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.e−0.5In9
C
time at which the material has decayed to half of its initial mass (in hours) is (ℓn12)(−12ℓn0.9)
D
time at which the material has decayed to half of its initial mass (in hours) is 2ln12ln0.9
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Solution
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dNdt∝N dNdt=−λN dNN=−λdt ℓnN=−λt+c t=0,N=50 ℓnN=−λt+ℓn50 N=50e−λt ......(1) At t=2,N=45 45=50e−λ.2 910=e−2λ √910=e−λ ∴N=50(910)t/2 ......(2) At t=4 N=50(0.9)2 When N=25kg25=50(0.9)t/2 ⇒12=(0.9)t/2t=2ℓn12ℓn0.9
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