Question

Write balanced nuclear equations for (a) the loss of an alpha particle by Th-230. (b) the loss of a beta particle by \(\mathrm{Pb}-210\). (c) the fission of \(\mathrm{U}-235\) to give \(\mathrm{Ba}-140\), another nucleus, and an excess of two neutrons. (d) the \(\mathrm{K}\) -capture of \(\mathrm{Ar}-37\).

Short answer

Question: Write balanced nuclear equations for the four given nuclear reactions. (a) Loss of an alpha particle by Th-230 Answer: \(\mathrm{Th}-230 \rightarrow \mathrm{He}-4 + \mathrm{Ra}-226\) (b) Loss of a beta particle by \(\mathrm{Pb}-210\) Answer: \(\mathrm{Pb}-210 \rightarrow \beta + \mathrm{Bi}-210\) (c) Fission of \(\mathrm{U}-235\) to give \(\mathrm{Ba}-140\), another nucleus, and an excess of two neutrons Answer: \(\mathrm{U}-235 \rightarrow \mathrm{Ba}-140 + \mathrm{Kr}-93 + 2\mathrm{n}\) (d) The \(\mathrm{K}\) -capture of \(\mathrm{Ar}-37\) Answer: \(\mathrm{Ar}-37 + \mathrm{e^-} \rightarrow \mathrm{Cl}-37\)

Step-by-step solution

(a) Loss of an alpha particle by Th-230

An alpha particle is a helium nucleus, which consists of 2 protons and 2 neutrons (mass number 4). The reaction can be written as: \(\mathrm{Th}-230 \rightarrow \mathrm{He}-4 + \mathrm{X}\) Now we balance the equation by finding the element X: \(\mathrm{Th}(90\text{_protons},230-90\text{_neutrons}) \rightarrow \mathrm{He}(2\text{_protons},4-2\text{_neutrons}) + \mathrm{X}(90-2\text{_protons},230-90-4\text{_neutrons})\) Thus, X is Radium (Ra) with atomic number 88, and the equation is: \(\mathrm{Th}-230 \rightarrow \mathrm{He}-4 + \mathrm{Ra}-226\)

(b) Loss of a beta particle by \(\mathrm{Pb}-210\)

A beta particle is an electron, which has a mass number of 0 and an atomic number of -1. The reaction can be written as: \(\mathrm{Pb}-210 \rightarrow \beta + \mathrm{X}\) Balancing the equation gives: \(\mathrm{Pb}(82\text{_protons},210-82\text{_neutrons}) \rightarrow \beta(-1\text{_protons},0\text{_neutrons}) + \mathrm{X}(82-(-1)\text{_protons},210-82\text{_neutrons})\) Thus, X is Bismuth (Bi) with atomic number 83, and the equation is: \(\mathrm{Pb}-210 \rightarrow \beta + \mathrm{Bi}-210\)

(c) Fission of \(\mathrm{U}-235\) to give \(\mathrm{Ba}-140\), another nucleus, and two neutrons

The reaction can be represented as: \(\mathrm{U}-235 \rightarrow \mathrm{Ba}-140 + \mathrm{X} + 2\mathrm{n}\) Balancing the equation gives: \(\mathrm{U}(92\text{_protons},235-92\text{_neutrons}) \rightarrow \mathrm{Ba}(56\text{_protons},140-56\text{_neutrons}) + \mathrm{X}(92-56\text{_protons},235-92-140\text{_neutrons}) + 2\mathrm{n}(0\text{_protons},1\text{_neutrons})\) Thus, X is Krypton (Kr) with atomic number 36, and the equation is: \(\mathrm{U}-235 \rightarrow \mathrm{Ba}-140 + \mathrm{Kr}-93 + 2\mathrm{n}\)

(d) The \(\mathrm{K}\) -capture of \(\mathrm{Ar}-37\)

In a \(\mathrm{K}\)-capture, an electron from the innermost shell (K-shell) is captured by the nucleus, decreasing the atomic number by 1 while keeping the mass number constant. The reaction can be written as: \(\mathrm{Ar}-37 + \mathrm{e^-} \rightarrow \mathrm{X}\) Balancing the equation gives: \(\mathrm{Ar}(18\text{_protons},37-18\text{_neutrons}) + \mathrm{e^-}(-1\text{_protons},0\text{_neutrons}) \rightarrow \mathrm{X}(18-1\text{_protons},37-18\text{_neutrons})\) Thus, X is Chlorine (Cl) with atomic number 17, and the equation is: \(\mathrm{Ar}-37 + \mathrm{e^-} \rightarrow \mathrm{Cl}-37\)