Question

Write a nuclear equation for the alpha decay of 231
91 Pa.

Answer 1

Answer:

231-91Pa>4-2a+227-89Ac

Explanation:

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Answer 2

The equation for the alpha decay of  $^{231} _{91}Pa$ is written as $^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac$.

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle.

The emission of this alpha particle reduces the mass number of the original atom by 4 and the atomic number by 2.

The alpha decay equation is written as;

$^a_zX\ ---> \ ^4_2\alpha \ + \ ^{a-4} _{z-2}Y$

When the given atom ($^{231} _{91}Pa$) under goes alpha decay, the resulting element is written using the following equation;

$^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac$

Thus, the equation for the alpha decay of  $^{231} _{91}Pa$ is written as $^{231}_{91}Pa\ ---> \ ^4_2\alpha \ + \ ^{227} _{89}Ac$.

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