Activity of a radioactive sample decreases to ( 3 1 ) r d of its original value in 3 days. Then in 9 days its activity reduces to

Activity after time t given that half-life is T : A = 2 t / T A 0 3 A 0 = 2 3/ T A 0 Activity in 9 days, A = 2 t / T A 0 = 2 9/ T A 0 = ( 2 3/ T ) 3 A 0 = 27 A 0

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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2022

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Activity of a radioactive sample decreases to $(\frac{1}{3})^{r d}$ of its original value in 3 days. Then in 9 days its activity reduces to

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Activity after time $t$ given that half-life is $T$ : $A = \frac{A _{0}}{2 ^{t / T}}$ $\frac{A _{0}}{3} = \frac{A _{0}}{2 ^{3/ T}}$ Activity in 9 days, $A = \frac{A _{0}}{2 ^{t / T}} = \frac{A _{0}}{2 ^{9/ T}} = \frac{A _{0}}{( 2 ^{3/ T} ) ^{3}} = \frac{A _{0}}{27}$

Stream:MHT_CETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: PYQ_2022

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