Problem 24 - Chemistry An Atoms First Approach

Question

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear equations, which have been used to synthesize elements. a. \(\quad+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow$$\quad$$+6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+$$\quad$$\rightarrow \frac{260}{105} D b+4 \frac{1}{6} n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\)__________

Step-by-Step Solution

Answer

The completed nuclear reactions are: a. \(^{239}_{95}Am+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\ ^{244}_{98}Cf +6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+^{11}_{7}N \rightarrow \ ^{260}_{105} D b+4 \frac{1}{6}n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \ ^{257}_{153} \mathrm{Lr}+2_{0}^{1}n\)

Step 1: Find the atomic and mass numbers of the missing element

We need to find the atomic and mass numbers for the missing element in the reaction. Since we know the atomic and mass numbers of other particles involved in the reaction, we can use conservation of atomic and mass numbers to find the missing element. To conserve the atomic numbers, we must have: missing element's atomic number + 2 = 97 To conserve the mass numbers, we must have: missing element's mass number + 4 = 243 Now, let's solve these equations to find the missing element's atomic and mass numbers.

Step 2: Solve the equations

Solving the equations from Step 1, we get the missing element's atomic and mass numbers as: missing element's atomic number = 97 - 2 = 95 missing element's mass number = 243 - 4 = 239 Thus, the missing element in the reaction is \(^{239}_{95}Am\). The completed reaction for (a) is: \(a. \quad ^{239}_{95}Am+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\quad+6_{0}^{1} n\)

Step 3: Find the atomic and mass numbers of the missing element

Following the same procedure as before, we need to conserve the atomic and mass numbers in the reaction. To conserve the atomic numbers, we must have: 92 + 6 = missing element's atomic number To conserve the mass numbers, we must have: 238 + 12 = missing element's mass number + 6

Step 4: Solve the equations

Solving the equations from Step 3, we get the missing element's atomic and mass numbers as: missing element's atomic number = 92 + 6 = 98 missing element's mass number = 238 + 12 - 6 = 244 Thus, the missing element in the reaction is \(^{244}_{98}Cf\). The completed reaction for (b) is: \(b. \quad ^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\ ^{244}_{98}Cf +6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+\quad\rightarrow \frac{260}{105} D b+4 \frac{1}{6}n\)

Step 5: Find the atomic and mass numbers of the missing element

Following the same procedure, we need to conserve the atomic and mass numbers in the reaction. To conserve the atomic numbers, we must have: 98 + missing element's atomic number = 105 To conserve the mass numbers, we must have: 249 + missing element's mass number = 260

Step 6: Solve the equations

Solving the equations from Step 5, we get the missing element's atomic and mass numbers as: missing element's atomic number = 105 - 98 = 7 missing element's mass number = 260 - 249 = 11 Thus, the missing element in the reaction is \(^{11}_{7}N\). The completed reaction for (c) is: \(c. \quad ^{249} \mathrm{Cf}+^{11}_{7}N \rightarrow \ ^{260}_{105} D b+4 \frac{1}{6}n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\text{__________}\)

Step 7: Find the number of neutrons emitted

Since we are given both atomic and mass numbers for both elements in the reaction, we can directly find the number of neutrons emitted by conserving the mass numbers. To conserve the mass numbers, we must have: 249 + 10 = 257 + missing neutrons

Step 8: Solve for the number of neutrons emitted

Solving the equation from Step 7, we get the missing neutrons as: missing neutrons = 249 + 10 - 257 = 2 The completed reaction for (d) is: \(d. \quad ^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \ ^{257}_{153} \mathrm{Lr}+2_{0}^{1}n\)

Key Concepts

Particle Accelerators

Particle accelerators are sophisticated machines designed to propel charged particles, such as protons or electrons, to high speeds. This process involves using strong electric and magnetic fields that guide these particles along a precise path. Once accelerated, these high-energy particles can be aimed at a target, often a fixed atomic nucleus. The primary goal of particle accelerators in nuclear chemistry is to collide these high-speed particles with atomic nuclei, producing nuclear reactions that can transform elements.

Key features of particle accelerators include:

• **Acceleration**: Particles achieve high speeds due to powerful electric fields, often close to the speed of light.
• **Containment**: Magnetic fields keep the particles on a prescribed path, preventing them from scattering.
• **Control**: Sophisticated systems ensure precise aiming of particles towards the target nucleus.

There are different types of particle accelerators, such as linear accelerators (linacs) and circular ones like cyclotrons and synchrotrons. Each type is suited for specific tasks within scientific research, including the synthesis of new elements.

Nuclear Reactions

Nuclear reactions involve the altering of the components within an atom's nucleus. This is distinct from chemical reactions, where only the electron cloud surrounding the nucleus is affected. In nuclear reactions, atomic nuclei can split, combine, or emit particles to form new elements or isotopes.

Crucial aspects of nuclear reactions include:

• **Conservation Laws**: Both mass number and atomic number are conserved during a nuclear reaction. This means that the total of these numbers before and after the reaction remains constant.
• **Energy Release**: Often, nuclear reactions release significant amounts of energy, which can manifest as heat or radiation.
• **Particle Emission**: Common emissions in nuclear reactions include alpha particles (\(^4_2He\)), beta particles (\(\beta\)), and neutrons (\(^1_0n\)).

In the context of element synthesis, nuclear reactions may involve bombarding heavy atoms with particles accelerated by particle accelerators to create new, often unstable, elements that decay into more stable forms.

Element Synthesis

Element synthesis is the creation of new elements through nuclear reactions. This process often involves the bombardment of target nuclei with accelerated particles in a particle accelerator. The interaction between the high-speed particles and atomic nuclei can form elements that do not naturally occur on Earth.

Highlights of element synthesis include:

• **Target Nuclei**: Elements chosen for bombardment are typically heavy, with large nuclei that can capture additional protons or other particles.
• **Reaction Types**: These are fusion reactions where two nuclei combine to form a heavier nucleus.
• **Formation of Rare Elements**: Many synthesized elements are highly unstable, existing only for a brief moment before decaying into other elements.

Many of the elements listed on the periodic table beyond uranium (atomic number 92) have been synthesized in this manner, showcasing the power of human ingenuity to extend our understanding of the fundamental constituents of matter.