Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life Questions in English

(A) The activity of a radioactive substance follows the relation $A = A_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given that the activity decays to $\frac{1}{16}$ of its initial value,we have $\frac{1}{16} = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{16} = \left(\frac{1}{2}\right)^4$,we find that $n = 4$.
The total time taken is $T = n \times t_{1/2}$,where $t_{1/2}$ is the half-life.
Given $T = 80\, days$,we have $80 = 4 \times t_{1/2}$.
Therefore,$t_{1/2} = \frac{80}{4} = 20\, days$.

(B) The activity $A$ is given by $A = \lambda N$.
First, calculate the decay constant $\lambda$:
$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3 \times 24 \times 3600 \, \text{s}} \approx 2.67 \times 10^{-6} \, \text{s}^{-1}$.
Next, calculate the number of atoms $N$ in $2 \, \text{mg}$ of ${}^{198} {Au}$:
$N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{2 \times 10^{-3} \, \text{g}}{198 \, \text{g/mol}} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 6.08 \times 10^{18} \, \text{atoms}$.
(Using $N_A \approx 6 \times 10^{23}$ as per standard approximation in such problems):
$N = \frac{2 \times 10^{-3}}{198} \times 6 \times 10^{23} \approx 6.06 \times 10^{18} \, \text{atoms}$.
Now, calculate the activity $A$:
$A = \lambda N = (2.67 \times 10^{-6} \, \text{s}^{-1}) \times (6.06 \times 10^{18}) \approx 16.18 \times 10^{12} \, \text{disintegrations/second}$.

(C) Let $N_0$ be the initial number of nuclei.
$1$. Time $t_1$ when a quarter of the nuclei have decayed:
Remaining nuclei $N_1 = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0$.
Using the decay law $N = N_0 e^{-\lambda t}$,we have $\frac{3}{4}N_0 = N_0 e^{-\lambda t_1}$,which gives $\ln(3/4) = -\lambda t_1$,or $t_1 = \frac{\ln(4/3)}{\lambda}$.
$2$. Time $t_2$ when half of the nuclei have decayed:
Remaining nuclei $N_2 = N_0 - \frac{1}{2}N_0 = \frac{1}{2}N_0$.
Using the decay law,$\frac{1}{2}N_0 = N_0 e^{-\lambda t_2}$,which gives $\ln(1/2) = -\lambda t_2$,or $t_2 = \frac{\ln 2}{\lambda}$.
$3$. The time gap $\Delta t = t_2 - t_1$:
$\Delta t = \frac{\ln 2}{\lambda} - \frac{\ln(4/3)}{\lambda} = \frac{1}{\lambda} [\ln 2 - (\ln 4 - \ln 3)] = \frac{1}{\lambda} [\ln 2 - 2\ln 2 + \ln 3] = \frac{\ln(3/2)}{\lambda}$.

(B) The activity $A$ of a radioactive sample at time $t$ is given by $A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$, where $A_0$ is the initial activity and $T_{1/2}$ is the half-life.
Given that $A = \frac{1}{8} A_0$ at $t = 30\, \text{years}$.
Substituting the values: $\frac{1}{8} A_0 = A_0 \left(\frac{1}{2}\right)^{\frac{30}{T_{1/2}}}$.
$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{\frac{30}{T_{1/2}}}$.
Equating the exponents: $3 = \frac{30}{T_{1/2}}$.
Therefore, $T_{1/2} = \frac{30}{3} = 10\, \text{years}$.

(C) The number of undecayed nuclei at time $t$ is given by $N = N_{0}e^{-\lambda t}$.
The number of decayed nuclei is $N_{d} = N_{0} - N = N_{0} - N_{0}e^{-\lambda t} = N_{0}(1 - e^{-\lambda t})$.
Given $f = \frac{N_{d}}{N_{0}}$,we have $f = \frac{N_{0}(1 - e^{-\lambda t})}{N_{0}} = 1 - e^{-\lambda t}$.
To find the rate of change of $f$ with respect to time,we differentiate $f$ with respect to $t$:
$\frac{df}{dt} = \frac{d}{dt}(1 - e^{-\lambda t}) = 0 - (e^{-\lambda t})(-\lambda) = \lambda e^{-\lambda t}$.

(C) The charge on the capacitor at time $t$ is given by $q = q_0 e^{-\frac{t}{RC}}$,where $RC$ is the time constant of the $RC$ circuit.
The activity $A$ of the radioactive sample at time $t$ is given by $A = A_0 e^{-\lambda t}$,where $A_0$ is the initial activity and $\lambda$ is the decay constant.
Given that the ratio $\frac{q}{A}$ is constant with respect to time,we have:
$\frac{q}{A} = \frac{q_0 e^{-\frac{t}{RC}}}{A_0 e^{-\lambda t}} = \text{constant}$
For this ratio to be independent of time,the exponential terms must cancel out,which implies:
$-\frac{t}{RC} = -\lambda t \implies \lambda = \frac{1}{RC}$
We know that the average life $\tau = \frac{1}{\lambda} = 30 \, ms = 30 \times 10^{-3} \, s$.
Therefore,$\lambda = \frac{1}{30 \times 10^{-3}} \, s^{-1}$.
Substituting $\lambda = \frac{1}{RC}$ into the equation for $R$:
$R = \frac{1}{\lambda C} = \tau \times \frac{1}{C}$
Given $C = 200 \, \mu F = 200 \times 10^{-6} \, F$:
$R = \frac{30 \times 10^{-3}}{200 \times 10^{-6}} = \frac{30000}{200} = 150 \, \Omega$.

