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First Order Reaction

The activity ...

Question

The activity of a sample of radioactive material is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . Its half life is $1$ .no of atoms decade in time $(t_{2} - t_{1})$ is proportional to?

A_{1} - A_{2}

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\frac{A_{1} - A_{2}}{t_{2} - t_{1}} = c o n s t a n t

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A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}

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A_{2} = A_{1} e^{(t_{1} / T t_{2})}

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Solution

The correct option is C $A_{2} = A_{1} e^{(t_{1} - t_{2}) / T}$
$\begin{matrix} Given-* Activity at t = t_{1} is A_{1} * Activity at t = t_{2} is A_{2} we know - activity A = A_{0} e^{- λ t} = N_{0} \cdot λ e^{λ t} \Rightarrow A_{1} = A_{0} \cdot e^{- λ t_{1}} and A_{2} = A_{0} \cdot e^{- λ t_{2}} \Rightarrow A_{1} = N_{0} \cdot λ \cdot e^{- λ t_{1}} A_{2} = N_{0} \cdot λ \cdot e^{- λ t_{2}} \end{matrix}$

$\begin{matrix} we have half life, t_{1 / 2} = t_{2} - t_{1} A_{1} - A_{2} = N_{0} λ [e^{- λ t_{1}} - e^{- λ t_{2}}] we have- λ_{t / 2} = ln 2 \Rightarrow t_{1 / 2} = \frac{ln 2}{λ} \Rightarrow A_{1} - A_{2} = \frac{N_{0}}{(t_{2} - t_{1})} \cdot ln 2 \cdot [e^{- λ t_{1}} - e^{- λ t_{2}}] \end{matrix}$

$\begin{matrix} \frac{A_{1}}{A_{2}} & = \frac{N_{0} λ \cdot e^{λ t_{1}}}{N_{0} λ \cdot e^{- λ t_{2}}} = e^{λ t_{2} - λ t_{1}} \Rightarrow A_{2} = A_{1} e^{λ (t_{1} t_{2})} A_{2} & = A_{1} e^{(t_{1} - t_{2}) / T} Whre [T = \frac{1}{λ}] mean life. \end{matrix}$

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The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2. Its half life is 1.no of atoms decade in time (t2−t1) is proportional to?

The activity of a sample of radioactive material is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . Its half life is $1$ .no of atoms decade in time $(t_{2} - t_{1})$ is proportional to?