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Chapter 19: Problem 18

The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?

Short Answer

Expert verified
In the complete decay series, 10 alpha particles and 10 beta particles are produced.

Step by step solution

01

Determine the initial and final atomic numbers and mass numbers

We need to know the atomic numbers (Z) and mass numbers (A) of both \({ }^{247} \mathrm{Bk}\) (initial isotope) and \({ }^{207} \mathrm{Pb}\) (final isotope). For \({ }^{247} \mathrm{Bk}\): - Mass number (A) = 247 - Atomic number (Z) = 97 (since Bk is Berkelium) For \({ }^{207} \mathrm{Pb}\): - Mass number (A) = 207 - Atomic number (Z) = 82 (since Pb is Lead)
02

Calculate the changes in atomic number (Z) and mass number (A)

We need to find how many protons and neutrons transformed to reach the final isotope. Change in atomic number (ΔZ) = 97 - 82 = 15 Change in mass number (ΔA) = 247 - 207 = 40
03

Calculate the number of alpha particles

Each alpha particle contains 2 protons, so the number of alpha particles can be calculated by dividing the change in atomic number by 2. Number of alpha particles = ΔZ / 2 = 15 / 2 = 7.5 But, the number of alpha particles must be an integer. Since it is not possible to have a half alpha particle, the change in atomic number must actually be due to both alpha and beta particles.
04

Calculate the number of beta particles

Beta decay does not change the mass number, only the atomic number. We will find the number of beta particles by considering the change in the mass number caused solely by alpha particles. Let's assume k alpha (α) particles and m beta (β) particles are involved in the decay process. We rewrite the change in atomic number ΔZ as: ΔZ = 2k + m (as alpha particles change atomic number by 2 and beta particles by 1) Also, we rewrite the change in mass number ΔA as: ΔA = 4k (alpha particles change mass number by 4) Now, let's find the value of k from the mass number change equation: k = ΔA / 4 = 40 / 4 = 10
05

Calculate the number of beta particles (m)

Now that we know the value of k (10 alpha particles), we can find the number of beta particles (m) using the atomic number change equation: 15 = 2 * 10 + m m = 15 - 20 = -5 Since we have a negative value for beta particles, which is not possible, this indicates that we have a beta-minus decay. In beta-minus decay, a neutron transforms into a proton and emits a beta particle (electron). So, we need to consider the overall change in atomic number or the net gain of protons.
06

Adjust for beta-minus decay and calculate the net change in atomic number

To account for beta-minus decay, we need to add 2 beta-minus particles for each beta particle counted earlier (which had a negative value). In this case, m = -5, so now we need to add 10 extra beta-minus particles. New ΔZ = 2 * k + (m + 10) New ΔZ = 2 * 10 + (-5 + 10) = 20 + 5 = 25 Now, we have a total of 10 alpha particles and 25 - 15 = 10 beta-minus particles in the complete decay series. So, in the complete decay series, 10 alpha particles and 10 beta particles are produced.

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Most popular questions from this chapter

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \({ }_{1}^{3} \mathrm{H}, 3.01605 \mathrm{u}\).

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\). Why is plutonium one of the most toxic substances known?

Radioactive copper-64 decays with a half-life of \(12.8\) days. a. What is the value of \(k\) in \(\mathrm{s}^{-1}\) ? b. A sample contains \(28.0 \mathrm{mg}^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \({ }^{64} \mathrm{Cu}\) is \(64.0 \mathrm{u}\). c. A chemist obtains a fresh sample of \({ }^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

The most significant source of natural radiation is radon- 222 . \({ }^{222} \mathrm{Rn}\), a decay product of \({ }^{238} \mathrm{U}\), is continuously generated in the earth's crust, allowing gaseous \(\mathrm{Rn}\) to seep into the basements of buildings. Because \({ }^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of \(3.82\) days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \({ }^{238} \mathrm{U}\) decays to \({ }^{222} \mathrm{Rn}\) ? What nuclei are produced when 222 \(\mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \({ }^{222} \mathrm{Rn}\) ? c. Another problem associated with \({ }^{222} \mathrm{Rn}\) is that the decay of \({ }^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) \(3.11 \mathrm{~min}\) ) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) -particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \({ }^{222} \mathrm{Rn}\) levels not exceed 4 pCi per liter of air \(\left(1 \mathrm{Ci}=1\right.\) curie \(=3.7 \times 10^{10}\) decay events per second; \(\left.1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\right) .\) Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \({ }^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(222 \mathrm{Rn}\) per liter of air.

Radioactive cobalt-60 is used to study defects in vitamin \(\mathrm{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron- 58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \({ }^{60} \mathrm{Co}=\) 59.9338 u; \({ }^{1} \mathrm{H}=1.00782 \mathrm{u}\) )? What is the de Broglie wavelength of the emitted particle if it has a velocity equal to \(0.90 c\), where \(c\) is the speed of light?
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