Question

When californium- 249 is bombarded by a small isotope, rutherfordium, \({ }_{104}^{257} \mathrm{Rf},\) is formed together with 4 neutrons. The isotope formed quickly decays to Nobelium- \(253,{ }_{102}^{253} \mathrm{No} .\) (a) What small isotope is used in the bombardment? (b) What particle is emitted in the decay of Rf-257?

Short answer

Based on the given information and the law of conservation of mass numbers and atomic numbers: (a) The small isotope used in the bombardment of Californium-249 is Helium-4 (\({ }_{6}^{4}\mathrm{He}\)), also known as an alpha particle. (b) The particle emitted in the decay of Rutherfordium-257 into Nobelium-253 is also an alpha particle (\({ }_{2}^{4}\mathrm{He}\)).

Step-by-step solution

Identify the Initial and Final Nuclides and Reactions

Californium-249 (\({ }_{98}^{249} \mathrm{Cf}\)) is bombarded by an unknown small isotope to produce Rutherfordium-257 (\({ }_{104}^{257} \mathrm{Rf}\)) and 4 neutrons. The Rutherfordium-257 quickly decays to Nobelium-253 (\({ }_{102}^{253} \mathrm{No}\)) by emitting an unknown particle. We need to determine the unknown small isotope and the unknown particle.

Use the Law of Conservation of Mass Numbers and Atomic Numbers for Bombardment Reaction

For the bombardment reaction: \({ }_{98}^{249} \mathrm{Cf} + { }_{x}^{y} \mathrm{Unknown} \rightarrow { }_{104}^{257} \mathrm{Rf} +4\times{ }_{0}^{1} \mathrm{n}\) Using the law of conservation of mass numbers, we get: \(249 + y = 257 + 4(1) \Rightarrow y = 253 - 249 = 4\) Now, using the law of conservation of atomic numbers, we get: \(98 + x = 104 \Rightarrow x = 6\) The small isotope used in the bombardment is \({}_{6}^{4}\mathrm{He}\) (Helium-4), also known as an alpha particle.

Use the Law of Conservation of Mass Numbers and Atomic Numbers for Rf-257 Decay Reaction

For the decay reaction: \({}_{104}^{257} \mathrm{Rf} \rightarrow { }_{102}^{253} \mathrm{No} + { }_{a}^{b} \mathrm{Unknown}\) Using the law of conservation of mass numbers, we get: \(257 = 253 + b \Rightarrow b = 4\) Now, using the law of conservation of atomic numbers, we get: \(104 = 102 + a \Rightarrow a = 2\) The unknown particle emitted in the decay of Rf-257 is \({}_{2}^{4}\mathrm{He}\) (Helium-4), also known as an alpha particle.

Final Answer

(a) The small isotope used in the bombardment is Helium-4 (\({ }_{6}^{4}\mathrm{He}\)), also known as an alpha particle. (b) The particle emitted in the decay of Rutherfordium-257 is also an alpha particle (\({ }_{2}^{4}\mathrm{He}\)).

Key concepts

Law of Conservation of Mass

The Law of Conservation of Mass is a fundamental principle in chemistry stating that matter cannot be created or destroyed in an isolated system. During a nuclear reaction, such as the bombardment of californium-249 by a small isotope, it's essential to account for all the particles before and after the reaction.

In solving the textbook exercise, we apply this law to ensure that the total mass (represented by the mass number) and the total charge (represented by the atomic number) both sides of the reaction equation remain equal. This conservation is what allows nuclear chemists to track nuclear reactions precisely and predict the products formed.

For instance, when californium-249 is bombarded, and rutherfordium-257 is formed, this law dictates that the sum of the mass numbers and atomic numbers on the reactant side must equal those on the product side. As illustrated in the textbook solution, to find the unknown isotope, one must sum up the mass numbers and atomic numbers accordingly, resulting in a Helium-4 nucleus, or alpha particle, as the bombarding isotope.

Radioactive Decay

Radioactive decay is the spontaneous transformation of an unstable atomic nucleus into a lighter nucleus. During this process, a nucleus emits particles or electromagnetic radiation - or sometimes both - to achieve greater stability.

As seen in the step-by-step solution of our textbook exercise, rutherfordium-257 undergoes radioactive decay, resulting in the emission of a particle. This transformation illustrates how elements seek stability via decay, which often results in the emission of alpha particles, beta particles, or gamma rays.

Understanding radioactive decay is crucial for determining the type of particle emitted during a nuclear reaction, which is essential for predicting how a particular isotope will transform over time and its potential applications or dangers.

Alpha Particle Emission

Alpha particle emission is a type of radioactive decay where an alpha particle, which is made up of two protons and two neutrons, is emitted from the nucleus of an unstable atom. This process results in a decrease in both atomic mass and atomic number, as explained in the textbook solution for the decay of rutherfordium-257.

The textbook example clarifies that rutherfordium-257 emits an alpha particle resulting in nobelium-253. Alpha particles are relatively large and carry a double positive charge due to the two protons. Despite their mass and charge, alpha particles have limited penetration capability but can be highly damaging to biological tissues if emitted internally. This property leads to their use in certain medical therapies and also requires careful handling to avoid health risks.

Isotopes

Isotopes are forms of the same element that contain equal numbers of protons but differing numbers of neutrons, giving them different mass numbers. While all isotopes of an element exhibit the same chemical behavior, their nuclear properties can vary significantly.

In the context of the textbook exercise, californium-249, rutherfordium-257, and nobelium-253 are all isotopes, specifically radioactive isotopes, that highlight the diversity within an element's isotopes. Despite having the same number of protons, different isotopes can have vastly different stabilities, leading to the radioactive decay observed in elements like rutherfordium. Understanding isotopes is fundamental in nuclear chemistry because it facilitates the calculation of nuclear reactions and the proper handling of various nuclear materials.