Question

Consider the nuclear change-

${}_{93}{}^{237}\mathrm{Np}\stackrel{-\mathrm{\alpha }}{\to }\mathrm{A}\stackrel{-\mathrm{\beta }}{\to }\mathrm{B}\stackrel{‐\mathrm{\alpha }}{\to }\mathrm{C}$

Which of the following statements is not correct?

• A. Mass number of B is 233
• B. Atomic number of A is 91.
• C. C is an isotope of A.
• D. B is an isobar of A.

Answer 1

The correct option is C

C is an isotope of A.

Step 1: Introduction

The nuclear reaction is the reaction where unstable nuclei can split into smaller stable nuclei and emits alpha, beta and gamma particles.

Alpha particle is represented as ${}_2{}^4\mathrm{He}$ and beta particle is represented as ${}_{-1}{}^0\mathrm{e}$.

Step 2: Determining A, B and C.

When alpha particle emits from the $\mathrm{Np}$ then A element is produce.

${}_{93}{}^{237}\mathrm{Np}\stackrel{-{}_2{}^{4}He}{\to }{}_{91}{}^{233}\mathrm{A}$

So, the atomic number of A is 91. Hence, option B is correct

When beta particle emits from the A then B element is produce.

${}_{93}{}^{237}\mathrm{Np}\stackrel{-{}_2{}^{4}He}{\to }{}_{91}{}^{233}\mathrm{A}\stackrel{-{}_{-1}{}^0\mathrm{e}}{\to }{}_{92}{}^{233}\mathrm{B}$

So, the mass number of B is 233. Hence, option A is correct.

B is the isobar of A because they have the same mass number but a different atomic number. Hence, option D is also correct.

${}_{93}{}^{237}\mathrm{Np}\stackrel{-{}_2{}^{4}He}{\to }{}_{91}{}^{233}\mathrm{A}\stackrel{-{}_{-1}{}^0\mathrm{e}}{\to }{}_{92}{}^{233}\mathrm{B}\stackrel{-{}_2{}^{4}He}{\to }{}_{90}{}^{229}\mathrm{C}$

C is not an isotope of A because they do not have the same atomic number.

Final answer:

Hence, option C is correct.