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Q.

The half life of a sample of a radioactive substance is 2 hour. If 8×1010atoms are present at t=0,  then the number of atoms decayed in the duration t=2hours to t=6hours will be

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a

1×1010

b

2×109

c

4×1011

d

3×1010

answer is D.

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Detailed Solution

Decay Law, N=N012tT12 Case 1: No. of atoms at t=2 hr’s ⇒N1=8×10101222=4×1010 Case 2: No. of atoms at t=6 hr’s ⇒N2=8×10101262=8×101018=1×1010 Hence,  No of atoms decayed in given duration =(4-1) ×1010=3×1010
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