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Chapter 21: Problem 15

The naturally occurring radioactive decay series that begins with \({ }_{92}^{235} \mathrm{U}\) stops with formation of the stable \({ }_{82}^{20} \mathrm{~Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series?

Short Answer

Expert verified
In the naturally occurring radioactive decay series from Uranium-235 to Lead-207, there are 7 alpha particle emissions and 4 beta particle emissions.

Step by step solution

01

Identify Initial and Final Nuclei Elements

The naturally occurring radioactive decay series starts with: \(^{235}_{92}\text{U}\): Uranium-235, where the mass number is 235 (total protons and neutrons), and atomic number (protons) is 92. The decay series ends with: \(^{207}_{82}\text{Pb}\): Lead-207, where the mass number is 207 (total protons and neutrons), and atomic number (protons) is 82.
02

Calculate Change in Atomic Number and Mass Number

To calculate the changes in atomic number and mass number, subtract the initial values from the final values: Change in Atomic Number (鈭哯) = Final Atomic Number - Initial Atomic Number 鈭哯 = 82 - 92 = -10 Change in Mass Number (鈭咥) = Final Mass Number - Initial Mass Number 鈭咥 = 207 - 235 = -28
03

Understanding Alpha and Beta Decay

During an alpha decay, an alpha particle (helium nucleus) is emitted, which consists of 2 protons and 2 neutrons. Therefore, the atomic number decreases by 2, and the mass number decreases by 4. Represented as: \(\alpha : \Delta Z = -2, \Delta A = -4\) During a beta decay, either a neutron is converted into a proton (beta minus decay) with the emission of an electron, or a proton is converted into a neutron (beta plus decay) with the emission of a positron. In this case, we will consider beta minus decay, as we have a decrease in atomic number. In beta minus decay, the atomic number increases by 1, and the mass number remains unchanged. Represented as: \(\beta^{-} : \Delta Z = +1, \Delta A = 0\)
04

Calculate Number of Alpha and Beta Decays

Let x be the number of alpha decays, and y be the number of beta decays. Using the changes in atomic number and mass number, we can set up the following equations: 伪 Decay: \(x (-2) + y (+1) = -10\) 尾 Decay: \(x (-4) + y (0) = -28\) Solve this system of equations to find x and y. From the 尾 Decay equation, we get: \(x = 7\) Plug the value of x into the 伪 Decay equation: \(7 (-2) + y (+1) = -10\) \(-14 + y = -10\) \(y = 4\)
05

Conclusion

The number of alpha decays (x) is 7, and the number of beta decays (y) is 4. Thus, the naturally occurring radioactive decay series from Uranium-235 to Lead-207 involves 7 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 Decay
Alpha decay is a type of radioactive decay where an unstable atomic nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, essentially making it a helium nucleus. This process:
  • Reduces the atomic number by 2, as two protons are lost.
  • Reduces the mass number by 4, due to the loss of 2 protons and 2 neutrons.
Alpha decay is common among heavy elements like uranium and radium because it helps the nucleus lose mass and move towards a more stable state. It is a spontaneous process, governed by the laws of quantum mechanics.
Alpha particles are relatively heavy and carry a positive charge, making them less penetrating compared to other forms of radiation. They can be stopped by something as thin as a sheet of paper.
Beta Decay
Beta decay is another form of radioactive decay, in which a neutron in the nucleus is transformed into a proton, or vice versa, coupled with the emission of a beta particle. For our context, let's focus on beta minus decay:
  • Neutron is converted into a proton.
  • An electron (beta particle) and an antineutrino are emitted.
  • The atomic number increases by 1, as a new proton is formed.
  • The mass number remains unchanged.
This process increases the stability of the atom by converting an abundance of neutrons into protons, helping to balance the forces in the nucleus. Beta particles are much lighter than alpha particles and have a greater penetration ability. A thin metal sheet, such as aluminum, is usually required to stop them.
Uranium-235
Uranium-235 is a naturally occurring isotope of uranium with significance in nuclear physics and applications such as nuclear reactors and nuclear weapons. It has:
  • An atomic number of 92, indicating it has 92 protons.
  • A mass number of 235, meaning the sum of its protons and neutrons is 235.
Uranium-235 can undergo fission, splitting into smaller nuclei and releasing energy. This property is harnessed in nuclear power and weaponry. In the radioactive decay series, Uranium-235 is the starting point, gradually transforming into more stable elements, ending as lead-207. It undergoes several alpha and beta decays to release energy and move toward a more stable configuration.
Lead-207
Lead-207 is the end product of the Uranium-235 decay series. It is a stable isotope of lead:
  • Has an atomic number of 82, with 82 protons.
  • Has a mass number of 207, indicating the total count of protons and neutrons.
Reaching lead-207 signifies the completion of a chain of decay reactions where the unstable Uranium-235 becomes stable. Lead-207 is non-radioactive, marking the end of the decay process with no further emissions of alpha or beta particles. Stability is achieved as the internal forces of the nucleus are balanced, and lead serves as a stable base for various natural and man-made applications.

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Most popular questions from this chapter

It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions \({ }^{\infty} 0\) (Section \(\left.14.5\right) ?\) Discuss.

An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine\(131\left(t_{1 / 2}=8.02\right.\) days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a \(1.00-\mu \mathrm{L}\) sample of the water was 214 counts per minute. After 30 days the level of activity in a \(1.00-\mu \mathrm{L}\) sample was \(15.7\) counts per minute. Did the plants absorb iodide from the water? Explain.

Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9} \mathrm{yr}\). What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(4.2 ?\)

What particle is produced during the following decay processes: (a) sodium-24 decays to magnesium-24; (b) mercury-188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s} .(\mathrm{a})\) Calculate the rate of mass loss from the Sun in \(\mathrm{kg} / \mathrm{s}\). (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?
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