Question

Element 114 was recently named flerovium, symbol F. The reaction for its synthesis involves bombarding Pu-244 with Ca-48. Write the balanced nuclear equation for the synthesis of flerovium. $(5,2)$

Short answer

As a result, the potassium-42 sample's initial activity is$16.0\mu \mathrm{Ci}$

Step-by-step solution

Introduction

A radioisotope's half life is the amount of time it takes for one-half of a sample to decay.

Potassium-42 has a half-life of 24 hours. $12\mathrm{h}$.

The sample's activity is $2.0\mu \mathrm{Ci}$after $36\mathrm{h}$

Calculate a sample's starting activity in microcurie.

Conversion factors for half-life

Calculate half-life conversion factors.

1 half-life $=12\mathrm{h}$

$\frac{12\mathrm{h}}{1\text{half-life}}$and $\frac{1\text{half- life}}{12\mathrm{h}}$

Number of half-lives

Calculate the number of half-lives in the given amount of time.

After three half-lives, the amount of potassium-42 present is$2.0\mu \mathrm{Ci}$.

Substitute the value in the following scheme according to the definition of half-life period.

$16.0\mu \mathrm{CiK}-42\stackrel{1\text{lulflife}}{⟶}8.0\mu \mathrm{CiK}-42\stackrel{2\text{hall-lives}}{⟶}4.0\mu \mathrm{CiK}-42\stackrel{3\text{bafl-ive}}{⟶}2.0\mu \mathrm{CiK}-42$

As a result, the potassium-42 sample's initial activity is$16.0\mu \mathrm{Ci}$

Calculating the amount of sample

Calculate the amount of sample that was initially present:

The amount of potassium-42 after three half-lives is present $2.0\mu \mathrm{Ci}$.

Substitute the value in the following scheme according to the definition of half-life period.

$16.0\mu \mathrm{CiK}-42\stackrel{\text{I hufflife}}{⟶}8.0\mu \mathrm{CiK}-42\stackrel{2\text{hasllives}}{⟶}4.0\mu \mathrm{CiK}-42\stackrel{3\text{kalflive}}{⟶}2.0\mu \mathrm{CiK}-42$

As a result, the potassium-42 sample's initial activity is$16.0\mu \mathrm{Ci}$