Question

Are the following decays possible? If not, why not?

$$\begin{gathered} \text{a.}{}^{232}\mathrm{Th}(Z=90)\to {}^{236}\mathrm{U}(Z=92)+\alpha \\ \text{b.}{}^{238}\mathrm{Pu}(Z=94)\to {}^{236}\mathrm{U}(Z=92)+\alpha \\ \text{c.}{}^{11}\mathrm{B}(Z=5)\to {}^{11}\mathrm{B}(Z=5)+\gamma \\ \text{d.}{}^{33}\mathrm{P}(Z=15)\to {}^{32}\mathrm{S}(Z=16)+{\mathrm{e}}^{-} \end{gathered}$$

Short answer

Therefore, the decays are explained.

Step-by-step solution

Given information

The following decays are given:

$$\begin{gathered} \text{a.}{}^{232}\mathrm{Th}(Z=90)\to {}^{236}\mathrm{U}(Z=92)+\alpha \\ \text{b.}{}^{238}\mathrm{Pu}(Z=94)\to {}^{236}\mathrm{U}(Z=92)+\alpha \\ \text{c.}{}^{11}\mathrm{B}(Z=5)\to {}^{11}\mathrm{B}(Z=5)+\gamma \\ \text{d.}{}^{33}\mathrm{P}(Z=15)\to {}^{32}\mathrm{S}(Z=16)+{\mathrm{e}}^{-} \end{gathered}$$

Explanation (a)

(a)No, the alpha particle is inserted into the incorrect nucleus.

Explanation (b)

(b)No, the original nucleus would have to lose four nucleons rather than two.

Explanation (c)

(c)Yes.

Explanation (d)

(d) No, A does not change during beta decay.