The radio active nucleus A 90232 decays into a stable nucleus B 82208.Then number of α and β particles emitted are respectively.

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α=3,β=3

α=6,β=4

α=6,β=0

α=4,β=6

answer is B.

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Detailed Solution

Suppose number of emitted α and βparticles are x and y respectively.∴232-4yx=208 ⇒x =6. Also, 90-2x+y=82 ⇒90-2×6+y=82 ⇒y=4.

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