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Question

The number of α and β particles emitted in the conversion of 90Th232 to 82Pb208 are

A
6,4
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B
4,6
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C
8,6
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D
6,8
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Solution

The correct option is A 6,4
(i) During α emission:

90T23282Pb208+(2He4)x+y(1e0)

Equation of mass number,

232=208+4x

232208=4x

x=244=6

(ii) During β emission, mass number will not change, but atomic number will increase by 1,

Equating atomic number,

906(2)+y(1)=82

y=8278=4

Hence, (A) is the correct answer.

Why this Question?This type of problems are frequently asked in JEE Mains.

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