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(109)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 (−21ℓn0.9)(ℓn21)
D
time at which the material has decayed to half of its initial mass (in hours) is ln0.92ln21
Hard
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Updated on : 2022-09-05
Solution
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Correct option is D)
dtdN∝N dtdN=−λN NdN=−λdt ℓnN=−λt+c t=0,N=50 ℓnN=−λt+ℓn50 N=50e−λt ......(1) At t=2,N=45 45=50e−λ.2 109=e−2λ 109=e−λ ∴N=50(109)t/2 ......(2) At t=4 N=50(0.9)2 When N=25kg25=50(0.9)t/2 ⇒21=(0.9)t/2⇒t=ℓn0.92ℓn21
