Question

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear reactions, which have been used to synthesize elements. a. _______ \(+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{97}^{243} \mathrm{Bk}+{ }_{0}^{1} \mathrm{n}\) b. b. \({ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \rightarrow\) ______ \(+6{ }_{0}^{1} \mathrm{n}\) c. C. \({ }_{98}^{249} \mathrm{Cf}+\) _____ \(\longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}\) d. \({ }_{98}^{249} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \rightarrow{ }_{103}^{257} \mathrm{Lr}+\) ______.

Short answer

The short version of the answer is: a. \({ }_{95}^{240} \mathrm{Am} +{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{97}^{243} \mathrm{Bk}+{ }_{0}^{1} \mathrm{n}\) b. \({ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{98}^{244} \mathrm{Cf} + 6{ }_{0}^{1} \mathrm{n}\) c. \({ }_{98}^{249} \mathrm{Cf} +{ }_{7}^{15} \mathrm{N} \longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}\) d. \({ }_{98}^{249} \mathrm{Cf}+{ }_{5}^{10} \mathrm{B} \rightarrow{ }_{103}^{257} \mathrm{Lr}+2{ }_{0}^{1} \mathrm{n}\)

Step-by-step solution

Identify the missing reactant

We need to find the atomic number and mass number of the missing reactant.

Balance the atomic and mass numbers

To balance the mass numbers: X_mass + 4 = 243 + 1, thus X_mass = 243 + 1 - 4 = 240. To balance the atomic numbers: X_atomic + 2 = 97, thus X_atomic = 97 - 2 = 95.

Write the complete reaction

The complete reaction is: \({ }_{95}^{240} \mathrm{Am} +{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{97}^{243} \mathrm{Bk}+{ }_{0}^{1} \mathrm{n}\) b. \({ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \rightarrow\) ______ \(+6{ }_{0}^{1} \mathrm{n}\)

Identify the missing product

We need to find the atomic number and mass number of the missing product.

Balance the atomic and mass numbers

To balance the mass numbers: 238 + 12 = X_mass + 6, thus X_mass = 238 + 12 - 6 = 244. To balance the atomic numbers: 92 + 6 = X_atomic, thus X_atomic = 98.

Write the complete reaction

The complete reaction is: \({ }_{92}^{238} \mathrm{U}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{98}^{244} \mathrm{Cf} + 6{ }_{0}^{1} \mathrm{n}\) c. \({ }_{98}^{249} \mathrm{Cf}+\) _____ $\longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}$

Identify the missing reactant

We need to find the atomic number and mass number of the missing reactant.

Balance the atomic and mass numbers

To balance the mass numbers: 249 + X_mass = 260 + 4, thus X_mass = 260 + 4 - 249 = 15. To balance the atomic numbers: 98 + X_atomic = 105, thus X_atomic = 105 - 98 = 7.

Write the complete reaction

The complete reaction is: \({ }_{98}^{249} \mathrm{Cf} +{ }_{7}^{15} \mathrm{N} \longrightarrow{ }_{105}^{260} \mathrm{Db}+4{ }_{0}^{1} \mathrm{n}\) d. ${ }_{98}^{249} \mathrm{Cf}+{ }_{5}^{10} \mathrm{B} \rightarrow{ }_{103}^{257} \mathrm{Lr}+$ ______.

Identify the missing product

We need to find the atomic number and mass number of the missing product.

Balance the atomic and mass numbers

To balance the mass numbers: 249 + 10 = 257 + X_mass, thus X_mass = 249 + 10 - 257 = 2. To balance the atomic numbers: 98 + 5 = 103, thus no need for any additional atomic number.

Write the complete reaction

The complete reaction is: \({ }_{98}^{249} \mathrm{Cf}+{ }_{5}^{10} \mathrm{B} \rightarrow{ }_{103}^{257} \mathrm{Lr}+2{ }_{0}^{1} \mathrm{n}\)

Key concepts

Nuclear Reactions

At the core of understanding synthetic elements is the process called nuclear reactions. These are not like chemical reactions that involve electrons, but rather involve the nuclei of atoms. When nuclei are transformed into different nuclei, it's usually accompanied by the change in the number of protons and neutrons, or by the emission of gamma rays. In the textbook exercise, we glimpse nuclear reactions that synthesize new elements by bombarding heavy atoms with high-energy particles. These particles can either add to the nucleus, changing its mass number, or change the composition of the nucleus itself, affecting its atomic number, which is often followed by the emission of neutrons.

For instance, when an alpha particle (omenclature{He}{Helium} nucleus) strikes another atom, it can merge to form a heavier nucleus. The exercise provided includes balancing nuclear reactions, which is crucial because it follows the law of conservation of mass and atomic number. Additional particles like neutrons (omenclature{n}{neutron}) are often released to balance the equation.

Particle Accelerators

The textbook exercise mentions using high-energy particles to bombard atoms, which hints at the utilization of particle accelerators. These sophisticated machines accelerate charged particles, such as protons, electrons, or atomic nuclei, to very high speeds — close to the speed of light in some cases — and then collide them with target atoms. This immense energy can lead to the fusion of particles, creating new elements that aren't found in nature.

Particle accelerators are not just for synthesizing elements; they have a wide range of applications from medical treatments to fundamental research in physics. In the context of the exercise, it’s these high-speed collisions orchestrated by accelerators that allow us to observe and study the resulting nuclear reactions and the properties of the synthesized elements.

Atomic Number

Every element on the periodic table has a unique identifier known as its atomic number. This is equal to the number of protons in an atom's nucleus and is denoted by the letter omenclature{Z}{Atomic Number}. In a neutral atom, this also equals the number of electrons orbiting the nucleus. The atomic number is fundamental to the identity of an element — if you change it, you change the element itself.

In the reactions provided in the exercise, identifying and balancing atomic numbers is essential for determining the missing components in a nuclear reaction. For instance, when a Uranium-238 nucleus (omenclature{U}{Uranium}) captures a Carbon-12 nucleus (omenclature{C}{Carbon}), it transforms into a different element with atomic number 98 (Californium), showcasing how the identity of an element changes in a nuclear reaction.

Mass Number

Contrary to the atomic number, the mass number, denoted by omenclature{A}{Mass Number}, is the total count of protons and neutrons in an atom’s nucleus. While the atomic number determines the element’s identity, the mass number helps distinguish between the different isotopes of that element.

In the nuclear equations you're studying, the mass numbers must be balanced on both sides of the reaction. For instance, when Californium-249 (omenclature{Cf}{Californium}) is bombarded with nitrogen-15 (omenclature{N}{Nitrogen}), not only do we see a new element, Dubnium-260 (omenclature{Db}{Dubnium}), being formed, but we also note the release of additional neutrons to keep the mass balanced. Understanding how mass number plays a role in nuclear reactions is critical for predicting the outcomes of such experiments and is a key concept in the synthesis of new elements.