Chapter 9, Problem 70P

Interpretation Introduction

Interpretation:

Successive three radioactive decay products starting from ²²⁵₈₉ Ac radionuclide by alpha-emission should be determined.

Concept Introduction:

Radioactive decay is a process of converting a less stable radionuclide to a more stable nuclei by loss of energy. This can be achieved by the loss of gain of elementary particles such as alpha, beta or positron particles. Alpha particles are nothing but a Helium nuclei (⁴₂ He). When a nucleus undergoes an alpha emission process, which is a type of elementary radioactive particle, the resulting product nuclei will have an atomic mass 4 unit lower and atomic number of 2 unit lower.

Answer to Problem 70P

Successive three radioactive decay products starting from ²²⁵₈₉ Ac radionuclide by alpha-emission are:

$\begin{array}{l}{{}^{225}}_{89}\text{a}\text{C}\to {}^{221}{}_{87}\text{F}r{+}^4{}_2\text{H}\text{e}\left[\text{F}\text{i}rstd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{221}}_{87}\text{F}r \to {}^{217}{}_{85}\text{a}t{+}^4{}_2\text{H}\text{e}\left[S\text{e}\text{C}\text{O}ndd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{217}}_{85}\text{a}t \to {}^{213}{}_{83}B\text{i}{+}^4{}_2\text{H}\text{e}\left[T\text{H}\text{i}rdd\text{e}\text{C}\text{a}y\right]\hfill \end{array}$.

Explanation of Solution

A radioactive element may undergo radioactive decay by loss of an elementary particle such as an alpha particle. Alpha particle (⁴₂ He) has a mass of 4 units and atomic number of 2 units, so the resulting product nuclei will have an atomic mass 4 unit lower and atomic number of 2 unit lower.

Actinium is represented in short form as Ac.

Francium is represented in short form as Fr.

Astatine is represented in short form as At.

Bismuth is represented in short form as Bi.

Representation of an element in a radioactive decay equation: ^x_y A.

Where,

X represents atomic mass of an element

Y represents atomic number of an element

A is the element

Ac (Actinium) has an atomic mass of 225 and atomic number of 89. When it undergoes a radioactive decay by alpha emission (⁴₂ He), the resulting product Fr (Francium) has an atomic mass of 221 (lower by 4 units) and atomic number of 87 (lower by 2 units). Further, when Fr (Francium) undergoes a radioactive decay by alpha emission (⁴₂ He), the resulting product At (Astatine) has an atomic mass of 217 (lower by 4 units) and atomic number of 85 (lower by 2 units). In the third successive decay process, when At (Astatine) undergoes a radioactive decay by alpha emission (⁴₂ He), the resulting product Bi (Bismuth) has an atomic mass of 213 (lower by 4 units) and atomic number of 83 (lower by 2 units).

²²⁵₈₉ Ac undergoes a series of alpha decay to produce daughter nuclei. The successive decay equations are as follows:

$\begin{array}{l}{{}^{225}}_{89}\text{a}\text{C}\to {}^{221}{}_{87}\text{F}r{+}^4{}_2\text{H}\text{e}\left[\text{F}\text{i}rstd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{221}}_{87}\text{F}r \to {}^{217}{}_{85}\text{a}t{+}^4{}_2\text{H}\text{e}\left[S\text{e}\text{C}\text{O}ndd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{217}}_{85}\text{a}t \to {}^{213}{}_{83}B\text{i}{+}^4{}_2\text{H}\text{e}\left[T\text{H}\text{i}rdd\text{e}\text{C}\text{a}y\right]\hfill \end{array}$

So it forms Francium-221 after first decay, Astatine-217 after second decay and Bismuth-213 after third decay.

Conclusion

Successive three radioactive decay products starting from ²²⁵₈₉ Ac radionuclide by alpha-emission are:

$\begin{array}{l}{{}^{225}}_{89}\text{a}\text{C}\to {}^{221}{}_{87}\text{F}r{+}^4{}_2\text{H}\text{e}\left[\text{F}\text{i}rstd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{221}}_{87}\text{F}r \to {}^{217}{}_{85}\text{a}t{+}^4{}_2\text{H}\text{e}\left[S\text{e}\text{C}\text{O}ndd\text{e}\text{C}\text{a}y\right]\hfill \\ {{}^{217}}_{85}\text{a}t \to {}^{213}{}_{83}B\text{i}{+}^4{}_2\text{H}\text{e}\left[T\text{H}\text{i}rdd\text{e}\text{C}\text{a}y\right]\hfill \end{array}$.