Question

In the given reaction, x₁, x₂, x₃, x₄ are either particles or radiation.

${}_{92}{}^{238}\mathrm{U}\stackrel{{\mathrm{x}}_1}{\to }{}_{90}{}^{234}\mathrm{Th}\stackrel{{\mathrm{x}}_2}{\to }{}_{91}{}^{234}\mathrm{Pa}\stackrel{{\mathrm{x}}_3}{\to }{}_{}{}^{234}\mathrm{Z}\stackrel{{\mathrm{x}}_4}{\to }{}_{90}{}^{230}\mathrm{Th}$

Which of the given options is correct?

• A. ${\mathrm{x}}_1$ will deflect towards negatively charged plate
• B. ${\mathrm{x}}_2$ is $\mathrm{\beta }$- particles.
• C. ${\mathrm{x}}_3$ is $\mathrm{\gamma }$- radiation.
• D. ${}_{}{}^{234}\mathrm{Z}$ is isotope of ${}_{92}{}^{238}\mathrm{U}$

Answer 1

Answer: (A), (B), (D)

${}_{}{}^{234}\mathrm{Z}$ is isotope of ${}_{92}{}^{238}\mathrm{U}$

The explanation for the correct answer

(A)${\mathrm{x}}_1$ will deflect towards negatively charged plate.

1. The first radioactive decay equation can also be written as ${}_{92}{}^{238}U\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$
2. In this decay, $\mathrm{\alpha }$ particles are emitted and they are positively charged particles.
3. Thus, will deflected towards negatively charged plate.

(B) ${\mathrm{x}}_2$ is $\mathrm{\beta }$- particles.

1. The second radioactive decay equation can also be written as ${}_{90}{}^{234}\mathrm{Th}\to {}_{91}{}^{234}\mathrm{Pa}+{}_{-1}{}^0\mathrm{e}$
2. In this decay we can clearly see that $\mathrm{\beta }$ particles are emitted.
3. Thus, it is the correct answer.

(D) ${}_{}{}^{234}\mathrm{Z}$ is isotope of ${}_{92}{}^{238}\mathrm{U}$

1. The third radioactive decay equation can also be written as ${}_{91}{}^{234}\mathrm{Pa}\to {}_{}{}^{234}\mathrm{Z}+{}_{-1}{}^0\mathrm{e}$
2. ${}_{}{}^{234}\mathrm{Z}$ is isotope of ${}_{92}{}^{238}\mathrm{U}$ as the atomic number of ${}_{}{}^{234}\mathrm{Z}$ is 92 same as ${}_{92}{}^{238}\mathrm{U}$ and those elements having same atomic number but different mass are called isotopes.
3. Thus, this is the correct answer.

The explanation for the incorrect answer

(C) ${\mathrm{x}}_3$ is $\mathrm{\gamma }$- radiation.

1. In The third radioactive decay equation, ${}_{91}{}^{234}\mathrm{Pa}\to {}_{}{}^{234}\mathrm{Z}+{}_{-1}{}^0\mathrm{e}$
2. Here $\mathrm{\beta }$ particles is emitted not $\mathrm{\gamma }$ rays.
3. Thus, This is an incorrect option.

Therefore, A, B, D are correct.