Potassium-42 is used to locate brain tumors. Its half-life is \(12.5\) hours. Starting with \(15.4 \mathrm{mg}\), what fraction will remain after 100 hours? If it was necessary to have at least \(1 \mu \mathrm{g}\) for a particular procedure, could you hold the original sample for 200 hours before using it?

Short Answer

Expert verified
After 100 hours, the fraction remaining is \( \frac{1}{256} \). After 200 hours, 0.235 μg remains, which is less than 1 μg.

Step by step solution

01

Understand the half-life concept

The half-life of a substance is the time it takes for half of the substance to decay. For Potassium-42, the half-life is given as 12.5 hours.
02

Calculate the number of half-lives in 100 hours

To find the number of half-lives within 100 hours, divide 100 by the half-life duration of 12.5 hours.o. of half-lives = \( \frac{100}{12.5} = 8 \)
03

Calculate the remaining fraction after 100 hours

Each half-life reduces the remaining amount to half of the previous quantity. After 8 half-lives, the fraction remaining is calculated as:Remaining fraction = \( \left(\frac{1}{2}\right)^8 = \frac{1}{256} \)
04

Determine the remaining amount after 100 hours

Multiply the initial amount by the remaining fraction to find the amount of potassium-42 after 100 hours.Remaining amount = \( 15.4 \mathrm{mg} \times \frac{1}{256} = 0.060 \mathrm{mg} = 60 \mu \mathrm{g} \)
05

Calculate the number of half-lives in 200 hours

To find the number of half-lives within 200 hours, divide 200 by the half-life duration of 12.5 hours.o. of half-lives = \( \frac{200}{12.5} = 16 \)
06

Determine the remaining amount after 200 hours

Using the same method as in Step 4, after 16 half-lives, the remaining fraction is:Remaining fraction = \( \left(\frac{1}{2}\right)^{16} = \frac{1}{65536} \)Remaining amount = \( 15.4 \mathrm{mg} \times \frac{1}{65536} \approx 0.000235 \mathrm{mg} = 0.235 \mu \mathrm{g} \)
07

Compare the remaining amount with the required amount for the procedure

Since at least 1 μg is needed for the procedure, and the remaining amount after 200 hours (0.235 μg) is less than 1 μg, it will not be sufficient.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

radioactive decay
Radioactive decay is a natural process in which an unstable atomic nucleus loses energy by emitting radiation. This process results in the transformation of the original nuclide into a different one. There are three primary types of radioactive decay: alpha, beta, and gamma.
Alpha decay involves the emission of an alpha particle, which is composed of two protons and two neutrons. Beta decay follows from the transformation of a neutron into a proton or a proton into a neutron, accompanied by the emission of an electron or a positron. Gamma decay, on the other hand, involves the release of gamma rays, which are high-energy photons.
A vital aspect of radioactive decay is that the rate of decay is exponential, which is described by the half-life concept.
Potassium-42
Potassium-42 (K-42) is a radioactive isotope of potassium. It has a half-life of 12.5 hours and decays by beta emission. This makes it useful in medical applications, especially in locating brain tumors.
The significance of K-42 in medical procedures lies in its ability to emit beta particles, which can be detected by imaging technology to highlight abnormal tissues. When introduced into the body, K-42 accumulates in areas with high metabolism, such as tumors, making them easier to locate.
The short half-life of K-42 is beneficial as it allows for rapid imaging and reduces long-term radiation exposure to patients.
nuclear chemistry
Nuclear chemistry involves the study of the changes that occur within atomic nuclei. It is a crucial field for understanding both beneficial and detrimental nuclear reactions.
Key concepts in nuclear chemistry include radioactive decay, nuclear fission (splitting of a heavy nucleus into two lighter nuclei), and nuclear fusion (combining two light nuclei to form a heavier nucleus). These processes release large amounts of energy, which can be harnessed for power generation or medical applications.
Nuclear chemistry also explains the behavior of radioisotopes, their decay patterns, and their applications in fields such as medicine, agriculture, and archaeology.
half-life
Half-life is the time required for half of a radioactive substance to decay. It is a constant property for each specific isotope. For K-42, the half-life is 12.5 hours.
The half-life formula helps in calculating the remaining quantity of a substance over time: \[ N(t) = N_0 \times \frac{1}{2}^\frac{t}{t_{1/2}} \]Where:
  • \(N(t)\): Amount remaining after time \( t \)
  • \(N_0\): Initial amount
  • \(t\): Time elapsed
  • \(t_{1/2}\): Half-life of the substance
The idea is that with each passing half-life, the remaining quantity is halved, creating an exponential decay pattern.
radioisotopes
Radioisotopes are isotopes that exhibit radioactive decay. They have applications in numerous fields:
  • Medicine: Used in diagnosis and treatment of diseases (e.g., K-42 for brain tumor detection)
  • Agriculture: Employed to improve food preservation and pest control
  • Archaeology: Utilized in dating ancient artifacts through carbon dating
Each radioisotope has a unique half-life and decay mode, making them suitable for specific applications. The study of radioisotopes helps scientists understand atomic behaviors and invent technologies for various real-world problems.

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Most popular questions from this chapter

Starting with \(1 \mathrm{~g}\) of a radioactive isotope whose half-life is 10 days, sketch a graph showing the pattern of decay for that material. On the \(x\)-axis, plot time (you may want to simply show multiples of the half-life), and on the \(y\)-axis, plot mass of material remaining. Then after completing the graph, explain why a sample never really gets to the point where all of its radioactivity is considered to be gone.

Many home smoke detectors use americunium-241 as a radiation source. In the smoke detector, \({ }^{241}\) Am emits alpha particles. These alpha particles are detected by the detector unless smoke particles block their passage. Thus, as long as there are no smoke particles, the detector gets a continuous stream of alpha particles hitting it; smoke causes a disruption in this stream and activates the alarm. Write out the nuclear equation for the decay of \({ }^{241}\) Am by alpha emission.

When the nuclide \({ }_{98}^{249} \mathrm{Cf}\) was bombarded with \({ }_{7}^{15} \mathrm{~N}\), four neutrons and a new transuranium element were formed. Write the nuclear equation for this transmutation.

The nuclide \(\frac{223}{8}\) Fr emits three particles losing 8 mass units and 3 atomic number units. Propose a radioactive decay series and write the symbol for the resulting nuclide.

Consider the fission reaction $$ { }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{94} \mathrm{Sr}+{ }_{54}^{139} \mathrm{Xe}+3{ }_{0}^{1} \mathrm{n}+\text { energy } $$ Calculate the following using these mass data \((1.0 \mathrm{~g}\) is equivalent to \(9.0 \times 10^{13} \mathrm{~J}\) ): $$ \begin{array}{rlr} \mathrm{U}-235=235.0439 \mathrm{amu} & \text { Sr-94 }=93.9154 \mathrm{amu} \\\ \mathrm{Xe}-139=138.9179 \mathrm{amu} & \mathrm{n}=1.0087 \mathrm{amu} \end{array} $$ (a) the energy released in joules for a single event (one uranium atom splitting) (b) the energy released in joules per mole of uranium splitting (c) the percentage of mass lost in the reaction

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