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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