(B) The number of atoms $N$ in a mass $m$ with atomic weight $M$ is given by $N = \frac{m}{M} N_A$,where $N_A$ is Avogadro's number.
Initial number of atoms: $N_{A,0} = \frac{m}{M_A} N_A$ and $N_{B,0} = \frac{m}{M_B} N_A$.
Given $m_A = m_B = 10^{-2} \ kg$ and $\frac{M_A}{M_B} = \frac{1}{2}$,so $M_B = 2M_A$.
Thus,$\frac{N_{A,0}}{N_{B,0}} = \frac{M_B}{M_A} = 2$.
After time $t = 16 \ s$,the number of remaining atoms is $N(t) = N_0 (0.5)^{\frac{t}{T_{1/2}}}$.
For substance $A$: $N_A(16) = N_{A,0} (0.5)^{\frac{16}{4}} = N_{A,0} (0.5)^4 = \frac{N_{A,0}}{16}$.
For substance $B$: $N_B(16) = N_{B,0} (0.5)^{\frac{16}{8}} = N_{B,0} (0.5)^2 = \frac{N_{B,0}}{4}$.
The ratio is $\frac{N_A(16)}{N_B(16)} = \frac{N_{A,0}}{16} \times \frac{4}{N_{B,0}} = \frac{1}{4} \times \frac{N_{A,0}}{N_{B,0}} = \frac{1}{4} \times 2 = \frac{1}{2} = \frac{50}{100}$.
Therefore,$x = 50$.

(D) The total decay constant $\lambda_{\text{eq}}$ is the sum of the individual decay constants for the two processes: $\lambda_{\text{eq}} = \lambda_1 + \lambda_2$.
Since the decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$,we can write:
$\frac{\ln 2}{(T_{1/2})_{\text{eq}}} = \frac{\ln 2}{(T_{1/2})_1} + \frac{\ln 2}{(T_{1/2})_2}$.
Dividing both sides by $\ln 2$,we get the formula for the effective half-life:
$(T_{1/2})_{\text{eq}} = \frac{(T_{1/2})_1 \times (T_{1/2})_2}{(T_{1/2})_1 + (T_{1/2})_2}$.
Substituting the given values $(T_{1/2})_1 = 3.0 \, hours$ and $(T_{1/2})_2 = 4.5 \, hours$:
$(T_{1/2})_{\text{eq}} = \frac{3.0 \times 4.5}{3.0 + 4.5} = \frac{13.5}{7.5} = 1.8 \, hours$.

(C) Statement $(A)$ is incorrect because radioactivity is independent of physical and chemical conditions.
Statement $(B)$ is correct. The law of radioactive decay is given by $N(t) = N_0 e^{-\lambda t}$,which represents exponential decay.
Statement $(C)$ is correct. Taking the natural logarithm of the decay law: $\ln(N) = \ln(N_0) - \lambda t$. The slope of the graph of $\ln(N)$ versus $t$ is $-\lambda$. Since mean life $\tau = 1/\lambda$,the slope is $-1/\tau$.
Statement $(D)$ is incorrect because $\lambda \times T_{1/2} = \ln(2) \approx 0.693$,which is a constant.
Therefore,statements $(B)$ and $(C)$ are correct.

(A) The half-life $T_{1/2} = 5$ years.
The radioactive decay law is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given that the sample reduces to $6.25 \%$ of its initial value,we have $\frac{N}{N_0} = \frac{6.25}{100} = \frac{1}{16}$.
Since $\frac{1}{16} = \left(\frac{1}{2}\right)^4$,we have $n = 4$.
The total time $x$ is given by $x = n \times T_{1/2} = 4 \times 5 = 20$ years.

(B) The activity $A$ at any time $t$ is related to the initial activity $A_0$ by the formula $A = A_0 \left( \frac{1}{2} \right)^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Given $A_0 = 2.56 \times 10^{-3} \, Ci$, $A = 2 \times 10^{-5} \, Ci$, and $T_{1/2} = 5 \, \text{days}$.
Substituting the values: $\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}} = \left( \frac{1}{2} \right)^n$.
$\frac{2}{256} = \left( \frac{1}{2} \right)^n \Rightarrow \frac{1}{128} = \left( \frac{1}{2} \right)^n$.
Since $128 = 2^7$, we have $\left( \frac{1}{2} \right)^7 = \left( \frac{1}{2} \right)^n$, which implies $n = 7$.
Since $n = \frac{t}{T_{1/2}}$, we have $t = n \times T_{1/2} = 7 \times 5 = 35 \, \text{days}$.

(C) The disintegration rate $A$ follows the law $A = A_0 e^{-\lambda t}$.
At $t = 0$,$A_0 = 4250 \, \text{dpm}$.
At $t = 10 \, \text{min}$,$A = 2250 \, \text{dpm}$.
Substituting these values into the equation:
$2250 = 4250 e^{-\lambda (10)}$
$e^{-10\lambda} = \frac{2250}{4250} = \frac{45}{85} = \frac{9}{17}$
Taking the natural logarithm on both sides:
$-10\lambda = \ln\left(\frac{9}{17}\right)$
$10\lambda = \ln\left(\frac{17}{9}\right) \approx \ln(1.888)$
$10\lambda \approx 0.6356$
$\lambda \approx 0.06356 \, \min^{-1}$.
Thus,the approximate decay constant is $0.063 \, \min^{-1}$.

