A certain element x undergoes two successive alpha emission reactions followed by a beta emission reaction. The daughter nuclide formed is ac-230 (z=89). What is the identity of element x?

Respuesta :

An element is a pure material made up of atoms with the same amount of protons in their nuclei. The atomic number and the mass number of the element X are 92 and 238 respectively.

What is an element?

In chemistry, an element is a pure material made up of atoms with the same amount of protons in their nuclei. Chemical elements, unlike chemical compounds, cannot be broken down into simpler substances by any chemical process.

Let us consider, that z is the atomic no and A is the mass no of the parent element 'X'.

As per the question, X emits two successive alpha particles (²₁He) and one beta, i.e, particle and reached to daughter element (²³⁰₈₉Ac).

Now according to the question, we can write,

[tex]^A_ZX \overset{-\alpha}{\rightarrow}\ \ ^{A-4}_{Z-2}X\ +\ ^4_2He\ \ \overset{-\alpha}{\rightarrow}\ \ ^{A-8}_{Z-4}X\ +\ ^4_2He\ \ \overset{-\-\beta}{\rightarrow}\ \ ^{A-8}_{Z-3}X\ +\ ^0_{-1}He[/tex]

We see, [tex]^A_ZX[/tex] that after two α and one β of emission, we get the daughter element [tex]^{A-8}_{Z-3}X[/tex]. Again as given in the problem, the daughter element is [tex]^{230}_{89}AC[/tex].

∴We can write-

[tex]^{A-8}_{Z-3}X=^{290}_{89}AC[/tex]

comparing the two, we get,

Z-3=89

Z=92

A-8=230

A=238

Hence, the atomic number and the mass number of the element X is 92 and 238 respectively.

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