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

(C) The radioactive material decays through two channels. The effective mean life $\tau$ is given by the relation:
$\frac{1}{\tau} = \frac{1}{\tau_1} + \frac{1}{\tau_2}$
Substituting the given values $\tau_1 = 1620 \, \text{years}$ and $\tau_2 = 520 \, \text{years}$:
$\tau = \frac{\tau_1 \tau_2}{\tau_1 + \tau_2} = \frac{1620 \times 520}{1620 + 520} = \frac{842400}{2140} \approx 393.64 \, \text{years} \approx 394 \, \text{years}$.
The half-life $T_{1/2}$ is related to the mean life $\tau$ by the formula:
$T_{1/2} = 0.693 \times \tau$
$T_{1/2} = 0.693 \times 393.64 \approx 272.79 \, \text{years} \approx 273 \, \text{years}$.
Therefore,the correct option is $C$.

(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{6.92 \ s} \approx 0.1 \ s^{-1}$.
The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
The fractional change in the number of nuclei is given by $\frac{N_0 - N(t)}{N_0} = 1 - e^{-\lambda t}$.
Substituting the values $t = 10 \ s$ and $\lambda = 0.1 \ s^{-1}$:
Fractional change $= 1 - e^{-(0.1)(10)} = 1 - e^{-1}$.
Since $e^{-1} \approx 0.37$,the fractional change is $1 - 0.37 = 0.63$.

(B) Let the decay constants for the two modes be $\lambda_{\alpha}$ and $\lambda_{\beta}$ respectively.
Given that the sample decays by $\alpha$-decay $66.6 \%$ of the time and by $\beta$-decay $33.3 \%$ of the time,the ratio of the decay rates is $\frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{66.6}{33.3} = 2$.
Thus,$\lambda_{\alpha} = 2\lambda_{\beta}$ or $\lambda_{\beta} = \frac{\lambda_{\alpha}}{2}$.
The effective decay constant $\lambda_{eff}$ is given by $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
The effective half-life $T_{eff}$ is $60 \text{ years}$,so $\lambda_{eff} = \frac{\ln 2}{60}$.
Substituting the values: $\lambda_{\alpha} + \frac{\lambda_{\alpha}}{2} = \frac{\ln 2}{60} \Rightarrow \frac{3\lambda_{\alpha}}{2} = \frac{\ln 2}{60}$.
Therefore,$\lambda_{\alpha} = \frac{2 \ln 2}{180} = \frac{\ln 2}{90}$.
The half-life for $\alpha$-decay only is $T_{\alpha} = \frac{\ln 2}{\lambda_{\alpha}} = 90 \text{ years}$.

(C) The effective decay constant $\lambda_{eff}$ for simultaneous decay is given by $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
Since mean life $\tau = \frac{1}{\lambda}$,we have $\frac{1}{\tau_{eff}} = \frac{1}{\tau_{\alpha}} + \frac{1}{\tau_{\beta}}$.
Given $\tau_{\alpha} = 30$ years and $\tau_{\beta} = 60$ years,the effective mean life is $\frac{1}{\tau_{eff}} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$.
Thus,$\tau_{eff} = 20$ years.
The half-life $T_{1/2}$ is related to mean life by $T_{1/2} = \tau \ln(2) \approx 0.693 \tau$.
However,the problem asks for the time $t$ when the sample reduces to $1/4$ of its initial amount $N_0$.
Using the decay law $N(t) = N_0 e^{-\lambda_{eff} t}$,we set $N(t) = \frac{N_0}{4}$.
$\frac{N_0}{4} = N_0 e^{-\lambda_{eff} t} \Rightarrow e^{\lambda_{eff} t} = 4 = 2^2$.
Taking natural log on both sides: $\lambda_{eff} t = \ln(4) = 2 \ln(2)$.
Since $\lambda_{eff} = \frac{1}{\tau_{eff}}$,we have $t = \tau_{eff} \times 2 \ln(2) = 20 \times 2 \times 0.693 = 27.72$ years.
Wait,if the question implies half-lives are $30$ and $60$ years,then $T_{1/2, eff} = 20$ years. After $2$ half-lives,the amount is $(1/2)^2 = 1/4$. Thus,$t = 2 \times 20 = 40$ years.