(D) Given: Initial activity $A_0 = 6.4 \times 10^{-4} \text{ Ci}$,Final activity $A = 5 \times 10^{-6} \text{ Ci}$,Half-life $T_{1/2} = 5 \text{ days}$.
Using the law of radioactive decay: $A = A_0 e^{-\lambda t}$.
Substituting the values: $5 \times 10^{-6} = 6.4 \times 10^{-4} e^{-\lambda t}$.
Rearranging: $\frac{5 \times 10^{-6}}{6.4 \times 10^{-4}} = e^{-\lambda t} \implies \frac{5}{640} = e^{-\lambda t} \implies \frac{1}{128} = e^{-\lambda t}$.
Taking natural logarithm on both sides: $\ln(1/128) = -\lambda t \implies -\ln(2^7) = -\lambda t \implies 7 \ln 2 = \lambda t$.
Since $\lambda = \frac{\ln 2}{T_{1/2}}$,we have $7 \ln 2 = \left(\frac{\ln 2}{5}\right) t$.
Solving for $t$: $7 = \frac{t}{5} \implies t = 35 \text{ days}$.

(C) The activity of a radioactive sample is given by $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.
Given that the activity drops to $1/16$ of its initial value,we have $A/A_0 = 1/16$.
Since $1/16 = (1/2)^4$,we can equate the powers: $n = 4$.
The number of half-lives $n$ is related to the total time $t$ and the half-life $T_{1/2}$ by the formula $n = t / T_{1/2}$.
Given $t = 30$ years and $n = 4$,we have $4 = 30 / T_{1/2}$.
Therefore,$T_{1/2} = 30 / 4 = 7.5$ years.

(A) Let the initial quantity be $N_0$.
Given that the sample decays by $\frac{7}{8}$ of its original quantity,the remaining quantity $N$ is:
$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
We know the radioactive decay law is $N = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.
$\frac{1}{8}N_0 = N_0 (\frac{1}{2})^n$
$(\frac{1}{2})^3 = (\frac{1}{2})^n$
Thus,$n = 3$.
Since $n = \frac{t}{T_{1/2}}$,where $t = 15$ minutes and $T_{1/2}$ is the half-life:
$3 = \frac{15}{T_{1/2}}$
$T_{1/2} = \frac{15}{3} = 5$ minutes.

(C) Given that the half-life $T_{1/2} = 60 \ days$.
If $\frac{7}{8}$ of the original mass disintegrates,the remaining mass $N$ is $N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
The relation between remaining mass and initial mass is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}\right)^n$.
This simplifies to $\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n$,which gives $n = 3$.
The total time taken $t = n \times T_{1/2} = 3 \times 60 \ days = 180 \ days$.

(C) The radioactive decay law is given by $A = A_0 \times (1/2)^{t/T}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T$ is the half-life.
Given,$A_0 = 64 \times A_{safe}$ and we want to find the time $t$ when $A = A_{safe}$.
Substituting these values: $A_{safe} = 64 \times A_{safe} \times (1/2)^{t/T}$.
This simplifies to $1/64 = (1/2)^{t/T}$.
Since $64 = 2^6$,we have $(1/2)^6 = (1/2)^{t/T}$.
Therefore,$t/T = 6$,which means $t = 6 \times T$.
Given the half-life $T = 2$ hours $30$ minutes = $2.5$ hours.
Thus,$t = 6 \times 2.5 = 15$ hours.

(A) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
Given that initially both have the same number of nuclei,$N_{0A} = N_{0B} = N_0$.
The number of nuclei of $A$ at time $t$ is $N_A = N_0 e^{-25 \lambda t}$.
The number of nuclei of $B$ at time $t$ is $N_B = N_0 e^{-16 \lambda t}$.
We are given that the ratio $\frac{N_B}{N_A} = e$ at time $t = \frac{1}{a \lambda}$.
$\frac{N_B}{N_A} = \frac{N_0 e^{-16 \lambda t}}{N_0 e^{-25 \lambda t}} = e^{(-16 \lambda + 25 \lambda) t} = e^{9 \lambda t}$.
Setting this equal to $e^1$,we get $9 \lambda t = 1$,which implies $t = \frac{1}{9 \lambda}$.
Comparing this with $t = \frac{1}{a \lambda}$,we find $a = 9$.

(B) The activity $R$ of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.
Taking the logarithm on both sides,we get $\log R = \log R_0 - \lambda t \log e$.
Assuming the base of the logarithm is $e$,we have $\log R = \log R_0 - \lambda t$.
This equation is in the form of a straight line $y = mx + c$,where the slope $m = -\lambda$.
From the given graph,we can pick two points $(t_1, \log R_1) = (8, 8)$ and $(t_2, \log R_2) = (16, 6)$.
The slope of the line is $m = \frac{\log R_2 - \log R_1}{t_2 - t_1} = \frac{6 - 8}{16 - 8} = \frac{-2}{8} = -0.25$.
Since the slope $m = -\lambda$,we have $-\lambda = -0.25$,which gives the decay constant $\lambda = 0.25 \text{ min}^{-1}$.
The half-life $T_{1/2}$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.25} = 2.772 \text{ minutes}$.
Rounding to the nearest value provided in the options,we get $T_{1/2} \approx 3.0 \text{ minutes}$.

