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Chapter 19: Problem 14

In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ? b. \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\) d. \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)

Short Answer

Expert verified
a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+ e⁻\) b. \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+ \alpha\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+ e⁺\) d. \({ }^{241} \mathrm{Cm}+ n \rightarrow{ }^{241} \mathrm{Am}\)

Step by step solution

01

a. Identify the missing particle in \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ?#

We are given the decay process: \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ?. To find the missing particle, we have to ensure the atomic and mass numbers are conserved in the process. Since both Ga and Ge have the same mass number (73), the missing particle must have a mass number of 0. The atomic number of Ga is 31 and the atomic number of Ge is 32. Thus, the missing particle must have an atomic number of -1 to conserve atomic numbers. This particle, with a mass number of 0 and an atomic number of -1, is an electron which is emitted in beta-minus decay (β⁻). So, the missing particle is an electron (e⁻). The decay process is: \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+ e⁻\).
02

b. Identify the missing particle in \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\)#

We are given the decay process: \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\). First, let's find the difference in the mass number (∆A) and the atomic number (∆Z) to ensure conservation: ∆A = 192 - 188 = 4 ∆Z = 78 (Pt) - 76 (Os) = 2 The missing particle must have a mass number of 4 and an atomic number of 2 to conserve the mass and atomic numbers. This particle, with a mass number of 4 and an atomic number of 2, is an alpha particle (α). So, the missing particle is an alpha particle. The decay process is: \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+ \alpha\).
03

c. Identify the missing particle in \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\)#

We are given the decay process: \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\). Both Bi and Pb have the same mass number (205), so the missing particle must have a mass number of 0. The atomic number of Bi is 83 and the atomic number of Pb is 82. Thus, the missing particle must have an atomic number of 1 to conserve the atomic numbers. This particle, with a mass number of 0 and an atomic number of 1, is a positron which is emitted in beta-plus decay (β⁺). So, the missing particle is a positron (e⁺). The decay process is: \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+ e⁺\).
04

d. Identify the missing particle in \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)#

We are given the decay process: \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\). Both Cm and Am have the same mass number (241), so the missing particle must have a mass number of 0. The atomic number of Cm is 96 and the atomic number of Am is 95. Thus, the missing particle must have an atomic number of 1 to conserve atomic numbers. This particle, with a mass number of 0 and an atomic number of 1, is a neutron which is absorbed in neutron capture. So, the missing particle is a neutron (n). The decay process is: \({ }^{241} \mathrm{Cm}+ n \rightarrow{ }^{241} \mathrm{Am}\).

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