To determine the half-life of a radioactive element, a student plots a graph of $l n$ $| \frac{dN (t)}{dt} |$ versus $t$ . Here, $| \frac{d N (t)}{d t} |$ is the rate of radioactive decay at time $t$ . If the number of radioactive nuclei of this element decreases by a factor of $p$ after 4.16 years, the value of $p$ is [2010]

Question

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(8)

$We know that$ $N = N_{0} e^{- λ t}$

$∴$ $\frac{d N}{d t} = N_{0} e^{- λ t} (- λ) = - λ N_{0} e^{- λ t}$

$Taking log on both sides,$

$\log_{e}$ $| \frac{d N}{d t} |$ $= \log_{e} (λ N_{0}) - λ t$

$Comparing it with the graph line,$

$Decay constant,$ $λ = \frac{1}{2} {yr}^{- 1}$ $[\frac{A C}{B C} = \frac{1}{2}]$

$∴$ $T_{1 / 2} = \frac{0.693}{λ} = 0.693 \times 2 = 1.386 years$

$n (t_{1 / 2}) = 4.16$

$∴$ $n = \frac{4.16}{1.386} \approx 3$

$∴$ $N = N_{0} {(\frac{1}{2})}^{3}$

$∴$ $\frac{1}{P} = \frac{1}{8}$

$∴$ $P = 8$

Solution

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Solution

Q. To determine the half-life of a radioactive element, a student plots a graph of ln|dN(t)dt| versus t. Here, |dN(t)dt| is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is [2010]

Ans. (8) We know that N=N0e-λt ∴ dNdt=N0e-λt(-λ)=-λN0e-λt Taking log on both sides, loge|dNdt|=loge(λN0)-λt Comparing it with the graph line, Decay constant, λ=12 yr-1 [ACBC=12] ∴ T1/2=0.693λ=0.693×2=1.386 years n(t1/2)=4.16 ∴ n=4.161.386≈3 ∴ N=N0(12)3 ∴ 1P=18 ∴ P=8

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