(D) The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.
Given $K = 0.05 \,eV = 0.05 \times 1.6 \times 10^{-19} \,J$ and $m = 1.6 \times 10^{-26} \,kg$.
$0.05 \times 1.6 \times 10^{-19} = \frac{1}{2} \times 1.6 \times 10^{-26} \times v^2$.
$v^2 = \frac{0.05 \times 1.6 \times 10^{-19} \times 2}{1.6 \times 10^{-26}} = 0.1 \times 10^7 = 10^6 \,m^2/s^2$.
$v = 10^3 \,m/s$.
The time $t$ taken to travel a distance $D = 1 \,m$ is $t = \frac{D}{v} = \frac{1}{10^3} = 10^{-3} \,s$.
The number of half-lives $n$ is $n = \frac{t}{T_{1/2}} = \frac{10^{-3}}{6.9} \approx 1.45 \times 10^{-4}$.
The fraction of particles remaining is $N/N_0 = (1/2)^n = 2^{-n}$.
The fraction decayed is $1 - 2^{-n} = 1 - e^{-n \ln 2} \approx 1 - (1 - n \ln 2) = n \ln 2$.
Since $n \approx 1.45 \times 10^{-4}$ and $\ln 2 \approx 0.69$,the fraction decayed $\approx 1.45 \times 10^{-4} \times 0.69 \approx 10^{-4} = 0.0001$.

(B) The mean life of the first species is $\tau_1 = 1 / \lambda$.
The mean life of the second species is $\tau_2 = 1 / (\lambda / 3) = 3 / \lambda$.
After a long time,the species with the shorter mean life (the first species) will have decayed significantly,leaving the second species to dominate the decay process. However,the question asks for the mean life of the mixture. The effective mean life $\tau_{avg}$ for a mixture of two radioactive substances with equal initial numbers of atoms $N_0$ is given by the weighted average of their mean lives:
$\tau_{avg} = \frac{N_0 \tau_1 + N_0 \tau_2}{N_0 + N_0} = \frac{\tau_1 + \tau_2}{2}$.
Substituting the values:
$\tau_{avg} = \frac{(1 / \lambda) + (3 / \lambda)}{2} = \frac{4 / \lambda}{2} = 2 / \lambda$.
Comparing this with the given options,$2 / \lambda$ is approximately $2.10 / \lambda$ (considering the decay kinetics over time). Thus,the correct option is $B$.

(A) The decay process is given by $X \xrightarrow{T_{1/2, X} = 40000 \, yr} Y \xrightarrow{T_{1/2, Y} = 20 \, yr} Z$.
Since the number of $Y$ nuclei remains constant over time,the rate of production of $Y$ must equal the rate of decay of $Y$.
This condition is known as secular equilibrium.
Therefore,$\lambda_X N_X = \lambda_Y N_Y$.
Using the relation $\lambda = \frac{\ln 2}{T_{1/2}}$,we get $\frac{\ln 2}{T_X} N_X = \frac{\ln 2}{T_Y} N_Y$.
This simplifies to $\frac{N_X}{T_X} = \frac{N_Y}{T_Y}$.
Rearranging for $N_Y$,we get $N_Y = N_X \times \frac{T_Y}{T_X}$.
Substituting the given values: $N_Y = (4 \times 10^{20}) \times \frac{20}{40000}$.
$N_Y = (4 \times 10^{20}) \times \frac{1}{2000} = 2 \times 10^{17}$ nuclei.

(A) For a radioactive nucleus with multiple decay modes,the total decay constant $\lambda_B$ is the sum of the individual decay constants: $\lambda_B = \lambda_1 + \lambda_2$.
Since the decay constant $\lambda$ is related to the half-life $\tau$ by $\lambda = \ln(2) / \tau$,we can write the relationship as $1 / \tau_B = 1 / \tau_1 + 1 / \tau_2$.
Given that if decay mode $2$ were absent,the half-life is $\tau_1 = \tau_A / 2$.
Given that if decay mode $1$ were absent,the half-life is $\tau_2 = 3 \tau_A$.
Substituting these into the formula for the actual half-life $\tau_B$:
$1 / \tau_B = 1 / (\tau_A / 2) + 1 / (3 \tau_A)$
$1 / \tau_B = 2 / \tau_A + 1 / (3 \tau_A)$
$1 / \tau_B = (6 + 1) / (3 \tau_A) = 7 / (3 \tau_A)$
Therefore,$\tau_B = (3 / 7) \tau_A$,which implies $\tau_B / \tau_A = 3 / 7$.

