Question

Which of the following elements is isodiaphere of ${}_{92}^{235}{\text{U}}$?
A. ${}_{83}^{209}{\text{Bi}}$
B. ${}_{82}^{212}{\text{Pb}}$
C. ${}_{90}^{231}{\text{Th}}$
D. ${}_{91}^{231}{\text{Pa}}$

Answer 1

Hint: Here, we will first define what is isodiaphere and then we will obtain the isodiaphere of the element ${}_{92}^{235}{\text{U}}$ by $\alpha$-emission. So, use this concept to reach the answer of the given problem.

Complete answer:
Definition of isodiaphers: Isodiapheres refers to nuclides which have different atomic numbers and mass numbers but the same neutron excess, which is the difference between numbers of neutrons and protons in the nucleus.
Given element is ${}_{92}^{235}{\text{U}}$
We know that isodiapheres are formed by $\alpha$-emission, where $\alpha$ is ${}_2^4{\text{He}}$.
So, we have to remove ${}_2^4{\text{He}}$ from the given element i.e.,
${}_{92}^{235}{\text{U}} - {}_2^4{\text{He}} \to {}_{90}^{231}{\text{Th}}$
Hence the isodiaphere of the element ${}_{92}^{235}{\text{U}}$ is ${}_{90}^{231}{\text{Th}}$.

Thus, the correct option is C. ${}_{90}^{231}{\text{Th}}$

Note: The modern view of atomic structure ${}_{\text{Z}}^{\text{A}}{\text{X}}$ represents the atomic symbol as ${\text{X}}$, the atomic number (number of protons in the atom) as ${\text{Z}}$ and the mass number (Number of protons and neutrons in the atom) as ${\text{A}}$.