The activity of a radioactive sample is A 1 at time t 1 and A 2 at time t 2- If t is average life of sample then the number of nuclei decayed in time t 2- t 1 isA- A 1 t 1- A 2 t 2B- A1-A2-2 TC- A 1- A 2 t 2- t 1D- A 1- A 2 T

The correct option is D (A_1 A_2)τLet A_0 bet the initial Activity, then A_1 = A_0 e^ λ t_1 and A_2 = nbsp; A_0 e^ λ t_2 Let N_0 bet the initial number of nu ...

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Byju's Answer

Standard XII

Physics

Half Life and Average Life

The activity ...

Question

The activity of a radioactive sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If $τ$ is average life of sample then the number of nuclei decayed in time $(t_{2} - t_{1})$ is

A_{1} t_{1} - A_{2} t_{2}

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\frac{(A_{1} - A_{2})}{2} τ

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(A_{1} - A_{2}) (t_{2} - t_{1})

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(A_{1} - A_{2}) τ

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Solution

The correct option is D $(A_{1} - A_{2}) τ$
Let $A_{0}$ bet the initial Activity, then
$A_{1} = A_{0} e^{- λ t_{1}}$

and $A_{2} = A_{0} e^{- λ t_{2}}$

Let $N_{0}$ bet the initial number of nuclei, then
$N_{1} = N_{0} e^{- λ t_{1}}$

and $N_{2} = N_{0} e^{- λ t_{2}}$

$∴$ Number of nuclei decayed = $N_{1} - N_{2} = N_{0} (e^{- λ t_{1}} - e^{- λ t_{2}})$

$= \frac{A_{0}}{λ} (e^{- λ t_{1}} - e^{- λ t_{2}}) = \frac{A_{1} - A_{2}}{λ} = (A_{1} - A_{2}) τ .$

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The activity of a radioactive sample is A1 at time t1 and A2 at time t2. If τ is average life of sample then the number of nuclei decayed in time (t2–t1) is

The correct option is D (A1−A2)τLet A0 bet the initial Activity, then A1=A0e−λt1 and A2=A0e−λt2 Let N0 bet the initial number of nuclei, then N1=N0e−λt1 and N2=N0e−λt2 ∴ Number of nuclei decayed = N1−N2=N0(e−λt1−e−λt2) =A0λ(e−λt1−e−λt2)=A1−A2λ=(A1−A2)τ.

The activity of a radioactive sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If $τ$ is average life of sample then the number of nuclei decayed in time $(t_{2} - t_{1})$ is