Question

Which of the following pairs are isodiaphers?
A.$_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ and $_{{\text{24}}}{\text{C}}{{\text{r}}^{{\text{55}}}}$
B.$_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ and $_{{\text{24}}}{\text{C}}{{\text{r}}^{{\text{52}}}}$
C.$_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ and $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$
D.$_{{\text{92}}}{{\text{U}}^{{\text{235}}}}$ and $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{232}}}}$

Answer 1

The nuclides have different atomic numbers and different mass numbers but the same number of excess neutrons are known as isodiaphers. The number of excess neutrons is the difference between the number of neutrons and the number of protons in the nucleus.

Complete step by step answer:
Isodiaphers are formed by the $\alpha$-emission, where $\alpha$ is $_{\text{2}}{\text{H}}{{\text{e}}^{\text{4}}}$.
The given element is $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$. The reaction when $_{\text{2}}{\text{H}}{{\text{e}}^{\text{4}}}$ is removed from $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ is,
$_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}{ - _{\text{2}}}{\text{H}}{{\text{e}}^{\text{4}}}{ \to _{{\text{27}}}}{\text{C}}{{\text{o}}^{{\text{61}}}}$
Thus, the isodiaphere of $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ is $_{{\text{27}}}{\text{C}}{{\text{o}}^{{\text{61}}}}$ and not $_{{\text{24}}}{\text{C}}{{\text{r}}^{{\text{55}}}}$.
Thus, option (A) is not correct.
The given element is $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$. The reaction when $_{\text{2}}{\text{H}}{{\text{e}}^{\text{4}}}$ is removed from $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ is,
$_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}{ - _{\text{2}}}{\text{H}}{{\text{e}}^{\text{4}}}{ \to _{{\text{27}}}}{\text{C}}{{\text{o}}^{{\text{61}}}}$
Thus, the isodiaphere of $_{{\text{29}}}{\text{C}}{{\text{u}}^{{\text{65}}}}$ is $_{{\text{27}}}{\text{C}}{{\text{o}}^{{\text{61}}}}$ and not $_{{\text{24}}}{\text{C}}{{\text{r}}^{{\text{52}}}}$.
Thus, option (B) is not correct.
The given element is $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$. The reaction when $_{\text{2}}{\text{H}}{{\text{e}}^{\text{4}}}$ is removed from $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ is,
$_{{\text{92}}}{{\text{U}}^{{\text{238}}}}{ - _{\text{2}}}{\text{H}}{{\text{e}}^{\text{4}}}{ \to _{{\text{90}}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$
Thus, the isodiaphere of $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ is $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$.
Thus, option (C) is correct.
The given element is $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$. The reaction when $_{\text{2}}{\text{H}}{{\text{e}}^{\text{4}}}$ is removed from $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ is,
$_{{\text{92}}}{{\text{U}}^{{\text{238}}}}{ - _{\text{2}}}{\text{H}}{{\text{e}}^{\text{4}}}{ \to _{{\text{90}}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$
Thus, the isodiaphere of $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ is $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$ and not $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{232}}}}$.
Thus, option (D) is incorrect.
Thus, the pair of isodiaphers is $_{{\text{92}}}{{\text{U}}^{{\text{238}}}}$ and $_{{\text{90}}}{\text{T}}{{\text{h}}^{{\text{231}}}}$.

Therefore, option C is the correct choice.

Note:
The general representation of elements is $_{\text{Z}}{{\text{X}}^{\text{A}}}$. Where ${\text{Z}}$ is the atomic number of the element, ${\text{A}}$ is the mass number of the element and ${\text{X}}$ is the atomic symbol of the element.
The number of protons in the nucleus of the atom or the number of electrons surrounding the nucleus of the atom of any element is known as the atomic number of the element.
The sum of the number of protons and the number of neutrons in the nucleus of an atom of an element is known as the mass number of the element.