Question

Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: (a) \({ }_{3}^{6} \mathrm{Li}+\frac{56}{25} \mathrm{Ni} \longrightarrow \longrightarrow ?\) (b) \({ }_{20}^{40} \mathrm{Ca}+{ }_{96}^{248} \mathrm{Cm} \ldots{ }_{62}^{147} \mathrm{Sm}+?\) (c) \(\frac{88}{38} \mathrm{Sr}+{ }_{36}^{84} \mathrm{Kr} \longrightarrow{\rightarrow} \rightarrow{ }_{46}^{116} \mathrm{Pd}+?\) (d) \({ }_{20}^{40} \mathrm{Ca}+{ }_{92}^{238} \mathrm{U} \ldots \longrightarrow \frac{70}{30} \mathrm{Zn}+4{ }_{0}^{1} \mathrm{n}+2 ?\)

Short answer

The short answers for the given reactions are: (a) \({}_{3}^{6}\textrm{Li} + {}_{25}^{56}\textrm{Ni} \longrightarrow {}_{28}^{62}\textrm{Ni}\) (b) \({}_{20}^{40}\textrm{Ca} + {}_{96}^{248}\textrm{Cm} \longrightarrow {}_{62}^{147}\textrm{Sm} + {}_{54}^{141}\textrm{Xe}\) (c) \({}_{38}^{88}\textrm{Sr} + {}_{36}^{84}\textrm{Kr} \longrightarrow {}_{46}^{116}\textrm{Pd} + {}_{28}^{56}\textrm{Ni}\) (d) \({}_{20}^{40}\textrm{Ca} + {}_{92}^{238}\textrm{U} \longrightarrow {}_{30}^{70}\textrm{Zn} + 4{_0^1}\textrm{n} + 2({_82}^{102}\textrm{Pb})\)

Step-by-step solution

Determine Atomic and Mass Numbers

For the first reaction, we are given: \({}_{3}^{6}\textrm{Li} + {}_{25}^{56}\textrm{Ni} \longrightarrow \ ? \) The total number of protons is 3 + 25 = 28, and the total mass number is 6 + 56 = 62.

Find the Unknown Element

Now search for the element in the periodic table with atomic number 28, which is Nickel (Ni).

Write the Balanced Equation

Now write the balanced nuclear equation: \({}_{3}^{6}\textrm{Li} + {}_{25}^{56}\textrm{Ni} \longrightarrow {}_{28}^{62}\textrm{Ni}\) (b)

Determine Atomic and Mass Numbers

For the second reaction, we are given: \({}_{20}^{40}\textrm{Ca} + {}_{96}^{248}\textrm{Cm} \longrightarrow {}_{62}^{147}\textrm{Sm} + \ ? \) First, find the difference in atomic and mass numbers: Protons difference: 20 + 96 - 62 = 54; Mass Number difference: 40 + 248 - 147 = 141.

Find the Unknown Element

Now search for the element in the periodic table with atomic number 54, which is Xenon (Xe).

Write the Balanced Equation

Now write the balanced nuclear equation: \({}_{20}^{40}\textrm{Ca} + {}_{96}^{248}\textrm{Cm} \longrightarrow {}_{62}^{147}\textrm{Sm} + {}_{54}^{141}\textrm{Xe}\) (c)

Determine Atomic and Mass Numbers

For the third reaction, we are given: \({}_{38}^{88}\textrm{Sr} + {}_{36}^{84}\textrm{Kr} \longrightarrow {}_{46}^{116}\textrm{Pd} + \ ? \) First, find the difference in atomic and mass numbers: Protons difference: 38 + 36 - 46 = 28; Mass Number difference: 88 + 84 - 116 = 56.

Find the Unknown Element

Now search for element in the periodic table with atomic number 28, which is Nickel (Ni).

