Chapter 30, Problem 94A

To determine

The rank of radioactive samples on the basis of remaining original mass.

Answer to Problem 94A

  $B>D>C>A>E$

Explanation of Solution

Given:

A. Original mass of $\text{P}\text{u}$

  $N_o=20$ g

Time $t=2.85$ years

Half-life $t_{1/2}=2.85$ years

B. Original mass of $\text{P}\text{u}$

  $N_o=20$ g

Time $t=7.9$ years

Half-life $t_{1/2}=373300$ years

C. Original mass of $\text{P}\text{b}$

  $N_o=5$ g

Time $t=2$ days $=48$ hours

Half-life $t_{1/2}=10.6$ hours

D. Original mass of $\text{C}$

  $N_o=5$ g

Time $t=100$ years

Half-life $t_{1/2}=5730$ years

E. Original mass of $\text{P}\text{o}$

  $N_o=500$ g

Time $t=1$ days $=86400$ sec

Half-life $t_{1/2}=0.392$ sec

Formula used:

Half-life formula,

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

Where,

  $N=$ Remaining mass

  $N_o=$ Original mass

  $t_{1/2}=$ Half-life

  $t=$ Time

Calculation:

A. For $Pu$ ,

  $\begin{array}{l}N=20\times {\left(\frac{1}{2}\right)}^{2.85/2.85}\\ N=10\text{ g}\end{array}$

Percentage of decay:

  $\begin{array}{l}P=\left|\frac{N_o-N}{N_o}\right|\times 100\\ P=\left|\frac{20-10}{20}\right|\times 100\\ P=50\%\end{array}$

Percentage of remaining mass= 50%

B. For $Pu$

  $\begin{array}{l}N=20\times {\left(\frac{1}{2}\right)}^{7.9/373300}\\ N=20\times {\left(\frac{1}{2}\right)}^{0.000021}\\ N=19.9998\text{ g}\end{array}$

Percentage of decay:

  $\begin{array}{l}P=\left|\frac{N_o-N}{N_o}\right|\times 100\\ P=\left|\frac{20-19.9998}{20}\right|\times 100\\ P=0.001\%\end{array}$

Percentage of remaining mass= 99.999%

C. For $Pb$

  $\begin{array}{l}N=5\times {\left(\frac{1}{2}\right)}^{48/10.6}\\ N=5\times {\left(\frac{1}{2}\right)}^{4.52}\\ N=0.217\text{ g}\end{array}$

Percentage of decay:

  $\begin{array}{l}P=\left|\frac{N_o-N}{N_o}\right|\times 100\\ P=\left|\frac{5-0.217}{5}\right|\times 100\\ P=95.6\%\end{array}$

Percentage of remaining mass= 4.4%

D. For $C$

  $\begin{array}{l}N=5\times {\left(\frac{1}{2}\right)}^{100/5730}\\ N=5\times {\left(\frac{1}{2}\right)}^{0.017}\\ N=4.94\text{ g}\end{array}$

Percentage of decay:

  $\begin{array}{l}P=\left|\frac{N_o-N}{N_o}\right|\times 100\\ P=\left|\frac{5-4.95}{5}\right|\times 100\\ P=1.2\%\end{array}$

Percentage of remaining mass= 98.8%

E. For $Po$

  $\begin{array}{l}N=500\times {\left(\frac{1}{2}\right)}^{86400/0.392}\\ N=500\times {\left(\frac{1}{2}\right)}^{220408.16}\\ N\sim 0\text{ g}\end{array}$

Percentage of decay:

  $\begin{array}{l}P=\left|\frac{N_o-N}{N_o}\right|\times 100\\ P=\left|\frac{5-0}{5}\right|\times 100\\ P=100\%\end{array}$

Percentage of remaining mass = 0%

Conclusion:

Hence, the rank is B > D > C > A > E.