Chapter 21, Problem 21.54QP

Interpretation Introduction

Interpretation: The activity of a 0.500-g sample of ${}_{\text{93}}^{\text{237}}\text{Np}$ should be shown & the equation should be balance written with proper abbreviation for the decay of ${}_{\text{93}}^{\text{237}}\text{Np}$.

Concept Introduction:

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejected which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

$\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.

Given:

${}_{\text{93}}^{\text{237}}\text{Np}$ Isotope decay by ${\text{α}}_{\text{2}}^{\text{4}}$ - particle.

To determine: The activity of 0.500-g sample of ${}_{93}^{237}\text{Np}$ is calculated by using the given data.