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Chapter 5: Q. 5.83 (page 166)

A $64 - μCi$ sample of $T l - 201$ decays to $4.0 μ Ci$ in $12$ days. What is the half-life, in days, of $T l - 201$ ? $(5.3, 5.4)$

Short Answer

Expert verified

As a result, a half-life of $T l - 201 = 3.0 d a y s .$

The $T I - 201$ has a $3.0 d a y s$ half-life.

Step by step solution

01

Introduction

Half-life:

A half life is the time it takes for half of a radioactive sample to decay. .In $12$ days, the $T I - 201$ sample drops from $64.0 μCi$ to $4.0 μCi .$

Calculate the number of half-lives required to reduce the sample size from $64.0 μ C i$ to $4.0 μCi$ .

02

Explanation

$64.0 μCi TI - 201 \overset{1 half - life}{\to} 32.0 μCi TI - 201 \overset{2 half - lives}{\to} 16.0 μCi TI - 201 \overset{3 half - lives}{\to} 8.0 μCi TI - 201 \overset{4 half - lives}{\to} 4.0 μCi TI - 201$

To reduce the $T l - 201$ sample from $64.0 μ C i$ to $4.0 μ C i$ , four half-lives are required.

03

Conclusion

Calculate the half-life of $T I - 201$ as follows:

$4 half - lives for (T I - 201) = 12 days$

$1 half - life of T I - 201 = 3.0 days$

As a result, a half-life of $T l - 201 = 3.0 d a y s$ .

The $T I - 201$ has a $3.0$ day half-life.

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Most popular questions from this chapter

All the elements beyond uranium, the trans uranium elements, have been prepared by bombardment and are not naturally occurring elements. The first trans uranium element neptunium, $Np$ , was prepared by bombarding $U - 238$ with neutrons to form a neptunium atom and a beta particle. Complete the following equation: $(5.2)$

data-custom-editor="chemistry" ${}_{0}^{1}n + {}_{92}^{238}U \to ? + ?$

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

Complete each of the following nuclear equations: $(5.2)$

a. ${}_{2}^{4}{He} + {}_{7}^{14}N ⟶ ? + {}_{1}^{1}H$

b. ${}_{2}^{4}{He} + {}_{13}^{27}{Al} ⟶ {}_{14}^{30}{Si} + ?$

c. ${}_{0}^{1}n + {}_{92}^{235}U ⟶ {}_{38}^{90}{Sr} + 3 {}_{0}^{1}n + ?$

d. ${}_{12}^{23 m}{Mg} ⟶$ $? + {}_{0}^{0}γ$

Carbon dating of small bits of charcoal used in cave paintings has determined that some of the paintings are from 10000 to 30000 yr old. Carbon- 14 has a half-life of 5730 yr. In a 1- $μ g$ sample of carbon from a live tree, the activity of carbon-14 is 6.4 $μ$ Ci. If researchers determine that 1 $μ g$ of charcoal from a prehistoric cave painting in France has an activity of 0.80 $μ C i$ , what is the age of the painting?

a. Technetium-99m emits only gamma radiation. Why would this type of radiation be used in diagnostic imaging rather than an isotope that also emits beta or alpha radiation?

b. A person with polycythemia vera (excess production of red blood cells) receives radioactive phosphorus- $32$ . Why would this treatment reduce the production of red blood cells in the bone marrow of the patient?

See all solutions

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