In a radioactive series
\( ^{235} _{92}L \to ^{235} _{92}M \to ^{231} _{90}N \to ^{231} _{91}\phi \)
The particles emitted are respectively
\( \beta , \beta , \alpha \)
\( \alpha , \gamma , \beta \)
\( \gamma , \alpha , \beta \)
\( \alpha , \alpha , \beta \)
\( \beta , \alpha , \beta \)
Explanation
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Discussions (7)
Option C is correct, because:
1. L represents a Gamma radiation, since there's no change in both the Mass number and Atomic number.
2. M represents an alpha disintegration, since the Mass number, 235 reduces by 4, to give 231, and the Atomic number, 92 reduces by 2, to give 90.
Hence, that's an alpha particle.
3. N represents a Beta decay, since there's no change in the Mass number, but its atomic number increases by 1.
That's a Beta decay.
Therefore, Option C, which represents Gamma, Alpha, Beta, is much appropriate.
