am

ightarrow^{4}_{2}He+^{237}_{93}Np$

1. $^{292}_{94}Pu

ightarrow^{4}_{2}He+^{288}_{92}U$

1. $^{226}_{88}Ra

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