Question

Uranium- 238 decays in a series of nuclear changes until stable lead-206 is produced. Complete the following nuclear equations that are part of the uranium- 238 decay series: a. \({ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+\) ? b. \({ }_{90}^{234} \mathrm{Th} \longrightarrow ?+{ }_{-1}^{0} e\) c. \(? \longrightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}\)

Short answer

(a) \({}_{92}^{238} \text{U} \rightarrow {}_{90}^{234} \text{Th} + {}_{2}^{4} \text{He}\) \(b) \({}_{90}^{234} \text{Th} \rightarrow {}_{91}^{234} \text{Pa} + {}_{-1}^{0} e\) \(c) \({}_{88}^{226} \text{Ra} \rightarrow {}_{86}^{222} \text{Rn} + {}_{2}^{4} \text{He}\)

Step-by-step solution

Understanding the Uranium-238 Decay Series

Uranium-238 undergoes a series of decays until it eventually becomes stable lead-206. Each decay involves either alpha decay or beta decay.

- Completing part (a)

In the first equation \({}_{92}^{238} \text{U} \rightarrow {}_{90}^{234} \text{Th} + ?\), notice that Uranium-238 loses an alpha particle. An alpha particle is represented by \({}_{2}^{4} \text{He}\). So, the missing particle is \({}_{2}^{4} \text{He}\).

- Completing part (b)

In the second equation \({}_{90}^{234} \text{Th} \rightarrow ? + {}_{-1}^{0} e\), Thorium-234 undergoes beta decay. A beta particle is represented by \({}_{-1}^{0} e\). The resulting element should have an atomic number of 91 and a mass number of 234. Thus, the missing element is Protactinium-234 (Pa), so \({}_{90}^{234} \text{Th} \rightarrow {}_{91}^{234} \text{Pa} + {}_{-1}^{0} e\).

- Completing part (c)

In the third equation \(? \rightarrow {}_{86}^{222} \text{Rn} + {}_{2}^{4} \text{He}\), the resulting element Rn-222 and an alpha particle are produced. For the initial nucleus, add the atomic number of radon (86) and the alpha particle (2) to get 88, and add their mass numbers (222 + 4) to get 226. The initial nucleus is Radium-226 (Ra), so the complete equation is \({}_{88}^{226} \text{Ra} \rightarrow {}_{86}^{222} \text{Rn} + {}_{2}^{4} \text{He}\).

Key concepts

nuclear equations

Nuclear equations are symbolic representations of nuclear reactions. These equations show the changes in the atomic nuclei during reactions. Unlike chemical equations, which involve electrons and chemical bonds, nuclear equations focus on the transformation of nuclei and the particles involved. In nuclear equations, both the sum of the atomic numbers (protons) and the sum of the mass numbers (protons + neutrons) must be balanced on both sides of the equation. For example, in alpha decay, an alpha particle (with 2 protons and 2 neutrons) is emitted. Therefore, the original nucleus loses 2 protons and 4 mass units. These equations help us understand processes like radioactive decay and nuclear fission and fusion. In the Uranium-238 decay series, we're dealing with both alpha and beta decay processes. Each type of decay has its unique representation in a nuclear equation, ensuring the balance of particles.

alpha decay

Alpha decay is a type of radioactive decay in which an unstable nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, symbolized as \(_{2}^{4} \text{He}\). This emission reduces the atomic number of the original nucleus by 2 and its mass number by 4. Alpha decay typically occurs in heavy elements like Uranium-238. For example, in the Uranium-238 decay series, Uranium-238 undergoes alpha decay to become Thorium-234: \(_{92}^{238} \text{U} \rightarrow _{90}^{234} \text{Th} + _{2}^{4} \text{He}\). The emitted alpha particle is the same as a helium nucleus. This change helps the original heavy nucleus become more stable. Understanding alpha decay is crucial for studying nuclear reactions and radioactive processes, contributing to fields like nuclear energy and radiometric dating.

beta decay

Beta decay is another type of radioactive decay where a nucleus emits a beta particle. A beta particle is a high-energy, high-speed electron or positron and is symbolized as \(_{-1}^{0} \text{e}\) or \(_{+1}^{0} \text{e}\). In beta decay, a neutron in the nucleus is transformed into a proton, or vice versa. In the Uranium-238 decay series, Thorium-234 undergoes beta decay to become Protactinium-234: \(_{90}^{234} \text{Th} \rightarrow _{91}^{234} \text{Pa} + _{-1}^{0} \text{e}\). Here, a neutron in Thorium's nucleus converts into a proton and emits an electron (beta particle). The atomic number increases by 1, but the mass number stays the same. Beta decay helps transform one element into another, moving towards a more stable configuration. This process is significant in fields like nuclear medicine and understanding stellar nucleosynthesis.

uranium decay chain

The Uranium-238 decay chain is a series of radioactive decays that start from Uranium-238 and end in stable Lead-206. This series involves multiple steps of alpha and beta decay. Each decay step results in the formation of a different isotope or element. The decay chain shows how a heavy, unstable nucleus can transform into a stable one. Step by step:

• Uranium-238 undergoes alpha decay to become Thorium-234 (\(_{92}^{238} \rightarrow _{90}^{234} \text{Th} + _{2}^{4} \text{He}\)).
• Thorium-234 then undergoes beta decay to become Protactinium-234 (\(_{90}^{234} \text{Th} \rightarrow _{91}^{234} \text{Pa} + _{-1}^{0} \text{e}\)).
• Further steps involve additional alpha and beta decays.

This decay chain is important in areas such as geology, where it helps with radiometric dating and understanding the age of rocks and minerals. It also informs safety protocols in nuclear energy production and handling radioactive materials.