SATHEE: Modern Physics 3 Question 27

Modern Physics 3 Question 27

28. For a radioactive material, its activity $A$ and rate of change of its activity $R$ are defined as $A = - \frac{d N}{d t}$ and $R = - \frac{d A}{d t}$ , where $N (t)$ is the number of nuclei at time $t$ . Two radioactive source $P ($ mean life $τ)$ and $Q$ (mean life $2 τ)$ have the same activity at $t = 0$ . Their rate of change of activities at $t = 2 τ$ are $R_{P}$ and $R_{Q}$ , respectively. If $\frac{R_{P}}{R_{Q}} = \frac{n}{e}$ , then the value of $n$ is

(2015 Adv.)

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Solution:

Let initial numbers are $N_{1}$ and $N_{2}$ .

$\frac{λ_{1}}{λ_{2}} = \frac{τ_{2}}{τ_{1}} = \frac{2 τ}{τ} = 2 = \frac{T_{2}}{T_{1}}$

$(T =$ Half life $)$

$A = \frac{- d N}{d t} = λ N$

Initial activity is same

$∴ λ_{1} N_{1} = λ_{2} N_{2}$

Activity at time $t$ ,

$\begin{aligned} A & = λ N = λ N_{0} e^{- λ t} \\ A_{1} & = λ_{1} N_{1} e^{- λ_{1} t} \\ \Rightarrow R_{1} - & = \frac{d A_{1}}{d t} = λ_{1}^{2} N_{1} e^{- λ_{1} t} \\ Similarly, R_{2} & = λ_{2}^{2} N_{2} e^{- λ_{2} t} \end{aligned}$

After $t = 2 τ$

$\begin{aligned} λ_{1} t = \frac{1}{τ_{1}} (t) = \frac{1}{τ} (2 τ) = 2 \\ λ_{2} t = \frac{1}{τ_{2}} (t) = 1 = \frac{1}{2 τ} (2 τ) = 1 \\ \frac{R_{P}}{R_{Q}} = \frac{λ_{1}^{2} N_{1} e^{- λ_{1} t}}{λ_{2}^{2} N_{2} e^{- λ_{2} t}} \\ \frac{R_{P}}{R_{Q}} = \frac{λ_{1}}{λ_{2}} \frac{e^{- 2}}{e^{- 1}} = \frac{2}{e} \end{aligned}$

Modern Physics 3 Question 27

Solution:

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Modern Physics 3 Question 27

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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