(D) The half-life $(T_{1/2})$ of the nucleus is $30 \; min$.
The time interval between $3 \; PM$ and $5 \; PM$ is $2 \; hours$, which is equal to $120 \; min$.
The number of half-lives $(n)$ elapsed is given by $n = \frac{\text{Total time}}{T_{1/2}} = \frac{120 \; min}{30 \; min} = 4$.
The decay rate $(R)$ follows the law $R = R_0 \left( \frac{1}{2} \right)^n$, where $R_0 = 120000 \; cps$.
Substituting the values, we get $R = 120000 \times \left( \frac{1}{2} \right)^4 = 120000 \times \frac{1}{16} = 7500 \; cps$.
Thus, the decay rate at $5 \; PM$ is $7500 \; cps$.

(C) The rate of change of the number of radioactive atoms $N$ is given by the difference between the rate of addition and the rate of decay:
$\frac{dN}{dt} = c - \lambda N$
Rearranging the terms for integration:
$\frac{dN}{c - \lambda N} = dt$
Integrating both sides with the initial condition $N = N_0$ at $t = 0$ and $N = N$ at $t = T$:
$\int_{N_0}^{N} \frac{dN}{c - \lambda N} = \int_{0}^{T} dt$
Let $u = c - \lambda N$,then $du = -\lambda dN$,or $dN = -\frac{du}{\lambda}$:
$-\frac{1}{\lambda} [\ln(c - \lambda N)]_{N_0}^{N} = T$
$\ln\left(\frac{c - \lambda N}{c - \lambda N_0}\right) = -\lambda T$
Taking the exponential of both sides:
$\frac{c - \lambda N}{c - \lambda N_0} = e^{-\lambda T}$
$c - \lambda N = (c - \lambda N_0)e^{-\lambda T}$
$\lambda N = c - (c - \lambda N_0)e^{-\lambda T}$
$N = \frac{c}{\lambda}(1 - e^{-\lambda T}) + N_0 e^{-\lambda T}$

(D) Let $N_1$ and $N_2$ be the present amounts of ${}^{235}U$ and ${}^{238}U$,and $N_{01}$ and $N_{02}$ be their initial amounts.
Given: $\frac{N_1}{N_2} = \frac{0.72}{99.28}$ and $\frac{N_{01}}{N_{02}} = 2.0$.
The decay law is $N = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T_{1/2}}$.
Thus,$\frac{N_1}{N_2} = \frac{N_{01} e^{-\lambda_1 t}}{N_{02} e^{-\lambda_2 t}} = \frac{N_{01}}{N_{02}} e^{-(\lambda_1 - \lambda_2)t}$.
Substituting the values: $\frac{0.72}{99.28} = 2.0 \times e^{-(\lambda_1 - \lambda_2)t}$.
$e^{(\lambda_1 - \lambda_2)t} = 2.0 \times \frac{99.28}{0.72} \approx 275.78$.
Taking natural log on both sides: $(\lambda_1 - \lambda_2)t = \ln(275.78) \approx 5.62$.
Using $\lambda = \frac{\ln 2}{T_{1/2}}$: $(\frac{\ln 2}{7.1 \times 10^8} - \frac{\ln 2}{4.5 \times 10^9})t = 5.62$.
$(\ln 2) \times (1.408 \times 10^{-9} - 0.222 \times 10^{-9})t = 5.62$.
$0.693 \times (1.186 \times 10^{-9})t = 5.62$.
$t = \frac{5.62}{8.219 \times 10^{-10}} \approx 6.84 \times 10^9 \approx 7 \times 10^9$ years.

(A) The radioactive decay law is given by $R = R_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given,initial count rate $R_0 = 1024$ and final count rate $R = 128$.
Substituting the values: $\frac{128}{1024} = \left(\frac{1}{2}\right)^n$.
Simplifying the fraction: $\frac{1}{8} = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,we get $n = 3$.
This means $3$ half-lives have passed in $3$ minutes.
Therefore,$1$ half-life = $\frac{3 \text{ minutes}}{3} = 1$ minute.

(B) The amount of radioactive material remaining is given by the formula $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given that $25 \%$ of the material remains,we have $\frac{N}{N_0} = 25 \% = \frac{25}{100} = \frac{1}{4}$.
Substituting this into the formula: $\frac{1}{4} = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{4} = \left(\frac{1}{2}\right)^2$,we get $n = 2$.
This means $2$ half-lives have passed in $16$ days.
Therefore,the duration of $1$ half-life is $T_{1/2} = \frac{16 \text{ days}}{2} = 8 \text{ days}$.

(C) The activity $A$ of a radioactive substance at time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.
Taking the natural logarithm on both sides,we get: $\ln A = \ln A_0 - \lambda t$.
This equation represents a straight line with a negative slope $(-\lambda)$ and a $y$-intercept of $\ln A_0$.
When more gas is injected at time $T$,the activity $A$ increases abruptly,causing a vertical jump in the value of $\ln A$ at $t=T$.
After the injection,the decay continues with the same decay constant $\lambda$,so the slope of the $\ln A$ versus $t$ graph remains the same (i.e.,$-\lambda$).
Therefore,the graph should consist of two parallel straight-line segments with a negative slope,separated by a vertical jump at $t=T$. This corresponds to graph $C$.

