In a radioactive material the activity at time t₁, is A₁ and at a later time t₂, it is A₂. If the decay constant of the material is λ, then

Correct Answer is : A1=A2e−λ(t1−t2)A₁ = A₂ e^{-λ (t₁ - t₂)}A1=A2e−λ(t1−t2)

A₁ = A₂ e^{-λ (t₁ - t₂)}

A₁ = A₂ e^{λ (t₁ - t₂)}

Correct Answer is : A1=A2e−λ(t1−t2)A₁ = A₂ e^{-λ (t₁ - t₂)}A1=A2e−λ(t1−t2)

Test Index

BITSAT 2022 Paper

Section: Physics

Question : 23 of 130

Marks: +1, -0

t_1

A_1

t_2

A_2

\lambda

Solution: 👈: Video Solution

From radioactive decay law,

-\frac{dN}{dt} \propto N \text{ or } -\frac{dN}{dt} = \lambda N

Thus,

A = -\frac{dN}{dt} \text{ or } A = \lambda N

\Rightarrow A = \lambda N_0 e^{-\lambda t}

.......(i)

Where,

A_0 = \lambda N_0

is the activity of the radioactive material at time,

t = 0

At time,

t_1 \dots A_1 = A_0 e^{-\lambda t_1} \qquad (ii)

At time,

t_2 \dots A_2 = A_0 e^{-\lambda t_2} \qquad (iii)

On didviding Eq. (ii) by Eq. (iii), we have

\frac{A_1}{A_2} = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{-\lambda (t_1 - t_2)}

\Rightarrow A_1 = A_2 e^{-\lambda (t_1 - t_2)}

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