Question

Exit signs Exit signs are coated with a paint containing phosphors. These phosphors are activated by the radioisotope tritium \(\left(\begin{array}{l}{3} \\\ {1}\end{array} \mathrm{H}\right)\) produced by bombarding lithium-6 with neutrons to produce lithium- \(7 .\) The lithium- 7 then undergoes alpha decay to produce the tritium. Write balanced nuclear equations for both steps.

Short answer

The balanced nuclear equations for the given steps are: 1. Bombarding lithium-6 with neutrons to produce lithium-7: \[_{3}^{6}Li + _{0}^{1}n \rightarrow _{3}^{7}Li\] 2. Alpha decay of lithium-7 to produce tritium: \[_{3}^{7}Li \rightarrow _{1}^{3}H + _{2}^{4}He\]

Step-by-step solution

Bombarding Lithium-6 with Neutrons

For the first step, we will consider the interaction of lithium-6 with a neutron to produce lithium-7. Lithium-6 can be represented as: \[_{3}^{6}Li\] A neutron can be represented as: \[_{0}^{1}n\] The nuclear equation for this step will be: \[_{3}^{6}Li + _{0}^{1}n \rightarrow _{3}^{7}Li\] This equation is already balanced, as the mass numbers on both sides of the equation are equal (6 + 1 = 7) and the atomic numbers are also equal (3 = 3).

Alpha Decay of Lithium-7 to Produce Tritium

In the second step, lithium-7 undergoes alpha decay to produce tritium. An alpha particle can be represented as: \[_{2}^{4}He\] The nuclear equation for this step will be: \[_{3}^{7}Li \rightarrow _{1}^{3}H + _{2}^{4}He\] This equation is also balanced, as the mass numbers on both sides of the equation are equal (7 = 3+4), and the atomic numbers are also equal (3 = 1+2). So, the balanced nuclear equations for the two steps are: 1. \[_{3}^{6}Li + _{0}^{1}n \rightarrow _{3}^{7}Li\] 2. \[_{3}^{7}Li \rightarrow _{1}^{3}H + _{2}^{4}He\]

Key concepts

Lithium-6

Lithium-6 is a stable isotope of lithium. It contains three protons and three neutrons, giving it an atomic mass of six. As an element, lithium sits in the alkali metal group on the periodic table and is known for its reactivity. Lithium-6 itself plays a crucial role in nuclear reactions, especially in contexts where it is bombarded with neutrons.

In the nuclear reaction specified in the exercise, lithium-6 undergoes a transformation when interacted with a neutron. When a neutron is introduced to lithium-6, it absorbs the neutron, changing to lithium-7. This reaction is often used in applications where lithium isotopes are important, such as nuclear reactors or experimental physics.

• Element: Lithium
• Isotope: Lithium-6
• Protons: 3
• Neutrons: 3

Understanding this process helps in recognizing how isotopic changes in elements can lead to different nuclear properties and applications.

Tritium

Tritium is a radioactive isotope of hydrogen, which is represented as hydrogen-3. It has one proton and two neutrons, giving it an atomic mass of three. Unlike regular hydrogen, tritium is unstable and eventually decays.

Tritium naturally decays into helium-3 through beta decay. However, in the scenario described, it is produced through an alpha decay reaction of lithium-7. The creation of tritium is vital in industries such as nuclear power and research, as well as in production of self-luminous materials for things like exit signs.

• Element: Hydrogen
• Isotope: Tritium (Hydrogen-3)
• Protons: 1
• Neutrons: 2

This understanding of tritium helps illustrate the nuclear transformations that allow for its production and the significance of isotopic variations in practical applications.

Alpha Decay

Alpha decay is a type of radioactive decay where an atomic nucleus emits an alpha particle and transforms into a new atom. This process decreases the mass number by four and the atomic number by two.

An alpha particle consists of two protons and two neutrons, which is the same as a helium nucleus \(_{2}^{4}He\). In the original exercise, lithium-7 undergoes alpha decay to form tritium and an alpha particle. This illustrates a process where one element changes to another through emission of particles. It's a common decay mode for heavy elements, providing a pathway for certain isotopes to reach a more stable state.

• Type: Radioactive Decay
• Particle Released: Alpha (\(_{2}^{4}He\))
• Results in: Change in both mass and atomic number

Understanding alpha decay highlights the significance of nuclear reactions in changing one element into another, which is fundamental in the study of nuclear physics and chemistry.