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Chapter 18: Problem 23

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.

Short Answer

Expert verified
The resulting nuclide is \( \frac{215}{84} \)Po.

Step by step solution

01

- Identify Initial Nuclide

The initial nuclide is \(\frac{223}{87} \)Fr where the atomic number (Z) is 87 and the mass number (A) is 223.
02

- Account for Mass Loss

The nuclide loses 8 mass units. Subtract 8 from the original mass number: \[ 223 - 8 = 215 \]
03

- Account for Atomic Number Loss

The nuclide loses 3 atomic number units. Subtract 3 from the original atomic number: \[ 87 - 3 = 84 \]
04

- Determine Resulting Nuclide

Using the updated mass number (215) and atomic number (84), determine the resulting element. Element 84 in the periodic table is polonium (Po). Thus, the resulting nuclide is \( \frac{215}{84} \)Po.

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Key Concepts

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

Nuclide
A nuclide is a distinct species of an atom characterized by the number of protons (atomic number, Z) and the number of neutrons (mass number, A - Z) in its nucleus. Nuclides are represented in the form \(\frac{A}{Z}\)\(X\), where X is the element symbol.
Mass Number
The mass number (A) is the total number of protons and neutrons in an atom's nucleus. In the example \(\frac{223}{87}\) Fr, the mass number is 223. When the nuclide undergoes radioactive decay, it loses mass units, which alters the mass number.
Atomic Number
The atomic number (Z) is the number of protons in the nucleus of an atom. It determines the element's identity. For \(\frac{223}{87}\) Fr, the atomic number is 87, meaning it has 87 protons. Losing 3 atomic number units during decay would give us a new atomic number.
Radioactive Decay Series
A radioactive decay series describes the steps a radioactive nuclide takes as it transforms into a stable nuclide. Each step involves emission of particles and energy and results in a new nuclide with a different mass number and atomic number.
Element Identification
Identifying an element involves matching its atomic number with the corresponding element symbol on the periodic table. After accounting for changes in both mass and atomic number during decay, we check the updated numbers to identify the resulting element.

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Most popular questions from this chapter

Barium-141 is a beta emitter. What is the half-life if a \(16.0-\mathrm{g}\) sample of the nuclide decays to \(0.500 \mathrm{~g}\) in 90 minutes?

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.

Rubidium-87, a beta emitter, is the product of positron emission. Identify (a) the product of rubidium- 87 decay (b) the precursor of rubidium- 87

Calculate (a) the mass defect and (b) the binding energy of \({ }_{3} \mathrm{Li}\) using the mass data: $$ \begin{array}{ll} { }_{3}^{7} \mathrm{Li}=7.0160 \mathrm{~g} & \mathrm{n}=1.0087 \mathrm{~g} \\ \mathrm{p}=1.0073 \mathrm{~g} & \mathrm{e}^{-}=0.00055 \mathrm{~g} \\ 1.0 \mathrm{~g}=9.0 \times 10^{13} \mathrm{~J} & \left(\text { from } E=m c^{2}\right) \end{array} $$

Clearly distinguish between fission and fusion. Give an example of each.
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