Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The number of \beta particles emitted by a radioactive substance is twice the number of \alpha particles emitted by it. The resulting daughter is an:

isobar of parent

isomer of parent

isotone of parent

isotope of parent

In Alpha ( $\alpha$ ) decay, the nucleus emits a helium nucleus ( ${}_{2}^{4}\mathrm{He}$ ). The atomic number ( $Z$ ) decreases by 2 and the mass number ( $A$ ) decreases by 4 .
In Beta ( $\beta^-$ ) decay, the nucleus emits an electron ( ${}_{-1}^{0}e$ ). The atomic number ( $Z$ ) increases by 1 and the mass number ( $A$ ) remains unchanged .

Let the number of $\alpha$ particles emitted be $n$ . The change in $Z$ due to $\alpha$ emission is $n \times (-2) = -2n$ .
The number of $\beta$ particles emitted is given as twice the number of $\alpha$ particles, i.e., $2n$ . The change in $Z$ due to $\beta$ emission is $2n \times (+1) = +2n$ .
Net change in atomic number ( $Z$ ): $\Delta Z = -2n + 2n = 0$ .

Conclusion: The final daughter nucleus has the same atomic number ( $Z$ ) as the parent nucleus but a different mass number (since $A$ decreases by $4n$ ). Atoms with the same atomic number but different mass numbers are defined as isotopes .

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