Question

The naturally occurring radioactive isotope \(^{{\rm{232}}}{\rm{Th}}\) does not make good fission fuel, because it has an even number of neutrons; however, it can be bred into a suitable fuel (much as \(^{{\rm{238}}}{\rm{U}}\) is bred into\(^{239}P\)).

(a) What are Z and N for\(^{{\rm{232}}}{\rm{Th}}\)?

(b) Write the reaction equation for neutron captured by \(^{{\rm{232}}}{\rm{Th}}\) and identify the nuclide \(^AX\)produced in\(n{ + ^{232}}Th{ \to ^A}X + \gamma \).

(c) The product nucleus \({\beta ^ - }\)decays, as does its daughter. Write the decay equations for each, and identify the final nucleus.

(d) Confirm that the final nucleus has an odd number of neutrons, making it a better fission fuel.

(e) Look up the half-life of the final nucleus to see if it lives long enough to be a useful fuel.

Short answer

(a) The atomic number of 232 Th is \({\rm{Z = 90}}\)and its neutron number is \(N = 142\).

(b) The product nucleus \(^{\rm{A}}{\rm{X}}\)is the isotope of Thorium, which is \(_{90}^{233}Th\).

(c) The equations for decay are,

\(\begin{array}{l}_{90}^{233}Th\,\, \to \,_{ - 1}^0e + \,_{91}^{233}\;Pa\\_{91}^{233}\;Pa\,\, \to \,_{ - 1}^0e + \,_{92}^{233}U\end{array}\)

The final nucleus is\(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

(d) The number of neutrons in the nucleus \(_{92}^{233}U\) is \({\rm{141}}\) , which is an odd number.

(e) The half-life of \(_{92}^{233}U\) is \(160,000y\)and hence the nuclides live for a long time and can be used as a nuclear fuel.

Step-by-step solution

Atomic number

The charge number of an atomic nucleus is known as the atomic number or nuclear charge number of a chemical element. This is equal to the proton number, or the number of protons found in the nucleus of every atom of that element, for ordinary nuclei.

The atomic number of \(^{{\rm{232}}}{\rm{Th}}\) and its neutron number

Considering the given information:

The nucleus \(^{{\rm{232}}}{\rm{Th}}\).

Apply the formula:

The atomic number Z and the neutron number N are related to the mass number A as follows:

\(F = \frac{{\Delta P}}{{\Delta t}}\)

Where \(\Delta P = \)Change in Momentum

\(\Delta t = \)Change in time

The photon's momentum is calculated as,

\(A = Z + N\)

The symbol for a nucleus is \(_Z^AX\). The element's chemical symbol is X. The

nucleus \(^{{\rm{232}}}{\rm{Th}}\)is represented by the symbol \(_{90}^{233}Th\).

Accordingtotheaboveexpression,A = 232, Z = 90.

To compute N, the formula used is,

\(\begin{array}{c}N = A - Z\\ = 232 - 90\\ = 142\end{array}\)

Therefore, the required atomic number of\(^{{\rm{232}}}{\rm{Th}}\)is\(Z = 90\)and its neutron number is\(N = 142\).

The reaction equation for neutron

Considering the given information:

The equation for the reaction:

\(n{ + ^{232}}{\rm{Th}}{ \to ^A}{\rm{X}} + \gamma \)

Rewrite the given equation using the atomic numbers of the reactants.

\(_0^1n + _{90}^{232}{\rm{Th}} \to _Z^A{\rm{X}} + \gamma \)

On both sides of the equation, the total atomic number remains constant.

Therefore,

\(\begin{array}{c}Z = 90 + 0\\ = 90\end{array}\)

On both sides of the equation, the total mass number is a constant.

Therefore,

\(\begin{array}{c}A = 1 + 232\\ = 233\end{array}\)

Because the atomic number of the product \(^{\rm{A}}{\rm{X}}\)is the same as the nucleus \(_{90}^{233}Th\), the nucleus\(^{\rm{A}}{\rm{X}}\)isThorium.

The product \(^{\rm{A}}{\rm{X}}\) is written as \(_{90}^{233}Th\).

Therefore, the required product nucleus\(^{\rm{A}}{\rm{X}}\)is the isotope of Thorium, which is\(_{90}^{233}Th\).

Find the final nucleus

Considering the given information:

The nucleus \(_{90}^{233}Th\) and its daughter nucleus undergo \(\beta \) decay.

Write the equation for the \({\rm{\beta }}\) decay of \(_{90}^{233}Th\). When a nucleus decays, its atomic number increases by one while its mass number remains constant.

\(_{90}^{233}{\rm{Th}} \to _{ - 1}^0e + _{91}^{233}{\rm{X}}\)

Protactinium is the element with the atomic number \({\rm{91}}\) in the periodic table (Pa).

As a result, the reaction's daughter nucleus is \(_{91}^{233}\;Pa\).

Theequationcanbewrittenasfollows:

\(_{90}^{233}Th \to _{ - 1}^0e + _{91}^{233}\;Pa\)

Further, the nucleus \(_{{\rm{91}}}^{{\rm{233}}}{\rm{\;Pa}}\) undergoes \({\rm{\beta }}\) decay, whose equation is as follows:

\(_{91}^{233}\;Pa \to _{ - 1}^0e + _{92}^{233}X\)

Uranium is the element with the atomic number \({\rm{92}}\) in the periodic table (U).

As a result, the final product is \(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

Thefollowingisanexampleofanequation:

\(_{91}^{233}{\rm{ }}\;Pa{\rm{ }} \to _{ - 1}^0e + _{92}^{233}U\)

Therefore, the required final product of the decay is the isotope of Uranium\(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

Find the final nucleus has an odd number of neutrons

Considering the given information:

The nucleus \(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

Apply the formula;

The atomic number Z and the neutron number N are related to the mass number A as follows:

\(A = Z + N\)

The atomic number is \(Z = 92\), and the mass number is \(A = 233\), as can be seen from the expression\(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

Calculate the nucleus' neutron number, which is \(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\).

\(\begin{array}{c}N = A - Z\\ = 233 - 92\\ = 141\end{array}\)

Therefore, the required number of neutrons in the nucleus\(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\)is\({\rm{141}}\), which is an odd number.

Explain the nuclides live for a long time and can be used as nuclear fuel

Thehalflifeofaradioactivenuclideistheamountoftimeittakesforhalfofthenuclidestodecay.

The half-life of the nucleus \(_{{\rm{92}}}^{{\rm{233}}}{\rm{U}}\)is \({\rm{160,000}}\)years.Thismeansthatthenucleusdecaysslowly,makingitsuitableforuseasanuclearfuel.