Q.5.32 - An Introduction to General, Organic, and Biological Chemistry

Question

A sample of sodium-24 with an activity of 12 mCi was used to study the rate of blood flow in the circulatory system. If sodium-24 has a half-life of 15h, what is the activity after each of the following intervals?

a. one half-life

b. 30h

c. three half-lives

d. 2.5 days

Step-by-Step Solution

Answer

Part (a). $6 mCi$

Part (b). $3 mCi$

Part (c).$1.5 mCi$

Part (d). $0.75 mCi$ 

Step 1: Given Information Part (a)

Half-life is the time expected for an amount to decrease to half of its underlying worth. The term is ordinarily utilized in atomic physical science to portray how rapidly unsteady particles go through radioactive rot or how long stable iotas get by.

Step 2: Explanation Part (a)

Given,

Initial activity = $12 mCi$

Calculating the half life,

$N\left(t\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Activity after one half-life = $\frac{12}{2}=6 mCi$ 

Step 3: Explanation Part (b)

Given,

Initial activity = $12 mCi$

Calculating the half-life,

$N\left(t\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Activity after $30 h$ or $\frac{30}{15}=2$half-lives,

$\Rightarrow \frac{12}{2^2}=3 mCi$

Step 4: Explanation Part (c)

Given,

Initial activity = $12 mCi$

Calculating the half-life,

$N\left(t\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Activity after three half-lives = $\frac{12}{2^3}=1.5 mCi$

Step 5: Explanation Part (d)

Given,

Initial activity =$12 mCi$

Calculating the half-life,

$N\left(t\right)=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$

Activity after $2.5$days or $\frac{60}{15}=4$half lifes,

$\Rightarrow \frac{12}{2^4}=0.75 mCi$