Answer: ²³⁵₉₂U + ¹₀n → ⁹²₃₈Sr + ¹⁴²₅₄Xe + 2¹₀n + energy

Answering the Question:

Equation 2 will behave like any other chemical equation. That is, the total mass of reactant will be equal to the total mass of product. In equation 2, uranium 235 undergoes nuclear fission to produce strontium 92, xenon 142, two neutrons and energy. Analyzing the mass number shows the mass number of the reactants is equal to the mass number of the products.

²³⁵₉₂U + ¹₀n → ⁹²₃₈Sr + ¹⁴²₅₄Xe + 2¹₀n + energy

235 + 1 → 92+142+(2x1), therefore

236 → 236, so not mass is lost.

The same holds true for the number of protons distributed throughout the reaction.

²³⁵₉₂U + ¹₀n → ⁹²₃₈Sr + ¹⁴²₅₄Xe + 2¹₀n + energy

92+0 → 38+54+0, therefore

92 → 92

So the number of protons also remains the same. To determine the possible products of nuclear fission, it is important to remember that both sides of the chemical reaction must be equal. This means that the mass number and total number of protons can be calculated from the given data.

The mass number would be calculated by:

²³⁵₉₂U + ¹₀n → ⁹²₃₈Sr + ______ + 2¹₀n + energy

235U + 1n → 92Sr+ unknown +(2x1) therefore

236 = 94 + unknown

236 -94 = unknown

Unknown = 142.

The same method can be used to calculate the number of protons (bottom number).

²³⁵₉₂U + ¹₀n → ⁹²₃₈Sr + _______ + 2¹₀n + energy

U + 0 → Sr + unknown + 0

92+ 0 → 38 + unknown + 0

92 → 38 + unknown

Therefore, unknown = 54.

This means the unknown has a proton count or 54 and a mass number of 142. The only element with 54 protons is Xenon (Xe). Therefore, based on calculations the resulting atom could be drawn

¹⁴²₅₄Xe.

Answer: Fission Reaction

Why? Equation 1 represents a fission reaction; this reaction type usually involves the splitting of a heavy atom into two smaller atoms and resulting neutrons. This reaction is very exothermic and is the type of reaction used to power nuclear power plants.

There are a number of heavy atoms, which are used, in nuclear fission; among them are Uranium, Plutonium and Thorium.

Answering the Question:

Answer: (1)

Why? All known living organisms contain the element carbon. Carbon is considered the element of life and is most abundant in the form ¹²C. In this form there are an equivalent number of protons and neutrons, which make up the nucleus of the atom. Atoms however, sometimes have the same number of protons (used to define the element), but differing numbers of neutrons (used to define the nuclide). In the case of carbon, it is defined by having 6 protons and 6 neutrons, in the case of ¹²C. There are other nuclides of carbon, which although they occur naturally, are present in much smaller quantities; one such is ¹⁴C, which has 6 protons and 8 neutrons. Carbon 14 is slightly radioactive and has a half-life of approximately 5740±40 years. Carbon 14 is present in living organisms, starting in plants with a ratio to ¹²C similar to what is present in nature. This information along with the half life of carbon was exploited by Willard Libby et al in 1949 to determine the approximate age of dead organisms, a process termed carbon dating. Carbon dating is determined mathematically by determining the ratio of ¹⁴C to ¹²C in the dead organism. Radioactive dating methods are dependent on knowing the half-life of the materials used to carry out the dating. The half-life of the material is the time it takes for the material to decay and decrease by half.

Answering the Question:

To answer this question it helps to understand the role of carbon in living organisms. However, if that were not known the concept of carbon dating tends to be a familiar one.  Answer (1) suggests the use of C-14 and C-12 for dating remains of once living organisms. Once-living organisms would be composed mostly of carbon. This would make answer (1) with the 2 nuclides of carbon a good guess for use as a means for dating organisms.

Answer (2) suggests nuclides of cobalt be used for carbon dating. Cobalt is used in some living organisms, but not all and is quite harmful in large quantities. Co-60 is radioactive and releases gamma radiation, making it very harmful in all but the minutest quantities. Co-60 also has a half-life of approximately six years; this would not make it very suitable for dating.

Answer (3) suggests the use of iodine (I-131) and xenon (Xe-131), both of which are highly unlikely. Xenon in particular, is a noble gas and, therefore, although present, is very stable in terms of chemical bonding and half-life. This would not make it suitable for dating, as mathematically there would be very little basis on which to carry out dating calculations. This would not make answer (3) suitable for dating.

Answer: (4)

Why? A radioactive isotope is usually the result of a difference in the number of protons and neutrons, which causes instability within the nucleus of atoms, hence the term nuclear. When an element undergoes radioactive decay, it does so to achieve a more stable nuclide.

Elements that undergo alpha decay (α) lose two protons and two neutrons, or a ⁴₂He²⁺ ion plus energy. This decay helps to bring the neutron to proton ratio down and hence makes the nuclide more stable.

Beta decay occurs in one of two forms, β⁺ and β^-. β^- results from having a high ratio of neutrons to protons. It occurs when one of the neutrons in the nucleus splits to form a proton, which remains in the nucleus and an electron, which is emitted at high velocity. In effect for beta minus you lose a neutron gain a proton. A β⁺ emission occurs when the atom has a low neutron to proton ratio, in this decay a proton splits to form a neutron, which remains in the nucleus, and a positron, this form of decay is not as common as beta minus decay. In effect beta plus decay loses a proton and gains a neutron.

Gamma radiation (γ) is short wave high-energy electromagnetic radiation. A change in the nuclide of the atom does not occur with the release of gamma radiation. Gamma rays are high in energy and are produced from processes such as nuclear fusion and fission or when an electron and a positron come in contact with each other.

With this knowledge the answers can be analyzed to determine if any of the forms of decay would produce a more stable nuclide.

Answering the Question:

To answer this question, it is important to understand what it is that makes an atom radioactive. Answer (1) is the hydrogen isotope tritium (³H), which is an unstable isotope of hydrogen. However, as γ rays do not result in a decrease in the number of neutrons to create a more stable form of hydrogen, γ decay is unlikely.

³₁H → ³₁H + γ radiation

Answer (2) is potassium 42 and is said to undergo β⁺ decay, which involves the loss on a proton and the gaining of a neutron. This would result in Argon 42 being formed, which would probably be less stable that potassium 42.

⁴²₁₉K → ⁴²₁₈Ar + ⁰₊₁e (β⁺ particle)

Answer (3) is nitrogen 16; α decay would result in the loss of two neutrons and two protons, resulting in Boron 12, which would probably be unstable.

¹⁶₇N → ¹²₅B + ⁴₂He (α particle)

Answer (4) is phosphorous 32; β^- decay would result in the loss of a neutron, and the gaining of a proton. This would result in the formation of sulfur 32, which is the naturally occurring form of sulfur, and would, therefore, have a stable nuclide.