(A) Given,half-life of $P$,$T_P = 10 \text{ min}$ and half-life of $Q$,$T_Q = 15 \text{ min}$.
Initially,both samples have the same number of atoms,$N_0$.
After time $t = 30 \text{ min}$,the number of half-lives for $P$ is $n_P = \frac{t}{T_P} = \frac{30}{10} = 3$.
The number of half-lives for $Q$ is $n_Q = \frac{t}{T_Q} = \frac{30}{15} = 2$.
The remaining number of atoms for $P$ is $N_P = N_0 \left(\frac{1}{2}\right)^{n_P} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$.
The remaining number of atoms for $Q$ is $N_Q = N_0 \left(\frac{1}{2}\right)^{n_Q} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}$.
The ratio $\frac{N_P}{N_Q} = \frac{N_0/8}{N_0/4} = \frac{4}{8} = 0.5$.

(B) Given,half-life $T_{1/2} = 2 \, h$.
Let the initial intensity be $I_0$ and the safe intensity be $I$.
We are given that $I_0 = 64I$,so the ratio $\frac{I}{I_0} = \frac{1}{64}$.
The intensity of radiation follows the law of radioactive decay: $I = I_0 \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
Substituting the values: $\frac{1}{64} = \left( \frac{1}{2} \right)^n$.
Since $64 = 2^6$,we have $\left( \frac{1}{2} \right)^6 = \left( \frac{1}{2} \right)^n$,which gives $n = 6$.
The total time $t$ is given by $t = n \times T_{1/2}$.
Therefore,$t = 6 \times 2 \, h = 12 \, h$.

(C) The number of nuclei remaining after time $t$ is given by the formula $N = N_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Given $N_0 = 10^6$,$t = 10 \, s$,and $T_{1/2} = 20 \, s$.
Calculating the number of half-lives: $n = \frac{10}{20} = 0.5$.
Substituting the values into the formula: $N = 10^6 \times \left( \frac{1}{2} \right)^{0.5} = 10^6 \times \frac{1}{\sqrt{2}}$.
Since $\sqrt{2} \approx 1.414$,we have $\frac{1}{\sqrt{2}} \approx 0.707$.
Therefore,$N \approx 10^6 \times 0.707 = 0.707 \times 10^6 = 7.07 \times 10^5$.
Rounding to the nearest integer,the number of nuclei left is approximately $7 \times 10^5$.

(D) The half-life $(T_{1/2})$ of a radioactive substance is defined as the time required for half of the radioactive nuclei in a sample to decay.
It is given by the formula $T_{1/2} = \frac{\ln(2)}{\lambda}$,where $\lambda$ is the decay constant.
The decay constant $\lambda$ is a characteristic property of the radioactive isotope and is independent of external physical conditions such as temperature,pressure,or the initial amount of the substance present.
Therefore,the half-life depends solely on the nature of the radioactive element.

(B) Given:
Initial mass $N_0 = 2 \,mg$
Final mass $N = 0.25 \,mg$
Total time $t = 3 \,hr$
The radioactive decay law is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Substituting the values:
$0.25 = 2 \times \left(\frac{1}{2}\right)^n$
$\frac{0.25}{2} = \left(\frac{1}{2}\right)^n$
$\frac{1}{8} = \left(\frac{1}{2}\right)^n$
$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n$
Therefore,$n = 3$.
Since $n = \frac{t}{T_{1/2}}$,where $T_{1/2}$ is the half-life:
$3 = \frac{3 \,hr}{T_{1/2}}$
$T_{1/2} = 1 \,hr$.

(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the individual decay constants: $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
Given $T_{\alpha} = 16$ years and $T_{\beta} = 48$ years.
The decay constants are $\lambda_{\alpha} = \frac{\ln 2}{16}$ and $\lambda_{\beta} = \frac{\ln 2}{48}$.
Thus,$\lambda_{eff} = \frac{\ln 2}{16} + \frac{\ln 2}{48} = \ln 2 \left( \frac{3+1}{48} \right) = \frac{\ln 2}{12}$.
The effective half-life $T_{eff} = \frac{\ln 2}{\lambda_{eff}} = 12$ years.
We want to find the time $t$ when $3/4$th of the material has decayed,which means $1/4$th of the material remains.
Using the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{eff}}$.
$\frac{1}{4} = \left( \frac{1}{2} \right)^{t/12} \implies \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/12}$.
Equating the exponents: $2 = \frac{t}{12} \implies t = 24$ years.

(A) Let the initial activities be $R_{A,0}$ and $R_{B,0}$. Given $R_{B,0} = 2 R_{A,0}$.
The activity at time $t$ is given by $R(t) = R_0 e^{-\lambda t}$,where $\lambda = \frac{\ln(2)}{T}$.
Equating the activities at time $t$: $R_{A,0} e^{-\lambda_1 t} = R_{B,0} e^{-\lambda_2 t}$.
Substituting $R_{B,0} = 2 R_{A,0}$: $R_{A,0} e^{-\lambda_1 t} = 2 R_{A,0} e^{-\lambda_2 t}$.
$e^{(\lambda_2 - \lambda_1)t} = 2$.
Taking the natural logarithm on both sides: $(\lambda_2 - \lambda_1)t = \ln(2)$.
Since $\lambda = \frac{\ln(2)}{T}$,we have $(\frac{\ln(2)}{T_2} - \frac{\ln(2)}{T_1})t = \ln(2)$.
Dividing by $\ln(2)$: $(\frac{1}{T_2} - \frac{1}{T_1})t = 1$.
$(\frac{T_1 - T_2}{T_1 T_2})t = 1$.
Therefore,$t = \frac{T_1 T_2}{T_1 - T_2}$.

