Q25PE - College Physics (Urone)

Question

: β decay producing ¹³⁷Ba. The parent nuclide is a major waste product of reactors and has chemistry similar to potassium and sodium, resulting in its concentration in your cells if ingested.

Step-by-Step Solution

Answer

The equation is obtained as: \[_{55}^{137}C{s_{82}} \to _{56}^{137}B{a_{81}}{ + _{ - 1}}\beta  + {\bar v_e}\].

Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Explanation

The term β^- decay" refers to the emission of a beta particle. The beta particle has an atomic mass of 0 and an atomic number of - 1. As a result, the general equation may be expressed as follows:

\[_Z^AX \to _{z + 1}^A{Y_{z - 1}}{ + _{ - 1}}\beta  + {\bar v_e}\]

The atomic number of ¹³⁷Ba is 56. 

As a result, the parent nucleus will have an atomic number of 56 - 1 = 55 before beta decay. The atomic mass of ¹³⁷Ba will be 137. 

As a result, the parent nucleus will have an atomic mass of 137 + 0 = 137 before beta decay.

As a result, the parent nucleus is \[_{55}^{137}Cs\]. 

As a result, the decay equation will be:

\[_{55}^{137}C{s_{82}} \to _{56}^{137}B{a_{81}}{ + _{ - 1}}\beta  + {\bar v_e}\]

Therefore, the equation is: \[_{55}^{137}C{s_{82}} \to _{56}^{137}B{a_{81}}{ + _{ - 1}}\beta  + {\bar v_e}\].