In the disintegration of a radioactive element, `alpha`- and `beta`-particles are evolved from the nucleus.
`._(0)n^(1) rarr ._(1)H^(1) + ._(-1)e^(0) +` Antineutrino + Energy
`4 ._(1)H^(1) rarr ._(2)He^(4) + 2 ._(+1)e^(0) +` Energy
Then, emission of these particles changes the nuclear configuration and results into a daughter nuclide. Emission of an `alpha`-particles results into a daughter element having atomic number lowered by 2 and mass number by 4, on the other hand, emission of a `beta`-particle yields an element having atomic number raised by 1.
Which of the following combinations give finally an isotope of the parent element?

Options

A. `alpha, alpha, beta`
B. `alpha, gamma, alpha`
C. `alpha, beta, beta`
D. `beta, gamma, alpha`

Answer

C

Explanation

Let us consider a reaction
`._(Z)b^(A) overset(-alpha)rarr ._(Z - 2)B^(A) overset(-beta)rarr ._(Z - 2 + 1)B^(A) overset(-beta)rarr`
`._(Z - 1 + 1)B^(A - 4)` or `._(Z)B^(A - 4)`
Hence, the parent element and daughter elements have same atomic number, i.e., isotopes