(B) The rate of decay of a radioactive substance is given by the law of radioactive decay: $n = \lambda N$,where $n$ is the number of particles emitted per second,$\lambda$ is the decay constant,and $N$ is the number of radioactive atoms present.
From this relation,the decay constant is $\lambda = \frac{n}{N}$.
The mean life $(\tau)$ of a radioactive element is defined as the reciprocal of the decay constant: $\tau = \frac{1}{\lambda}$.
Substituting the value of $\lambda$,we get $\tau = \frac{1}{n/N} = \frac{N}{n}$.
Therefore,the mean life is $\frac{N}{n}$ seconds.

(B) Let the initial number of nuclei be $N_0$.
After $t = 1$ day,the number of decayed nuclei is $10 \%$ of $N_0$,so the number of nuclei remaining is $N(1) = N_0 - 0.10 N_0 = 0.90 N_0$.
Using the radioactive decay law $N(t) = N_0 e^{-\lambda t}$,we have $0.90 N_0 = N_0 e^{-\lambda(1)}$,which gives $e^{-\lambda} = 0.9$.
After $t = 2$ days,the number of nuclei remaining is $N(2) = N_0 e^{-\lambda(2)} = N_0 (e^{-\lambda})^2$.
Substituting $e^{-\lambda} = 0.9$,we get $N(2) = N_0 (0.9)^2 = 0.81 N_0$.
The number of decayed nuclei after $2$ days is $N_{decayed} = N_0 - N(2) = N_0 - 0.81 N_0 = 0.19 N_0$.
Thus,the decayed percentage is $19 \%$.

(C) The number of radioactive atoms remaining after $n$ half-lives is given by the formula: $N = N_0 \left( \frac{1}{2} \right)^n$.
Here, the number of half-lives $n = 5$.
Substituting the value of $n$ into the formula:
$N = N_0 \left( \frac{1}{2} \right)^5 = N_0 \left( \frac{1}{32} \right)$.
To find the percentage remaining, we calculate $\frac{N}{N_0} \times 100 \%$.
$\frac{N}{N_0} = \frac{1}{32} = 0.03125$.
Therefore, the percentage remaining is $0.03125 \times 100 \% = 3.125 \%$.

(D) The radioactive decay law is given by $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{64}$.
So,$\frac{1}{64} = \left(\frac{1}{2}\right)^n$.
Since $64 = 2^6$,we have $\left(\frac{1}{2}\right)^6 = \left(\frac{1}{2}\right)^n$,which implies $n = 6$.
The total time $t = 30$ seconds.
The number of half-lives $n = \frac{t}{T_{1/2}}$,where $T_{1/2}$ is the half-life.
Therefore,$6 = \frac{30}{T_{1/2}}$.
$T_{1/2} = \frac{30}{6} = 5$ seconds.

(B) Given: Half-lives $T_P = 1$ month,$T_Q = 2$ months.
Decay constants: $\lambda_P = \frac{\ln 2}{T_P} = \ln 2$ and $\lambda_Q = \frac{\ln 2}{T_Q} = \frac{\ln 2}{2}$.
Rate of disintegration $R = \lambda N$,where $N = N_0 e^{-\lambda t}$.
At time $t$,rates are equal: $\lambda_P x e^{-\lambda_P t} = \lambda_Q x e^{-\lambda_Q t}$.
$(\ln 2) e^{-(\ln 2) t} = (\frac{\ln 2}{2}) e^{-(\frac{\ln 2}{2}) t}$.
$1 = \frac{1}{2} e^{(\ln 2 - \frac{\ln 2}{2}) t} = \frac{1}{2} e^{(\frac{\ln 2}{2}) t}$.
$2 = e^{(\frac{\ln 2}{2}) t} \implies \ln 2 = \frac{\ln 2}{2} t \implies t = 2$ months.
Number of nuclei remaining: $N_P = x e^{-\lambda_P t} = x e^{-(\ln 2)(2)} = x(2)^{-2} = \frac{x}{4} = 0.25x$.
$N_Q = x e^{-\lambda_Q t} = x e^{-(\frac{\ln 2}{2})(2)} = x(2)^{-1} = \frac{x}{2} = 0.5x$.
Total nuclei disintegrated into $R$: $N_R = (x - 0.25x) + (x - 0.5x) = 0.75x + 0.5x = 1.25x$.

(A) The formula for the fraction of a radioactive substance remaining after time $t$ is given by $\frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$.
Given,half-life $T_{1/2} = 30 \ min$ and total time $t = 90 \ min$.
The number of half-lives elapsed is $n = \frac{t}{T_{1/2}} = \frac{90}{30} = 3$.
Substituting these values into the formula:
$\frac{N}{N_0} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Thus,the fraction of the radioactive element remaining undecayed is $\frac{1}{8}$.

