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Longest Chain Rule

The radioacti...

Question

The radioactivity of a sample is $R_{1}$ at a time $T_{1}$ and $R_{2}$ at a time $T_{2}$ . If the half-life of the specimen is $T$ , the number of atoms that have disintegrated in the time $(T_{2} - T_{1})$ is proportional to:

R_{1} T_{1} - R_{2} T_{2}

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R_{1} - R_{2}

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\frac{(R_{1} - R_{2})}{T}

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(R_{1} - R_{2}) T

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Solution

The correct option is A $(R_{1} - R_{2}) T$
Radioactivity $= \frac{d N}{d t} = λ N$
(Since $N (t) = N_{o} e^{- λ t})$
Given :
$R_{1} = λ N (T_{1})$
$R_{2} = λ N (T_{2})$
No. of disintegrated atoms $= N (T_{2}) - N (T_{1})$ $= \frac{R_{2} - R_{1}}{λ} = \frac{(R_{2} - R_{1}) T}{l n 2}$

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The radioactivity of a sample is R1 at a time T1 and R2 at a time T2. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time (T2−T1) is proportional to:

The correct option is A (R1−R2)TRadioactivity =dNdt=λN(Since N(t)=Noe−λt)Given :R1=λN(T1)R2=λN(T2)No. of disintegrated atoms =N(T2)−N(T1)=R2−R1λ=(R2−R1)Tln2

The radioactivity of a sample is $R_{1}$ at a time $T_{1}$ and $R_{2}$ at a time $T_{2}$ . If the half-life of the specimen is $T$ , the number of atoms that have disintegrated in the time $(T_{2} - T_{1})$ is proportional to:

The correct option is A $(R_{1} - R_{2}) T$
Radioactivity $= \frac{d N}{d t} = λ N$
(Since $N (t) = N_{o} e^{- λ t})$
Given :
$R_{1} = λ N (T_{1})$
$R_{2} = λ N (T_{2})$
No. of disintegrated atoms $= N (T_{2}) - N (T_{1})$ $= \frac{R_{2} - R_{1}}{λ} = \frac{(R_{2} - R_{1}) T}{l n 2}$