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Question

The number of α and β emitted during the radioactive decay chain starting from 22688Ra and endjg at 20682Pb is

A
3α and 6β
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B
4α and 5β
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C
5α and 4β
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D
6αand 6β
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Solution

The correct option is C 5α and 4β
Let no of α particle emitted =x no of β panticle emitted =y
Now, 22688Ra20682 Pb+x42α+y01β
balancing atomic mass 226=206+4x+04x=20x=5
balancing atomic no. 88=82+2x+(1)y6=10yy=4
So, 5α and 4β

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