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Chapter 19: Problem 2

Lead-210 is used to prepare eyes for corneal transplants. Its decay product is bismuth-210. Identify the emission from lead-210.

Short Answer

Expert verified
Answer: Beta-minus particle (an electron).

Step by step solution

01

Understanding the Radioactive Decay Process

Radioactive decay is a random process where unstable atomic nuclei lose energy by emitting radiation. There are several types of radioactive decay, including alpha decay, beta decay (beta-minus and beta-plus), and gamma decay. In each decay process, a nucleus will release particles or energy in order to become more stable.
02

Writing the Decay Reaction

To determine the type of emission from lead-210 (Pb-210) as it decays into bismuth-210 (Bi-210), we can write the decay reaction as follows: \[ ^{210}_{82}Pb → ^{210}_{83}Bi + X \] In this reaction, the lead-210 nucleus is converted to a bismuth-210 nucleus and some emission X. To identify X, we need to compare the atomic numbers and mass numbers of the initial nucleus (Pb-210) and the final nucleus (Bi-210).
03

Identifying the Emission Type

Let's examine the atomic numbers and mass numbers of the initial and final nuclei: - Initial nucleus: Lead-210 (Pb-210): atomic number (Z) = 82, mass number (A) = 210 - Final nucleus: Bismuth-210 (Bi-210): atomic number (Z) = 83, mass number (A) = 210 From the reaction, we can observe that the mass number remains constant (A = 210), whereas the atomic number increases by one (83 - 82 = 1). This increase in atomic number corresponds to the emission of a beta-minus particle (an electron), which can be represented by the symbol \(β^-\). Therefore, the decay reaction for lead-210 turning into bismuth-210 is: \[ ^{210}_{82}Pb → ^{210}_{83}Bi + β^- \]
04

Conclusion

The emission from lead-210 as it decays into bismuth-210 is a beta-minus particle (an electron).

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Most popular questions from this chapter

One of the causes of the explosion at Chernobyl may have been the reaction between zirconium, which coated the fuel rods, and steam. $$\mathrm{Zr}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{ZrO}_{2}(s)+2 \mathrm{H}_{2}(g)$$ If half a metric ton of zirconium reacted, what pressure was exerted by the hydrogen gas produced at \(55^{\circ} \mathrm{C}\) in the containment chamber, which had a volume of \(2.0 \times 10^{4} \mathrm{~L} ?\)

The amount of oxygen dissolved in a sample of water can be determined by using thallium metal containing a small amount of the isotope Tl- 204\. When excess thallium is added to oxygen-containing water, the following reaction occurs. $$2 \mathrm{Tl}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Tl}^{+}(a q)+2 \mathrm{OH}^{-}(a q)$$ After reaction, the activity of a 25.0-mL water sample is 745 counts per minute (cpm), caused by the presence of \(\mathrm{Tl}^{+}-204\) ions. The activity of Tl-204 is \(5.53 \times 10^{5} \mathrm{cpm}\) per gram of thallium metal. Assuming that \(\mathrm{O}_{2}\) is the limiting reactant in the above equation, calculate its concentration in moles per liter.

For how many years could all the energy needs of the world be supplied by the fission of \(\mathrm{U}-235\) ? Use the following assumptions: The world has about \(1.0 \times 10^{7}\) metric tons of uranium ore, which are about 0.75\% U-235. The energy consumption of the world is about \(4.0 \times 10^{15} \mathrm{~kJ} / \mathrm{y}\) and does not change with time. The fission of U-235 releases about \(8.0 \times 10^{7} \mathrm{~kJ} / \mathrm{g}\) of \(\mathrm{U}-235\).

It is possible to estimate the activation energy for fusion by calculating the energy required to bring two deuterons close enough to one another to form an alpha particle. This energy can be obtained by using Coulomb's law in the form \(E=8.99 \times 10^{9} q_{1} q_{2} / r\), where \(q_{1}\) and \(q_{2}\) are the charges of the deuterons \(\left(1.60 \times 10^{-19} \mathrm{C}\right), r\) is the radius of the He nucleus, about \(2 \times 10^{-15} \mathrm{~m}\), and \(E\) is the energy in joules. (a) Estimate \(E\) in joules per alpha particle. (b) Using the equation \(E=m v^{2} / 2\), estimate the velocity (meters per second) each deuteron must have if a collision between the two of them is to supply the activation energy for fusion \((m\) is the mass of the deuteron in kilograms).

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of \(1.2 \times 10^{5}\) disintegrations \(/ \mathrm{min}\). How many grams of Br-82 were present in the sample? Assume that there were no other radioactive nuclides in the sample.
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