Find the number of elements which are isodiaphere of ${ }_{89}{\mathrm{Ac}}^{227}$. An element is represented as (A, Z}.

$\begin{array}{l}(231,91);(223,87);(227,90);(223,88);(219,85);\\ (215,83);(215,84);(207,82);(211,83);\end{array}$

The number of elements in the form of (A, Z}. which are isodiapher of ${ }_{89}{\mathrm{Ac}}^{227}$ are $(231,91)(223,87),(219,85), (215,83)$.

Isodiaphers are a class of nuclides where the protons and neutrons are distinct in number. The pair having having the distinct number is found by the difference of mass number and twice the atomic number.

$\mathrm{A}-2\mathrm{Z}$  values for the given pair is,

$$\begin{gathered} =A - 2 Z \\ =231-2\times 91 \\ =49 \end{gathered}$$

Hence there are 4  pair of elements in the form of (A, Z}. which are isodiapher of ${ }_{89}{\mathrm{Ac}}^{227}$ .