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A radioactive decay series that begins with 90 Th ends with formation of the stable nuclide \({ }^{208} \mathrm{~Pb} .\) How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Short Answer

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The radioactive decay series that begins with 90 Th and ends with the stable nuclide \({ }^{208} \mathrm{~Pb} \) involves 6 alpha-particle emissions and 4 beta-particle emissions.

Step by step solution

01

Identify initial and final nuclides

Initially, the decay series begins with 90 Th, and the final stable nuclide is \({ }^{208} \mathrm{~Pb} \). Let us denote the number of alpha-particle emissions as "a" and the number of beta-particle emissions as "b". The initial atomic number (Z) of Th is 90, and its mass number (A) is not given. The final atomic number (Z) of Pb is 82, and its mass number (A) is 208.
02

Write equations for changes in atomic and mass numbers

Considering the changes in atomic and mass numbers due to alpha and beta emissions, we can write the following equations: \( Z_{initial} - 2a + b = Z_{final} \) \( A_{initial} - 4a = A_{final} \)
03

Calculate the initial mass number

We can calculate the initial mass number (A) of Thorium by using the given final mass number (208) of Lead. From the equation "A_initial - 4a = A_final", we have: \( A_{initial} - 4a = 208 \) Since we are not given the initial mass number, we'll assume the known isotope of Thorium, Th-232: \( 232 - 4a = 208 \)
04

Determine the number of alpha emissions

Rearrange and calculate the number of alpha emissions: \( 4a = 232 - 208 \) \( 4a = 24 \) Therefore, \(a = 6\). There are 6 alpha-particle emissions.
05

Determine the number of beta emissions

Using the equation for changes in atomic number due to alpha and beta emissions ("Z_initial - 2a + b = Z_final"), plug in the given Z values and the calculated number of alpha emissions: \( 90 - 2(6) + b = 82 \) Now, solve for the number of beta emissions: \( 90 - 12 + b = 82 \) \( b = 82 - 90 + 12 \) Therefore, \(b = 4\). There are 4 beta-particle emissions.
06

State the results

The radioactive decay series that begins with 90 Th and ends with the formation of the stable nuclide \({ }^{208} \mathrm{~Pb} \) involves 6 alpha-particle emissions and 4 beta-particle emissions.

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Key Concepts

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

Alpha-Particle Emission
An essential concept in radioactive decay is the alpha-particle emission. This type of emission occurs when an unstable atomic nucleus releases an alpha particle, which consists of two protons and two neutrons. This effectively means it discharges a helium nucleus. Alpha particles are relatively large and have a charge of +2 due to their protons. As a result, when an alpha-particle emission occurs, the original atom loses:
  • 2 protons
  • 2 neutrons
This causes a decrease in both the atomic number, by 2, and the mass number, by 4, of the resulting element. This change transforms the atom into a different element lower in the periodic table. Because alpha particles are relatively bulky, they do not penetrate materials deeply, making them less hazardous when contained. However, they can cause significant damage to living tissue if ingested or inhaled due to their high energy. Their emission is a common mode of decay in heavy nuclides, as seen with Thorium-232, which undergoes such decay to eventually form Lead-208.
Beta-Particle Emission
Beta-particle emission is another significant process in radioactive decay. It involves the transformation of a neutron into a proton inside the nucleus, with the release of a beta particle, commonly an electron, and an antineutrino. This process increases the atomic number of the element by one, while the mass number remains unchanged:
  • Atomic Number increases by 1
  • Mass Number remains the same
This change develops the atom into a new element higher in the periodic table. Beta particles are much smaller and lighter than alpha particles, carrying a -1 charge due to the electron ejection. They have a medium penetration power, able to pass through skin but not dense materials like lead or thick glass. Beta decay plays a crucial role in balancing the charges in an unstable nucleus, helping it reach a more stable state. In the decay series from Thorium to Lead, beta emissions are vital for adjusting the atomic number that results from reducing through alpha emissions.
Decay Series
A decay series is the sequence of radioactive decays that certain isotopes go through to reach a stable end-state. Starting from a radioactive parent isotope, it undergoes a series of emissions, transforming into various elements until reaching a non-radioactive nucleus. In the decay series from Thorium-232 to Lead-208:
  • The series begins with Thorium-232
  • Undergoes 6 alpha-particle emissions
  • Completes with 4 beta-particle emissions
  • Ends as Lead-208, a stable element
This specific sequence illustrates how elements can drastically change during decay as they strive for stability. Comprehending a decay series helps us understand how isotopes evolve and the processes they go through before stabilizing. It's pivotal in fields such as radioactive dating, nuclear physics, and understanding natural radioactivity. By identifying each step in this journey, we decode the path that elements follow, unlocking knowledge about their ultimate roles in nature. The detailed balance of decay modes plays a key role in determining the end isotopes in decay paths, like that beginning with Thorium.

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Most popular questions from this chapter

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Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

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