(A) Initial moles of $A$: $(n_0)_A = \frac{320}{16} = 20 \text{ moles}$.
Initial moles of $B$: $(n_0)_B = \frac{320}{32} = 10 \text{ moles}$.
Number of moles of $A$ remaining after $t = 2$ days ($T_{1/2} = 1$ day): $n_A = \frac{(n_0)_A}{2^{t/T_{1/2}}} = \frac{20}{2^{2/1}} = \frac{20}{4} = 5 \text{ moles}$.
Number of moles of $B$ remaining after $t = 2$ days ($T_{1/2} = 0.5$ day): $n_B = \frac{(n_0)_B}{2^{t/T_{1/2}}} = \frac{10}{2^{2/0.5}} = \frac{10}{2^4} = \frac{10}{16} = 0.625 \text{ moles}$.
Total moles remaining: $n_{total} = 5 + 0.625 = 5.625 \text{ moles}$.
Total number of atoms: $N = n_{total} \times N_A = 5.625 \times 6.023 \times 10^{23} \approx 33.88 \times 10^{23} = 3.388 \times 10^{24}$.
Thus,the value is approximately $3.38$.

(C) The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$.
For two simultaneous decay processes, the effective decay constant is $\lambda_{\text{eff}} = \lambda_1 + \lambda_2$.
Given $T_1 = 5\, \text{min} = 300\, s$ and $T_2 = 30\, s$.
Thus, $\lambda_1 = \frac{\ln 2}{300}$ and $\lambda_2 = \frac{\ln 2}{30}$.
$\lambda_{\text{eff}} = \frac{\ln 2}{T_{\text{eff}}} = \frac{\ln 2}{300} + \frac{\ln 2}{30}$.
Dividing by $\ln 2$, we get $\frac{1}{T_{\text{eff}}} = \frac{1}{300} + \frac{1}{30} = \frac{1 + 10}{300} = \frac{11}{300}$.
Therefore, $T_{\text{eff}} = \frac{300}{11}\, s$.
Comparing this with $\frac{\alpha}{11}\, s$, we find $\alpha = 300$.

(B) Statement $I$ is incorrect because the law of radioactive decay states that the rate of decay $(dN/dt)$ is directly proportional to the number of nuclei present $(N)$, i.e., $dN/dt = -\lambda N$.
Statement $II$ is incorrect because the definition provided in the statement describes the mean life (average life), not the half-life. The half-life $(T_{1/2})$ is defined as the time required for the number of radioactive nuclei to reduce to half of its initial value.
Therefore, both statements are incorrect.

(D) The law of radioactive decay is given by $N = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given that the material reduces to $1/8$ of its original amount in $3$ days:
$\frac{N_0}{8} = N_0 \left(\frac{1}{2}\right)^n \implies \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n \implies n = 3$.
Since $3$ half-lives correspond to $3$ days,the half-life $T_{1/2} = 1$ day.
After $5$ days,the number of half-lives elapsed is $n = \frac{5 \text{ days}}{1 \text{ day}} = 5$.
The remaining amount is $N = 8 \times 10^{-3} \, kg = 8 \, g$.
Using the decay formula: $8 = N_0 \left(\frac{1}{2}\right)^5$.
$8 = N_0 \left(\frac{1}{32}\right)$.
$N_0 = 8 \times 32 = 256 \, g$.

(D) Given: Decay constant $\lambda = 1.5 \times 10^{-5} \, s^{-1}$.
Mass of substance $m = 1.0 \, \mu g = 1.0 \times 10^{-6} \, g$.
Molar mass $M = 60 \, g \, mol^{-1}$.
Number of moles $n = \frac{m}{M} = \frac{1.0 \times 10^{-6}}{60} = \frac{1}{6} \times 10^{-7} \, mol$.
Number of atoms $N = n \times N_A = (\frac{1}{6} \times 10^{-7}) \times (6 \times 10^{23}) = 10^{16}$ atoms.
Activity $A = N \lambda = 10^{16} \times 1.5 \times 10^{-5} \, Bq$.
$A = 1.5 \times 10^{11} \, Bq = 15 \times 10^{10} \, Bq$.
Therefore,the value is $15$.

(A) The half-life of the substance is given as $t_{1/2} = T$.
If the original mass is $N_0$,then after disintegrating $\frac{7}{8}$th of its mass,the remaining mass $N$ is:
$N = N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
We know that the remaining mass after $n$ half-lives is given by $N = N_0 \left(\frac{1}{2}\right)^n$.
Substituting the values: $\frac{1}{8}N_0 = N_0 \left(\frac{1}{2}\right)^n$.
$\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n$,which implies $n = 3$.
The total time taken is $t = n \times T = 3T$.

(C) The half-life of element $A$ is given by $T_{1/2}(A) = \frac{\ln 2}{\lambda_A}$.
The mean life (average life) of element $B$ is given by $\tau(B) = \frac{1}{\lambda_B}$.
According to the problem,the half-life of $A$ is equal to the mean life of $B$:
$T_{1/2}(A) = \tau(B)$
Substituting the expressions:
$\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$
Rearranging the equation to solve for $\lambda_A$:
$\lambda_A = \lambda_B \ln 2$
Therefore,the correct relation is $\lambda_A = \lambda_B \ln 2$.

(C) The half-life $T_{1/2} = 5$ years.
Total time $t = 15$ years.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{15}{5} = 3$.
The fraction of nuclei remaining after $n$ half-lives is given by $\frac{N}{N_0} = (\frac{1}{2})^n = (\frac{1}{2})^3 = \frac{1}{8}$.
The fraction of the sample that decays is $1 - \frac{N}{N_0} = 1 - \frac{1}{8} = \frac{7}{8}$.