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Chapter 19: Problem 21

Uranium-2355 undergoes a series of \(\alpha\) -particle and \(\beta\) -particle productions to end up as lead-207. How many \(\alpha\) particles and \(\beta\) particles are produced in the complete decay series?

Short Answer

Expert verified
Uranium-2355 decays to lead-207 by undergoing 5 alpha particle emissions, which reduce the atomic number by 2 each time, and 8 beta particle emissions, which increase the atomic number by 1 each time. Therefore, the complete decay series produces 5 alpha particles and 8 beta particles.

Step by step solution

01

Determine the initial and final atomic and mass numbers

Uranium has an atomic number (Z) of 92, while lead has an atomic number (Z) of 82. Uranium-2355 has a mass number (A) of 235, whereas lead-207 has a mass number of 207.
02

Analyze alpha and beta particle emissions

Alpha particles consist of two protons and two neutrons, causing a decrease in atomic number by 2 and a decrease in mass number by 4. Beta particles result in a neutron-to-proton conversion, which increases the atomic number by 1 and leaves the mass number unchanged.
03

Determine the number of alpha particle decays

Since alpha particle decay reduces the atomic number by 2, we can determine the number of alpha particle emissions (α) by finding the difference in atomic numbers between uranium and lead: α = (92 - 82) / 2 = 10 / 2 = 5.
04

Determine the number of beta particle decays

Since beta particle decay does not change the mass number, we can't directly determine the number of beta particle emissions (β). However, we can determine the number of neutron changes (Δn) that occur during both alpha and beta decay as follows: Δn = initial mass number - final mass number - 4 * alpha_decay, where 4 * alpha_decay represents the change in mass number due to alpha decay. Therefore, Δn = 235 - 207 - 4 * 5 = 28 - 20 = 8. Now recall that beta decay increases the atomic number by 1, so n_beta = Δn = 8. So, uranium-2355 undergoes a series of radioactive decays, releasing 5 alpha particles and 8 beta particles, to ultimately form lead-207.

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Most popular questions from this chapter

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and $235 \mathrm{U},\( with relative abundances of 99.28\)\%\( and \)0.72 \%,$ respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for 235 \(\mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the 28 \(\mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

Estimate the temperature needed to achieve the fusion of deuterium to make an \(\alpha\) particle. The energy required can be estimated from Coulomb's law [use the form \(E=9.0 \times 10^{9}\) \(\left(Q_{1} Q_{2} / r\right),\) using \(Q=1.6 \times 10^{-19} \mathrm{C}\) for a proton, and \(r=2 \times\) \(10^{-15} \mathrm{m}\) for the helium nucleus; the unit for the proportionality constant in Coloumb's law is J \(\cdot \mathrm{m} / \mathrm{C}^{2} ]\)

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{24} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am}\) ?

What are transuranium elements and how are they synthesized?

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, discoverer of the transuranium elements. a. \(^{263}\) Sg was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) g decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg}\) ?
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