Q169 / 180 · KCET 2012 · Physics

In a radioactive decay, an element $\mathrm{z}^{\mathrm{X}}$ emits four $\alpha$-particles, three $\beta$-particles and eight gamma photons. The atomic number and mass number of the resulting final nucleus are

  • A
    $\mathrm{Z}-8, \quad \mathrm{~A}-13$
  • B
    $Z-11, A-16$
  • C
    $\mathrm{Z}-5, \quad \mathrm{~A}-13$
  • Correct answer
    D
    $\mathrm{Z}-5, \quad \mathrm{~A}-16$

Solution

After $\mathrm{n}_{\alpha} \quad \alpha$-decays and $\mathrm{n}_{\beta} \beta$-decays, the respective atomic number and mass number of resulting nucleus are given by and $\quad \mathrm{Z}^{\prime}=\mathrm{Z}-2 \mathrm{n}_{\alpha}+\mathrm{n}_{\beta}=\mathrm{Z}-8+3=\mathrm{Z}-5$ We have, $\quad \mathrm{A}-4 \mathrm{n}_{\alpha}=\mathrm{A}-16$

Hence, the correct answer is option: 'D'.

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