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Q.

In the case of Thorium (A = 232 and z = 90), we obtain an isotope of Lead (A = 208 and z = 82) after some radio active disintegrations. The number of β particles emitted are respectively

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answer is 2.

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Detailed Solution

Decrease of mass number = 232 - 208 = 24Number of α-particles emitted=244=6Due to emission of 6 particles, decrease in charge number is 12But actual decrease in charge number is 8.Clearly 12 - 8 = 4 β -particles are emitted
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In the case of Thorium (A = 232 and z = 90), we obtain an isotope of Lead (A = 208 and z = 82) after some radio active disintegrations. The number of β particles emitted are respectively