Chapter 10, Problem 52P

Interpretation Introduction

(a)

Interpretation:

The amount of each isotope present after 14 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 52P

After 14.0 days,

Amount of Iodine-131 left = 62 mg

Amount of Xenon-131 formed = 62 mg

Explanation of Solution

Given Information:

N₀ = 124 mg

t_1/2 = 14 days

Calculation:

After 14.0 days, the initial concentration of phosphorus-32 reduces to half of its initial concentration and converts to sulfur-32.

Thus,

  $\begin{array}{l}N(t=14.0\text{ days})=\frac{N_0}{2}=\frac{124\text{ mg}}{2}\\ \therefore N(t=14.0\text{ days})=62\text{ mg}\end{array}$

Hence,

Amount of phosphorus-32 left = 62 mg

Amount of sulfur-32 formed = 124 mg − 62 mg = 62 mg

Interpretation Introduction

(b)

Interpretation:

The amount of each isotope present after 28 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 52P

After 28.0 days,

Amount of Phosphorus-32 left = 32 mg

Amount of Sulfur-32 formed = 92 mg

Explanation of Solution

Given Information:

N₀ = 124 mg

t_1/2 = 14 days

t = 28.0 days

Calculation:

After 28 days, amount of phosphorus-32 would be defined by N(t),where t is 28.0 days, as

  $\begin{array}{l}N(t)=N_0{(}^{\frac{1}{2}}\\ N(t)=(124\text{ mg) }{(}^{\frac{1}{2}}\\ N(t)=(124\text{ mg) }{(}^{\frac{1}{2}}\\ N(t)=(124\text{ mg) }(\frac{1}{4})\\ N(t)=32\text{ mg}\end{array}$

Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,

Amount of Phosphorus-32 left after 28.0 days = 32 mg

Amount of sulfur-32 formed after 28.0 days = 124 mg − 32 mg = 92 mg

Interpretation Introduction

(c)

Interpretation:

The amount of each isotope present after 42 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 52P

After 42.0 days,

Amount of Phosphorus-32 left = 15.5 mg

Amount of Sulfur-32 formed = 108.5 mg

Explanation of Solution

Given Information:

N₀ = 124 mg

t_1/2 = 14 days

t = 42.0 days

Calculation:

After42 days, amount of phosphorus-32 would be defined by N(t),where t is 42.0 days, as

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

  $N(t)=(124\text{ mg) }{(\frac{1}{2})}^{\frac{\text{42}\text{.0 days}}{\text{14}\text{.0 days}}}$

  $N(t)=(124\text{ mg) }{(\frac{1}{2})}^3$

  $N(t)=(124\text{ mg) }(\frac{1}{8})$

  $N(t)=15.5\text{ mg}$

Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,

Amount of Phosphorus-32 left after 42.0 days = 15.5 mg

Amount of sulfur-32 formed after 42.0 days = 124 mg − 15.5 mg = 108.5 mg

Interpretation Introduction

(d)

Interpretation:

The amount of each isotope present after 56 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 52P

After 56.0 days,

Amount of Phosphorus-32 left = 7.75 mg

Amount of Sulfur-32 formed = 116.25 mg

Explanation of Solution

Given Information:

N₀ = 124 mg

t_1/2 = 14 days

t = 56.0 days

Calculation:

After 56 days, amount of phosphorus-32 would be defined by N(t),where t is 56.0 days, as

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

  $N(t)=(124\text{ mg) }{(\frac{1}{2})}^{\frac{\text{56}\text{.0 days}}{\text{14}\text{.0 days}}}$

  $N(t)=(124\text{ mg) }{(\frac{1}{2})}^4$

  $N(t)=(124\text{ mg) }(\frac{1}{16})$

  $N(t)=7.75\text{ mg}$

Hence, the amount of phosphorus-32 decays and converts to sulfur-32. Therefore,

Amount of Phosphorus-32 left after 56.0 days = 7.75 mg

Amount of sulfur-32 formed after 56.0 days = 124 mg − 7.75 mg =116.25 mg