Question

Decay of which nucleus will lead to the following products: (a) bismuth- 211 by beta decay; (b) chromium 50 by positron emission; (c) tantalum-179 by electron capture; (d) radium-226 by alpha decay?

Short answer

The decaying nuclei leading to the given products are: (a) lead-211 (Pb-211) by beta decay; (b) manganese-50 (Mn-50) by positron emission; (c) tungsten-179 (W-179) by electron capture; and (d) thorium-230 (Th-230) by alpha decay.

Step-by-step solution

Case (a): Bismuth-211 by beta decay

To find the decaying nucleus that leads to bismuth-211 (Bi-211) by beta decay, we first analyze the decay process: $Bi-211\to X+e^{-}$ In beta decay, the atomic mass number (A) remains the same, but an atomic number (Z) increases by 1. We will use this to find the decaying nucleus: 1. Find the initial atomic number (Z) Bismuth-211 has an atomic number Z = 83. 2. Calculate the unknown nucleus In beta decay, Z increases by 1, so the parent nucleus has Z = 83 - 1 = 82, and A remains the same at 211. 3. Identify the element The element with Z = 82 is lead (Pb). Therefore, the decay of lead-211 (Pb-211) nucleus will lead to bismuth-211 by beta decay.

Case (b): Chromium-50 by positron emission

To find the decaying nucleus that leads to chromium-50 (Cr-50) by positron emission, we first analyze the decay process: $Cr-50\to X+e^{+}$ In positron emission, the atomic mass number (A) remains the same, but the atomic number (Z) decreases by 1. We will use this to find the decaying nucleus: 1. Find the initial atomic number (Z) Chromium-50 has an atomic number Z = 24. 2. Calculate the unknown nucleus In positron emission, Z decreases by 1, so the parent nucleus has Z = 24 + 1 = 25, and A remains the same at 50. 3. Identify the element The element with Z = 25 is manganese (Mn). Therefore, the decay of manganese-50 (Mn-50) nucleus will lead to chromium-50 by positron emission.

Case (c): Tantalum-179 by electron capture

To find the decaying nucleus that leads to tantalum-179 (Ta-179) by electron capture, we first analyze the decay process: $e^{-}+Ta-179\to X$ In electron capture, the atomic mass number (A) remains the same, but the atomic number (Z) decreases by 1. We will use this to find the decaying nucleus: 1. Find the initial atomic number (Z) Tantalum-179 has an atomic number Z = 73. 2. Calculate the unknown nucleus In electron capture, Z decreases by 1, so the parent nucleus has Z = 73 + 1 = 74, and A remains the same at 179. 3. Identify the element The element with Z = 74 is tungsten (W). Therefore, the decay of tungsten-179 (W-179) nucleus will lead to tantalum-179 by electron capture.

Case (d): Radium-226 by alpha decay

To find the decaying nucleus that leads to radium-226 (Ra-226) by alpha decay, we first analyze the decay process: $Ra-226\to X+\alpha$ In alpha decay, the atomic mass number (A) decreases by 4, and the atomic number (Z) decreases by 2. We will use this to find the decaying nucleus: 1. Find the initial atomic number (Z) Radium-226 has an atomic number Z = 88. 2. Calculate the unknown nucleus In alpha decay, Z decreases by 2, so the parent nucleus has Z = 88 + 2 = 90, and A decreases by 4, so the parent nucleus has A = 226 + 4 = 230. 3. Identify the element The element with Z = 90 is thorium (Th). Therefore, the decay of thorium-230 (Th-230) nucleus will lead to radium-226 by alpha decay.

Key concepts

Beta Decay

Beta decay is a type of radioactive decay in which an unstable atomic nucleus transforms into a slightly lighter nucleus by emitting a beta particle, which can be either an electron ($e^{-}$) or a positron ($e^{+}$). The atomic mass number (A) of the nucleus remains unchanged, but the atomic number (Z) increases by one, as a neutron is converted into a proton within the nucleus.

To illustrate, consider the decay of lead-211 ($Pb-211$) into bismuth-211 ($Bi-211$). Since a beta particle is an electron, and it is emitted from the nucleus, the atomic number of lead increases from 82 to 83, thereby becoming bismuth. The mass number remains 211 because this form of decay does not affect the number of nucleons (protons and neutrons) in the nucleus. The equation for this beta decay process is:
$Pb-211\to Bi-211+e^{-}$.

Positron Emission

Positron emission, also known as beta plus decay (${\beta }^{+}$ decay), is a process where an unstable nucleus emits a positron ($e^{+}$)—the antimatter counterpart of the electron. In this decay mode, a proton is converted into a neutron, leading to a decrease in the atomic number by one while the mass number stays the same.

For example, when manganese-50 ($Mn-50$) undergoes positron emission, it transforms into chromium-50 ($Cr-50$). The atomic number decreases from 25 to 24 as a result of the positron emission, and since the mass number doesn't change, it remains at 50. The reaction is represented as follows:
$Mn-50\to Cr-50+e^{+}$.

Electron Capture

Electron capture is a decay process where an inner-shell electron from the atom is captured by the nucleus, combining with a proton to form a neutron and emitting a neutrino ($u_e$). Similar to positron emission, this results in a decrease of the atomic number by one while the atomic mass number remains constant.

Take the decay of tungsten-179 ($W-179$) into tantalum-179 ($Ta-179$) as an example. An electron is captured by the nucleus, reducing the atomic number from 74 to 73 but keeping the mass number at 179. The resulting nucleus is tantalum-179, and the equation can be represented by:
$e^{-}+W-179\to Ta-179+u_e$.

Alpha Decay

Alpha decay is one of the most common forms of radioactive decay wherein an alpha particle, consisting of two protons and two neutrons (${}^{4}He$ nucleus), is emitted from an unstable parent nucleus. This results in a reduction of the atomic mass number by four and the atomic number by two.

An example of alpha decay is thorium-230 ($Th-230$) decaying into radium-226 ($Ra-226$). The thorium nucleus loses an alpha particle, leading to the new configuration with a decreased atomic number from 90 to 88 (radium) and a mass number reduced from 230 to 226. The alpha decay process is represented by:
$Th-230\to Ra-226{+}^{4}He$.