Q156 / 160 · TS EAMCET 2023 (12 May Shift 2) · Physics

In the following nuclear reaction $\mathrm{x}$ is ${ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \rightarrow$ ${ }_0^{n^1+\mathrm{X}}$

  • A
    ${ }_{15} \mathrm{P}^{31}$
  • B
    ${ }_{14} \mathrm{Si}^{30}$
  • Correct answer
    C
    ${ }_{15} \mathrm{P}^{30}$
  • D
    ${ }_{15} \mathrm{Si}^{31}$

Solution

${ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \longrightarrow{ }_0 \mathrm{n}^1+\mathrm{x}$ Mass no. remain conserve in nuclear reaction. $A_x=27+4-1=30$ Atomic number remain conserve in nuclear reaction, $\mathrm{Z}_{\mathrm{x}}=13+2-0=15$ $\therefore{ }_{15} \mathrm{X}^{30}$ the atom is $\mathrm{P}$. Hence option (3) is correct.

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

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