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PhysicsModern PhysicsRadioactivityEasy2 minPYQ_2020

PhysicsEasysingle choice

The activity of a radioactive element decreases to one-third of the original activity $I_{0}$ in a period of nine years. After a further lapse of nine years its activity will be

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Answer:

Solution:

$I = I_{0} e^{- λ t}$

$\Rightarrow \frac{I_{0}}{3} = I_{0} e^{- 9 λ}$

$e^{- 9 λ} = \frac{1}{3}$

After a further lapse of nine years, total time is eighteen years $I = I_{0} {e^{-}}^{λ \times 18}$

$\Rightarrow I = I_{0} {({e^{-}}^{λ \times 9})}^{2}$

$\Rightarrow I = I_{0} {(\frac{1}{3})}^{2}$

$\Rightarrow I = \frac{I_{0}}{9}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Radioactivity

⏱ 2mℹ️ Source: PYQ_2020

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