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Chapter 19: Problem 37

Write balanced nuclear equations for the following reactions and identify X: (a) \(\mathrm{X}(\mathrm{p}, \alpha){ }_{6}^{12} \mathrm{C}\) (b) \({ }_{13}^{27} \mathrm{Al}(\mathrm{d}, \alpha) \mathrm{X}\) (c) \(\frac{55}{25} \mathrm{Mn}(\mathrm{n}, \gamma) \mathrm{X}\).

Short Answer

Expert verified
(a) The nucleus X is Boron-9 (\(_{5}^{9}B\)). (b) The nucleus X is Magnesium-25 (\(_{12}^{25}Mg\)). (c) The nucleus X is Manganese-56 (\(_{25}^{56}Mn\)).

Step by step solution

01

Analyze the First Equation

In part (a) \(\mathrm{X}(\mathrm{p}, \alpha){ }_{6}^{12} \mathrm{C}\), proton (p) is captured by a nucleus X and an alpha particle (α) is emitted, resulting in a carbon-12 atom. Since proton (p) has an atomic number 1 and mass number 1 and alpha particle (α) has atomic number 2 and mass number 4, by principles of conservation of mass numbers and atomic numbers, the unknown nucleus X can be determined.
02

Calculate the Atomic and Mass Numbers for X in the First Equation

The mass number of X can be given as: \(Mass \: number \: of \: X = Mass \: number \: of \: Carbon - Mass \: number \: of \: Alpha + Mass \: number \: of \: Proton=> Mass \: number \: of \: X = 12 - 4 + 1 = 9 \).The atomic number of X can be given as: \(Atomic \: number \: of \: X = Atomic \: number \: of \: Carbon - Atomic \: number \: of \: Alpha + Atomic \: number \: of \: Proton=> Atomic \: number \: of \: X = 6 - 2 + 1 = 5 \).Therefore, \( Nucleus \: X =_{5}^{9}B \).
03

Analyze the Second Equation

In part (b) \({ }_{13}^{27} \mathrm{Al}(\mathrm{d}, \alpha) \mathrm{X}\), deuteron (d) is captured by an aluminum-27 nucleus and an alpha particle (α) is emitted to produce a nucleus X. Since deuteron (d) has an atomic number 1 and mass number 2 and alpha particle (α) has atomic number 2 and mass number 4, the unknown nucleus X can be determined similarly.
04

Calculate the Atomic and Mass Numbers for X in the Second Equation

The mass number of X can be given as: \(Mass \: number \: of \: X = Mass \: number \: of \: Aluminium - Mass \: number \: of \: Alpha + Mass \: number \: of \: Deuteron=> Mass \: number \: of \: X = 27 - 4 + 2 = 25 \).The atomic number of X can be given as: \(Atomic \: number \: of \: X = Atomic \: number \: of \: Aluminium - Atomic \: number \: of \: Alpha + Atomic \: number \: of \: Deuteron=> Atomic \: number \: of \: X = 13 - 2 + 1 = 12 \).Therefore, \( Nucleus \: X =_{12}^{25}Mg \).
05

Analyze the Third Equation

In part (c) \(_{25}^{55} \mathrm{Mn}(\mathrm{n}, \gamma) \mathrm{X}\), a neutron (n) is captured by a manganese-55 nucleus and a gamma photon (γ) is emitted, resulting in a nuclear transformation to X. Since neutron (n) has atomic number 0 and mass number 1, and a gamma ray photon (γ) has no atomic number nor mass number (0,0), X can be determined.
06

Calculate the Atomic and Mass Numbers for X in the Third Equation

The mass number of X can be given as: \(Mass \: number \: of \: X = Mass \: number \: of \: Manganese + Mass \: number \: of \: Neutron=> Mass \: number \: of \: X = 55 + 1 = 56 \).The atomic number of X can be given as: \(Atomic \: number \: of \: X = Atomic \: number \: of \: Manganese + Atomic \: number \: of \: Neutron=> Atomic \: number \: of \: X = 25 + 0 = 25 \).Therefore, \( Nucleus \: X =_{25}^{56}Mn \).

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Most popular questions from this chapter

Consider the decay series $$\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}$$ where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

Define nuclear binding energy, mass defect, and nucleon.

A freshly isolated sample of \({ }^{90} \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December 3,2003 . At 2: 15 P.M. on December \(17,2003,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \({ }^{90} \mathrm{Y}\).

In 2006 , an ex-KGB agent was murdered in London. Subsequent investigation showed that the cause of death was poisoning with the radioactive isotope \({ }^{210} \mathrm{Po},\) which was added to his drinks/food. (a) \({ }^{210} \mathrm{Po}\) is prepared by bombarding \({ }^{209} \mathrm{Bi}\) with neutrons. Write an equation for the reaction. (b) Who discovered the element polonium? (Hint: See an Internet source such as Webelements.com.) (c) The half-life of \({ }^{210} \mathrm{Po}\) is \(138 \mathrm{~d}\). It decays with the emission of an \(\alpha\) particle. Write an equation for the decay process. (d) Calculate the energy of an emitted \(\alpha\) particle. Assume both the parent and daughter nuclei to have zero kinetic energy. The atomic masses are \({ }^{210} \mathrm{Po}(209.98285 \mathrm{amu})\) \({ }^{206} \mathrm{~Pb}(205.97444 \mathrm{amu}),{ }_{2}^{4} \alpha(4.00150 \mathrm{amu}) .(\mathrm{e})\) Inges- tion of \(1 \mu \mathrm{g}\) of \({ }^{210} \mathrm{Po}\) could prove fatal. What is the total energy released by this quantity of \({ }^{210} \mathrm{Po} ?\)

(a) What is the activity, in millicuries, of a \(0.500-\mathrm{g}\) sample of \({ }_{93}^{237} \mathrm{~Np} ?\) (This isotope decays by \(\alpha\) -particle emission and has a half-life of \(2.20 \times 10^{6}\) yr. (b) Write a balanced nuclear equation for the decay of \({ }_{93}^{237} \mathrm{~Np}\).
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