Question

E. The element 231/90Th decays to 231/91 Pa. Use the laws of conservation of charge and nucleon number to determine the decay particle emitted. Explain your reasoning.

Answer 1

Answer:

A negatron emission

Explanation:

We know that radioactivity orginates from instability of the nucleus. When the nucleus is unstable, radioactive emissions are produced in the form of any of these rays:

> Alpha particle emisson

>Beta particles

> Gamma rays

These emissions create a balance for a radioactive decay.

In balancing nuclear reactions we make sure that the charges on both sides must be conserved and that the mass number and atomic numbers conserved too. This means that the sum of mass number and atomic numbers on both side of the reaction must be equal.

The nucleons are the protons and neutrons, they add up to give the mass number. The atomic number is the proton number.

For the given radioactive reaction:

²³¹₉₀Th → ²³¹₉₁Pa + ?

From this equation, we see that the mass number is conserved but the atomic number is not.

The mass number is the superscript whereas the atomic number is the subscript.

Let's say the decay produces an emission of a particle denoted by X

²³¹₉₀Th → ²³¹₉₁Pa + ᵃₙX

What would the nature of X be?

For the charges and masses to be conserved, X must have mass number of 0 and an atomic number of -1.

Checking:

Mass number:

231 = 231 + a ( a is the mass number)

a = 231 - 231 = 0

Atomic number:

90 = 91 + n

n = 90- 91 = - 1

With X having a mass number of 0 and an atomic number of -1, we have a beta particle emission. Specifically, a negatron has been emitted.

A negatron is denoted as ⁰₋₁β which perfectly makes the equation conserved and suits the description of X.

The complete equation is thus written as:

²³¹₉₀Th → ²³¹₉₁Pa + ⁰₋₁β + energy