Question

Write balanced equations for each of the processes described below. a. Chromium- 51 , which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- 32, which accumulates in the liver, decays by \(\beta\) particle production.

Short answer

a. The balanced equation for the electron capture decay of Chromium-51 is: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \] b. The balanced equation for the beta decay of Iodine-131 is: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \] c. The balanced equation for the beta decay of Phosphorus-32 is: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \]

Step-by-step solution

a. Chromium-51 (electron capture)

In electron capture, a proton in the nucleus captures an electron and becomes a neutron. The atomic number decreases by 1, and the mass number remains the same. The general formula for an electron capture reaction is: \[ ^A_Z X + e^- \rightarrow ^A_{Z-1}Y \] So for Chromium-51: 1. Identify the isotope and its atomic number (Z) and mass number (A). Chromium-51 (Cr): Z = 24, A = 51. 2. Write the balanced nuclear equation: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \] Thus, the balanced equation for the electron capture decay of Chromium-51 is: \[ ^{51}_{24}Cr + e^- \rightarrow ^{51}_{23}V \]

b. Iodine-131 (beta decay)

In beta decay, a neutron in the nucleus converts into a proton and releases a beta particle (electron). The atomic number increases by 1, and the mass number remains the same. The general formula for a beta decay reaction is: \[ ^A_Z X \rightarrow ^A_{Z+1} Y + \beta^- \] So for Iodine-131: 1. Identify the isotope and its atomic number (Z) and mass number (A). Iodine-131 (I): Z = 53, A = 131. 2. Write the balanced nuclear equation: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \] Thus, the balanced equation for the beta decay of Iodine-131 is: \[ ^{131}_{53}I \rightarrow ^{131}_{54}Xe + \beta^- \]

c. Phosphorus-32 (beta decay)

For Phosphorus-32, we have another beta decay: 1. Identify the isotope and its atomic number (Z) and mass number (A). Phosphorus-32 (P): Z = 15, A = 32. 2. Write the balanced nuclear equation: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \] Thus, the balanced equation for the beta decay of Phosphorus-32 is: \[ ^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- \]

Key concepts

Electron Capture

Electron capture is a fascinating type of nuclear process. It involves an atomic nucleus "capturing" an electron from its inner shell. This captured electron combines with a proton to form a neutron. This process decreases the atomic number by one, but the mass number remains unchanged because the total number of nucleons (protons and neutrons) stays the same.
To illustrate this with chromium-51, which decays via electron capture, we start by identifying its atomic and mass numbers. Chromium-51 has an atomic number, Z = 24, and a mass number, A = 51. Using the nuclear equation for electron capture:

• Before the process: \[^A_Z X = ^{51}_{24}Cr\]
  
• The equation is: \[^A_Z X + e^- \rightarrow ^A_{Z-1}Y\]
  

After electron capture, chromium-51 transforms into vanadium-51 (\[^A_{Z-1}Y = ^{51}_{23}V\]), with one less proton and one more neutron. The balanced equation becomes:

\[^A_Z Cr + e^- \rightarrow ^{51}_{23}V\]

Beta Decay

Beta decay is one of the most common nuclear reactions, involving the transformation of a neutron into a proton with the emission of a beta particle (an electron) and an antineutrino. During this process, the atomic number of the nucleus increases by one, while the mass number remains the same. This happens because a neutron is lost and a proton is gained.
For iodine-131, which undergoes beta decay, its atomic number is Z = 53 and mass number is A = 131. The nuclear equation for beta decay is:

• Before the process: \[^A_Z X = ^{131}_{53}I\]
  
• The equation becomes: \[^A_Z X \rightarrow ^A_{Z+1} Y + \beta^-\]
  

After beta decay, iodine-131 transforms into xenon-131 (\[^A_{Z+1}Y = ^{131}_{54}Xe\]). The balanced equation is:

\[^A_Z I \rightarrow ^A_{Z+1}Xe + \beta^-\]

Nuclear Equation Balancing

Balancing nuclear equations is crucial to understanding nuclear reactions. It ensures that the number of protons and neutrons is conserved during the reaction.
Here's how to balance nuclear equations involving beta decay, using phosphorus-32 as an example.
With phosphorus-32 decaying by beta emission, its atomic number is 15, and its mass number is 32. The balanced nuclear equation for this reaction is

• Initial isotope: \[^A_Z X = ^{32}_{15}P\]
  
• The equation becomes: \[^A_Z X \rightarrow ^A_{Z+1} Y + \beta^-\]
  

After transforming a neutron into a proton, phosphorus-32 becomes sulfur-32 (\[^A_{Z+1} Y = ^{32}_{16}S\]). Hence, the balanced equation is

\[^A_Z P \rightarrow ^A_{Z+1}S + \beta^-\]
This balancing maintains mass and charge, reflecting physical conservation laws in every nuclear reaction.