Question

Supply the missing particle and state the type of decay for each of the following nuclear processes.

Short answer

a. Alpha decay, ${}_{90}{}^{234}\mathrm{Th}$.

b. beta decay,${}_{-1}{}^0\mathrm{e}$.

Step-by-step solution

Missing particle and describing the type of decay for part (a)

In the first radioactive element, a helium atom is added to the product, this is an alpha decay.

In the second particle, the atomic number and the mass number are lesser by 2 and 4 respectively, as compared to the first particle. Hence, the resulting particle is ${}_{90}{}^{234}\mathrm{Th}$. The complete equation is:

${}_{92}{}^{238}\mathrm{Th}\to {}_2{}^4\mathrm{He}+{}_{90}{}^{234}\mathrm{Th}$

Missing particle and describing the type of decay for part (b)

In the second process, we can observe that the atomic mass remains the same and there is an increase in the atomic number by 1. So, the process is a beta decay.

Also, the particle missing is an electron $\left({}_{-1}{}^0\mathrm{e}\right)$. Thus, the complete equation is:

${}_{90}{}^{234}\mathrm{Th}\to {}_{91}{}^{234}\mathrm{Pa}+{}_{-1}{}^0\mathrm{e}$