Chapter 19, Problem 52A

Interpretation Introduction

Interpretation:

The nature of particles produced during decay of U-238 to Pb-206 must be explained.

Concept Introduction:

U-238 is a radioactive nuclide with very high n/p ratio and so it undergoes decay to finally produce a stable nuclide which is Pb-206.

Answer to Problem 52A

In the decay series of U-238, total 8 α and 6 β particles are produced along with Pb-206.

Explanation of Solution

The radioactive decay reactions are shown below.

  $\begin{array}{l}{\text{U}}_{\text{92}}^{\text{238}}\to {\text{Th}}_{\text{90}}^{\text{234}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Th}}_{\text{90}}^{\text{234}}\to {\text{Pa}}_{\text{91}}^{\text{234}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\\ {\text{Pa}}_{\text{91}}^{\text{234}}\to {\text{U}}_{\text{92}}^{\text{234}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\\ {\text{U}}_{\text{92}}^{\text{234}}\to {\text{Th}}_{\text{90}}^{\text{230}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Th}}_{\text{90}}^{\text{230}}\to {\text{Ra}}_{88}^{\text{226}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Ra}}_{88}^{\text{226}}\to {\text{Rn}}_{86}^{\text{222}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Rn}}_{86}^{\text{222}}\to {\text{Po}}_{84}^{\text{218}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Po}}_{84}^{\text{218}}\to {\text{Pb}}_{82}^{\text{214}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Pb}}_{82}^{\text{214}}\to {\text{Bi}}_{83}^{\text{214}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\\ {\text{Bi}}_{83}^{\text{214}}\to {\text{Po}}_{84}^{\text{214}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\end{array}$

And,

  $\begin{array}{l}{\text{Po}}_{84}^{\text{214}}\to {\text{Pb}}_{82}^{\text{210}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\\ {\text{Pb}}_{82}^{\text{210}}\to {\text{Bi}}_{83}^{\text{210}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\\ {\text{Bi}}_{83}^{\text{210}}\to {\text{Po}}_{84}^{\text{210}}{\text{+e}}_{\text{-1}}^0\text{(β)}\text{.}\\ {\text{Po}}_{84}^{\text{210}}\to {\text{Pb}}_{82}^{\text{206}}{\text{+He}}_{\text{2}}^{\text{4}}\text{(α)}\text{.}\end{array}$

From the reaction it is clear that, per each β emission there is 2 units loss of atomic number and 4 units loss of mass number. Similarly, per each ß emission, there is one unit increase in atomic number and no change in mass number.

As per law of conservation of mass, total mass number of reactants is equal to total mass number of products. Similarly, total atomic number of reactants is equal to total atomic number of products.

Conclusion

All the reactions are shown for the decay series of U-238 to Pb-206.

All the particles which are produced can be seen from these nuclear disintegration reactions.