Question

# A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is 50kg of the material present and after two hours it is observed that the material has lost 10%of its original mass, then the

A
mass of the material after four hours is 50(910)2
B
mass of the material after four hours is 50.e0.5In9
C
time at which the material has decayed to half of its initial mass (in hours) is (n12)(12n0.9)
D
time at which the material has decayed to half of its initial mass (in hours) is 2ln12ln0.9
Solution
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#### dNdt∝NdNdt=−λNdNN=−λdtℓnN=−λt+ct=0,N=50ℓnN=−λt+ℓn50N=50e−λt ......(1)At t=2,N=4545=50e−λ.2910=e−2λ√910=e−λ∴N=50(910)t/2 ......(2)At t=4N=50(0.9)2When N=25kg25=50(0.9)t/2⇒12=(0.9)t/2t=2ℓn12ℓn0.9

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