To convert 90Th232 into 90Th228- the least number of - and - particles that have to be emitted are and respectively-

The correct option is C 1, 2 _ 90 Th ^ 232 + _ 2 α #160; ^ 4 +2 _ 1 β #160; ^ o → _ 90 Th ^ 228 So, 1 α decay and 2 β decay ...

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Byju's Answer

Standard XII

Physics

Alpha Decay

To convert ...

Question

To convert $_{90} T h^{232}$ into $_{90} T h^{228}$ , the least number of $α$ and $β$ particles that have to be emitted are _ and _ respectively.

1, 1

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2, 1

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1, 2

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2, 2

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Solution

The correct option is C 1, 2
$_{90} {T h}^{232} +_{2} α^{4} + 2_{- 1} β^{o} \to_{90} {T h}^{228}$
So, 1 $α$ decay and 2 $β$ decay

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Similar questions

Q. When

_{90} {Th}^{228}

gets converted into

_{83} {Bi}^{212}

, the number of

α

and

β

particles emitted will respectively be-

Q. When

_{90} T h^{228}

transforms to

_{83} B i^{212}

, then the number of the emitted

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_{90} T h^{232}

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particles emitted during the transformation of

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are respectively :

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_{90} T h^{228} ⟶_{83} B i^{212}

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To convert 90Th232 into 90Th228, the least number of α and β particles that have to be emitted are _____ and _____ respectively.

The correct option is C 1, 290Th232+2α4+2−1βo→90Th228So, 1 αdecay and 2 βdecay

To convert $_{90} T h^{232}$ into $_{90} T h^{228}$ , the least number of $α$ and $β$ particles that have to be emitted are _ and _ respectively.

The correct option is C 1, 2
$_{90} {T h}^{232} +_{2} α^{4} + 2_{- 1} β^{o} \to_{90} {T h}^{228}$
So, 1 $α$ decay and 2 $β$ decay