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Question

When 90Th238 changes into 83Bi222, then the number of emitted α and β particles are ?

A
8α,7β
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B
4α,7β
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C
4α,4β
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D
4α,1β
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Solution

The correct option is D 4α,1β


balancing mass no. 238=222+4x+04x=16 which means 4α particle released
balaneing atomie no. 90=83+2x+y(1)7=8yy=1 So, 1 β particle released 4α,1β particle released

1995383_1140080_ans_6ae33c25f1164121a1d600ebcb71f90d.jpg

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