Chapter 10, Problem 51P

Interpretation Introduction

(a)

Interpretation:

The amount of each isotope present after 8.0 days needs to be determined.

Concept Introduction:

Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life ($t_{1/2}$ )

  $\text{Concentration of reactant}=\frac{\text{ Initial concentration}}{\text{2}}$

The decay of the radioactive element can be described by the following formula-

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

Where

N(t) − amount of reactant at time t

N₀ − Initial concentration of the reactant

t_1/2 − Half-life of the decaying reactant

Answer to Problem 51P

After 8.0 days,

Amount of Iodine-131 left = 32 mg

Amount of Xenon-131 formed = 32 mg

Explanation of Solution

Given Information:

N₀ = 64 mg

t_1/2 = 8 days

Calculation:

After 8.0 days, the initial concentration of Iodine -131 reduces to half of its initial concentration and converts to Xenon-131.

Thus,

  $\begin{array}{l}N(t=8.0\text{ days})=\frac{N_0}{2}=\frac{64\text{ mg}}{2}\\ \therefore N(t=8.0\text{ days})=32\text{ mg}\end{array}$

Hence,

Amount of Iodine-131 left = 32 mg

Amount of Xenon-131 formed = 64 mg − 32 mg = 32 mg

Interpretation Introduction

(b)

Interpretation:

The amount of each isotope present after 16 days needs to be determined.

Concept Introduction:

Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life ( $t_{1/2}$ )

  $\text{Concentration of reactant}=\frac{\text{ Initial concentration}}{\text{2}}$

The decay of the radioactive element can be described by the following formula-

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

Where

N(t) − amount of reactant at time t

N₀ − Initial concentration of the reactant

t_1/2 − Half-life of the decaying reactant

Answer to Problem 51P

After 16.0 days,

Amount of Iodine-131 left = 16 mg

Amount of Xenon-131 formed = 48 mg

Explanation of Solution

Given Information:

N₀ = 64 mg

t_1/2 = 8 days

t = 16.0 daysCalculation:

After 16 days, amount of iodine-131 would be defined by N(t),where t is 16.0 days, as

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

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^{\frac{\text{16}\text{.0 days}}{\text{8}\text{.0 days}}}$

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^2$

  $N(t)=(64\text{ mg) }(\frac{1}{4})$

  $N(t)=16\text{ mg}$

Hence, the amount of Iodine-131 decays and converts to Xenon. Therefore,

Amount of Iodine-131 left after 16.0 days = 16 mg

Amount of Xenon-131 formed after 16.0 days = 64 mg − 16mg = 48 mg

Interpretation Introduction

(c)

Interpretation:

The amount of each isotope present after 24 days needs to be determined.

Concept Introduction:

Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life ( $t_{1/2}$ )

  $\text{Concentration of reactant}=\frac{\text{ Initial concentration}}{\text{2}}$

The decay of the radioactive element can be described by the following formula-

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

Where

N(t) − amount of reactant at time t

N₀ − Initial concentration of the reactant

t_1/2 − Half-life of the decaying reactant

Answer to Problem 51P

After 24.0 days,

Amount of Iodine-131 left = 8 mg

Amount of Xenon-131 formed = 56 mg

Explanation of Solution

Given Information:

N₀ = 64 mg

t_1/2 = 8 days

t = 24.0 days

Calculation:

After 24.0 days, amount of iodine-131 would be defined by N(t),where t is 24.0 days, as

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

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^{\frac{\text{24}\text{.0 days}}{\text{8}\text{.0 days}}}$

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^3$

  $N(t)=(64\text{ mg) }(\frac{1}{8})$

  $N(t)=8\text{ mg}$

Hence, the amount of Iodine-131 decays and converts to Xenon. Therefore,

Amount of Iodine-131 left after 24.0 days = 8 mg

Amount of Xenon-131 formed after 24.0 days = 64 mg − 8 mg = 56 mg

Interpretation Introduction

(d)

Interpretation:

The amount of each isotope present after 32 days needs to be determined.

Concept Introduction:

Half-life − It is the time required by original radioactive element to reduce to the half of its initial concentration. Thus, at half-life ( $t_{1/2}$ )

  $\text{Concentration of reactant}=\frac{\text{ Initial concentration}}{\text{2}}$

The decay of the radioactive element can be described by the following formula-

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

Where

N(t) − amount of reactant at time t

N₀ − Initial concentration of the reactant

t_1/2 − Half-life of the decaying reactant

Answer to Problem 51P

After 32.0 days,

Amount of Iodine-131 left = 4 mg

Amount of Xenon-131 formed = 60 mg

Explanation of Solution

Given Information:

N₀ = 64 mg

t_1/2 = 8 days

t = 32.0 days

Calculation:

After 32 days, amount of iodine-131 would be defined by N(t),where t is 32.0 days, as

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

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^{\frac{\text{32}\text{.0 days}}{\text{8}\text{.0 days}}}$

  $N(t)=(64\text{ mg) }{(\frac{1}{2})}^4$

  $N(t)=(64\text{ mg) }(\frac{1}{16})$

  $N(t)=4\text{ mg}$

Hence, the amount of Iodine-131 decays and converts to Xenon. Therefore,

Amount of Iodine-131 left after 32.0 days = 4 mg

Amount of Xenon-131 formed after 32.0 days = 64 mg − 4 mg = 60 mg