Nuclear Chemistry

Here are the solutions to the questions:

1. Nuclear Chemistry

• (i) Identify the particles emitted in steps A and B.

  • Step A: $^{238}_{92}\text{Ur} \rightarrow ^{234}_{90}\text{Th}$ The mass number decreases by 4 ($238-234=4$) and the atomic number decreases by 2 ($92-90=2$). This corresponds to the emission of an alpha particle. Particle A: $\boxed{\text{Alpha particle}}$
  • Step B: $^{234}_{90}\text{Th} \rightarrow ^{234}_{91}\text{Pa}$ The mass number remains unchanged ($234-234=0$) and the atomic number increases by 1 ($91-90=1$). This corresponds to the emission of a beta particle. Particle B: $\boxed{\text{Beta particle}}$

• (ii) If a beta particle is emitted in stage C, give the atomic number and mass number of the isotope X.

  • Stage C: $^{234}_{91}\text{Pa} \rightarrow \text{X} + ^{0}_{-1}\text{e}$ (beta particle)
  • For the mass number of X: $234 - 0 = 234$
  • For the atomic number of X: $91 - (-1) = 91 + 1 = 92$
  • Atomic number: $\boxed{92}$
  • Mass number: $\boxed{234}$

• (iii) If the activity of $^{234}_{90}\text{Th}$ is reduced to 25% in 48 days. What is the half-life of $^{234}_{90}\text{Th}$?

  Step 1: Determine the number of half-lives.
  If the activity is reduced to 25%, it means it has undergone two half-lives:
  $100\% \xrightarrow{1 \text{ half-life}} 50\% \xrightarrow{1 \text{ half-life}} 25\%$
  So, $n = 2$ half-lives.

  Step 2: Calculate the half-life.
  Total time = $n \times \text{half-life}$
  $48 \text{ days} = 2 \times \text{half-life}$

  $$\text{Half-life} = \frac{48 \text{ days}}{2} = \boxed{24 \text{ days}}$$