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Law of Conservation of Energy

Two identical...

Question

Two identical samples (same material and same material and same amount initially.) P and Q of a radioactive substance having mean life T are observed to have activities $A_{p}$ & $A_{Q}$ respectively at the time of observation. If P is older than Q, then the difference in their ages is:

T

(\frac{A_{P} - A_{Q}}{ℓ n 2})

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T

(\frac{A_{P} + A_{Q}}{ℓ n 2})

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T

(\frac{A_{Q} - A_{P}}{ℓ n 2})

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None of these

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Solution

The correct option is A $T$ $(\frac{A_{P} - A_{Q}}{ℓ n 2})$
We know that
Average life is given by
$λ = \frac{1}{T}$
According to question we have
$A_{P} = A_{0} e^{- λ P}$ ---eq.1
$A_{Q} = A_{0} e^{- λ Q}$ ---eq.2
eq.1 / eq,2
$\frac{A_{P}}{A_{Q}} = e^{- λ (P - Q)}$
$ln \frac{A_{P}}{A_{Q}} = - λ (P - Q)$
$ln \frac{A_{P}}{A_{Q}} = - \frac{(P - Q)}{T}$
$P - Q = - T ln \frac{A_{P}}{A_{Q}}$
$P - Q = T ln \frac{A_{Q}}{A_{P}}$

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Two identical samples (same material and same material and same amount initially.) P and Q of a radioactive substance having mean life T are observed to have activities Ap & AQ respectively at the time of observation. If P is older than Q, then the difference in their ages is:

The correct option is A T (AP−AQℓn2)We know that Average life is given by λ=1TAccording to question we have AP=A0e−λP ---eq.1AQ=A0e−λQ ---eq.2eq.1 / eq,2APAQ=e−λ(P−Q)lnAPAQ=−λ(P−Q)lnAPAQ=−(P−Q)TP−Q=−TlnAPAQP−Q=TlnAQAP

Two identical samples (same material and same material and same amount initially.) P and Q of a radioactive substance having mean life T are observed to have activities $A_{p}$ & $A_{Q}$ respectively at the time of observation. If P is older than Q, then the difference in their ages is: