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Chapter 16: Problem 72

Complete this nuclear reaction, and name the decay process: \({ }_{4}^{8} \mathrm{Be}+? \rightarrow{ }_{3}^{8} \mathrm{Li}\)

Short Answer

Expert verified
The missing particle in the nuclear reaction is a positron (\( e^{+} \)), having the notation \( {}_{1}^{0}\mathrm{e^{+}} \). The complete nuclear reaction is \({}_{4}^{8}\mathrm{Be} + {}_{1}^{0}\mathrm{e^{+}} \rightarrow{}_{3}^{8}\mathrm{Li}\), and the decay process is Beta-plus Decay or Positron Emission.

Step by step solution

01

Analyze the given reaction

In the given nuclear reaction, we have an isotope of Beryllium: \({}_{4}^{8}\mathrm{Be}\) After the decay, an isotope of Lithium is produced: \({}_{3}^{8}\mathrm{Li}\) We need to find the missing particle (?) in this reaction.
02

Apply conservation of atomic number (Z)

The number of protons must be conserved during the decay process. Hence, we have: (Protons in the initial nucleus) = (Protons in the final nucleus) + (Protons in the missing particle) In our case, this conservation principle can be written as: \(4 = 3 + Z_{?}\) Solving for \(Z_{?}\), we find: \(Z_{?} = 4 - 3 = 1\)
03

Apply conservation of mass number (A)

The total number of nucleons (protons + neutrons) must also be conserved during the decay process. Hence, we have: (Mass number of the initial nucleus) = (Mass number of the final nucleus) + (Mass number of the missing particle) In our case, this conservation principle can be written as: \(8 = 8 + A_{?}\) Solving for \(A_{?}\), we find: \(A_{?} = 8 - 8 = 0\)
04

Identify the missing particle

Now that we have found both the atomic number (Z) and the mass number (A) of the missing particle, we can identify it. The missing particle has 1 proton and 0 neutrons (since the mass number equals the number of protons plus neutrons). This particle is a positron (e+), which has the following notation: \({}_{1}^{0}\mathrm{e^{+}}\) The complete nuclear reaction is: \({}_{4}^{8}\mathrm{Be} + {}_{1}^{0}\mathrm{e^{+}} \rightarrow{}_{3}^{8}\mathrm{Li}\)
05

Identify the decay process

The decay process that involves the emission of a positron is called "Beta-plus Decay" or "Positron Emission." This is a type of radioactive decay where a proton is transformed into a neutron, and a positron is emitted. In conclusion, the complete nuclear reaction is: \({}_{4}^{8}\mathrm{Be} + {}_{1}^{0}\mathrm{e^{+}} \rightarrow{}_{3}^{8}\mathrm{Li}\) And the decay process is Beta-plus Decay or Positron Emission.

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Most popular questions from this chapter

Suppose \({ }_{20}^{40} \mathrm{Ca}\) is the product of a beta-emission nuclear reaction. What was the parent isotope?

Past the "bismuth buoy," \(\beta\) radiation will not render a radioactive nucleus nonradioactive. Explain why.

The isotopes \({ }^{17} \mathrm{~F}^{20} \mathrm{~F}\), and \({ }^{21} \mathrm{~F}\) are all radioactive, decaying either by beta emission or by positron emission. Name the decay process for each isotope.

Polonium-210, an alpha emitter, has a halflife of \(138.4\) days. Suppose you were to collect the helium gas originating from the alpha particles. How many milliliters of helium gas at standard temperature and pressure would you collect from \(1.000 \mathrm{~g}\) of polonium dioxide, \(\mathrm{PoO}_{2}\), in a period of \(138.4\) days? [Assume all the polonium in the sample is \({ }^{210} \mathrm{Po}\), molar mass \(209.98287 \mathrm{~g} / \mathrm{mol}\). Alpha emission from polonium- 210 yields the nonradioactive isotope lead-206; see Problem 16.103.]

True or false? (a) The number of nucleons in an atom is equal to its atomic number. (b) The number of nucleons in an atom is equal to its mass number. (c) The number of protons in an atom is equal to its mass number minus its atomic number. (d) The number of neutrons in an atom is equal to its mass number minus its atomic number.
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