Write the Balanced Equation

Now write the balanced nuclear equation: \({}_{38}^{88}\textrm{Sr} + {}_{36}^{84}\textrm{Kr} \longrightarrow {}_{46}^{116}\textrm{Pd} + {}_{28}^{56}\textrm{Ni}\) (d)

Determine Atomic and Mass Numbers

For the fourth reaction, we are given: \({}_{20}^{40}\textrm{Ca} + {}_{92}^{238}\textrm{U} \longrightarrow {}_{30}^{70}\textrm{Zn} + 4{_0^1}\textrm{n} \ + \ 2 \ ? \) Count the protons and mass numbers on both sides: Protons difference: 20 + 92 - 30 - 0 = 82; Mass Number difference: 40 + 238 - 70 - 4 = 204 (since there are two unknown nuclei, divide this by 2 to have 102 for each).

Find the Unknown Element

Now search for the element in the periodic table with atomic number 82, which is Lead (Pb).

Write the Balanced Equation

Now write the balanced nuclear equation: \({}_{20}^{40}\textrm{Ca} + {}_{92}^{238}\textrm{U} \longrightarrow {}_{30}^{70}\textrm{Zn} + 4{_0^1}\textrm{n} + 2({_82}^{102}\textrm{Pb})\)

Key concepts

Heavy-Ion Collisions

Heavy-ion collisions involve the interaction of nuclei, like lithium or calcium, with others in an accelerated manner. These collisions are often conducted in particle accelerators, where ions are propelled at high speeds and energies.
This process mimics natural cosmic events on Earth and allows scientists to explore the properties of matter under extreme conditions. In our original exercise, reactions such as \({ }_{3}^{6} \mathrm{Li}+{}_{25}^{56} \mathrm{Ni}\longrightarrow \) depict a fusion process where light elements are bombarded with heavier ions.

• Heavy-ion bombardments lead to the fusion of atomic nuclei.
• This method enables scientists to simulate and study astrophysical processes.
• It plays a critical role in synthesizing new and artificial elements.

Through such collisions, nuclei can be transformed into new elements or isotopes, expanding our understanding and the periodic table itself.

Nuclear Synthesis

Nuclear synthesis, or nucleosynthesis, is the process of creating new atomic nuclei from pre-existing protons and neutrons. This is especially relevant to heavy-ion collisions, where the fusion of nuclei results in the formation of heavier elements. Nuclear synthesis is how elements heavier than helium develop in stars or during supernovae explosions.
In labs on Earth, this synthesis often occurs at a smaller scale using reactors or particle accelerators, leading to the creation of isotopes or transuranic elements - elements that are heavier than uranium.

• Nuclear synthesis helps us understand the origins of the elements in the universe.
• Laboratories can artificially reproduce these conditions to discover new isotopes.
• This process adds new elements to the periodic table, increasing its richness.

In our exercise, reactions like \({ }_{20}^{40} \mathrm{Ca}+{}_{96}^{248} \mathrm{Cm}\ldots \rightarrow {}_{62}^{147} \mathrm{Sm}+?\), demonstrate nuclear synthesis by forming potentially new or rare isotopes.

Mass Number Calculations

Mass number calculations play a vital role in understanding nuclear reactions. The mass number of an element is the sum of protons and neutrons in its nucleus. When balancing nuclear equations, calculating these numbers ensures conservation laws are observed, specifically the conservation of nucleon number and charge conservation.
For any given reaction, such as the one described in the original exercise \({}_{88}^{38}\mathrm{Sr} + {}_{36}^{84}\mathrm{Kr} \longrightarrow {}_{46}^{116}\mathrm{Pd} +?\), we calculate the total mass number before and after the reaction. This maintains the equation's balance; any discrepancy helps identify missing or newly formed particles.

• Add the atomic and mass numbers to find the missing component of the equation.
• Use these calculations to identify unknown elements formed during reactions.
• This process ensures that physical laws are satisfied in chemical equations.

The principle of maintaining the mass and atomic numbers through calculations allows us, like puzzle-solving, to identify newly synthesized or missing elements in nuclear reactions.