(C) Let $N_0$ be the initial quantity kept the first time. Then the quantity kept the second time is $2N_0$.
Let $t_1$ be the time elapsed for the first sample and $t_2$ be the time elapsed for the second sample. The age difference is $\tau = t_1 - t_2$.
The activity is given by $A = \lambda N = \lambda N_{initial} e^{-\lambda t}$.
For the first sample: $A_1 = \lambda N_0 e^{-\lambda t_1}$.
For the second sample: $A_2 = \lambda (2N_0) e^{-\lambda t_2}$.
Dividing the two equations: $\frac{A_1}{A_2} = \frac{\lambda N_0 e^{-\lambda t_1}}{2 \lambda N_0 e^{-\lambda t_2}} = \frac{1}{2} e^{-\lambda (t_1 - t_2)} = \frac{1}{2} e^{-\lambda \tau}$.
Rearranging for $\tau$: $e^{\lambda \tau} = \frac{A_2}{2A_1}$.
Taking the natural logarithm on both sides: $\lambda \tau = \ln \left( \frac{A_2}{2A_1} \right)$.
Since $\lambda = \frac{\ln 2}{T}$,we have $\tau = \frac{T}{\ln 2} \ln \left( \frac{A_2}{2A_1} \right)$.

(B) The rate of disintegration is given by $R = \lambda N$. Given $R_A = R_B$,we have $\lambda_A N_A = \lambda_B N_B$.
Since $\lambda = \frac{\ln 2}{T_{1/2}}$,we have $\left( \frac{\ln 2}{T} \right) N_A = \left( \frac{\ln 2}{2T} \right) N_B$,which simplifies to $N_B = 2N_A$.
Using the decay law $N = N_0 e^{-\lambda t}$,we have $x e^{-\lambda_B t} = 2 x e^{-\lambda_A t}$.
Substituting $\lambda_A = \frac{\ln 2}{T}$ and $\lambda_B = \frac{\ln 2}{2T}$,we get $e^{-\frac{\ln 2}{2T} t} = 2 e^{-\frac{\ln 2}{T} t}$.
Taking the natural logarithm on both sides: $-\frac{\ln 2}{2T} t = \ln 2 - \frac{\ln 2}{T} t$.
Dividing by $\ln 2$: $-\frac{t}{2T} = 1 - \frac{t}{T} \Rightarrow \frac{t}{2T} = 1 \Rightarrow t = 2T$.
At $t = 2T$,$A$ has undergone $2$ half-lives,so $N_A = \frac{x}{2^2} = \frac{x}{4}$. Disintegrated $A = x - \frac{x}{4} = 0.75x$.
At $t = 2T$,$B$ has undergone $1$ half-life,so $N_B = \frac{x}{2^1} = \frac{x}{2}$. Disintegrated $B = x - \frac{x}{2} = 0.5x$.
The total number of nuclei of $C$ formed is the sum of disintegrated $A$ and $B$: $0.75x + 0.5x = 1.25x$.

(B) The fraction of the sample that decays is $7/8$. Therefore, the fraction remaining is $N/N_0 = 1 - 7/8 = 1/8$.
Using the radioactive decay law, $N/N_0 = (1/2)^n$, where $n$ is the number of half-lives.
$1/8 = (1/2)^n$
$(1/2)^3 = (1/2)^n$
So, $n = 3$.
The number of half-lives $n$ is given by $n = t / T_{1/2}$, where $t = 15 \ minutes$.
$3 = 15 / T_{1/2}$
$T_{1/2} = 15 / 3 = 5 \ minutes$.
The half-life of the sample is $5 \ minutes$.

(A) The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{\ln 2}{T_{1/2}}$.
For $\alpha$-emission,the decay constant is $\lambda_{\alpha} = \frac{\ln 2}{T}$.
For $\beta$-emission,the decay constant is $\lambda_{\beta} = \frac{\ln 2}{2T}$.
Since the decays occur simultaneously,the total decay constant $\lambda_{total}$ is the sum of the individual decay constants:
$\lambda_{total} = \lambda_{\alpha} + \lambda_{\beta} = \frac{\ln 2}{T} + \frac{\ln 2}{2T}$.
Taking $\frac{\ln 2}{T}$ as a common factor:
$\lambda_{total} = \frac{\ln 2}{T} (1 + \frac{1}{2}) = \frac{\ln 2}{T} (\frac{3}{2}) = \frac{3}{2} \frac{\ln 2}{T}$.

(C) The activity $A$ of a radioactive sample is defined as the rate of decay,given by $A = |\frac{dN}{dt}| = \lambda N$,where $N$ is the number of radioactive nuclei present.
The number of moles in the sample is $n = \frac{m}{M}$.
The total number of nuclei $N$ is given by $N = n \times N_A = \frac{m}{M} \times N_A$.
Substituting this into the activity formula,we get $A = \lambda \times \frac{m}{M} \times N_A = \frac{\lambda m N_A}{M}$.

(D) The graph is a straight line plotted between $\ln A$ and $t$. The equation of a straight line is $y = mx + C$.
Here,$y = \ln A$ and $x = t$.
The intercept $C$ on the $\ln A$ axis is $2.6$.
The slope $m$ is calculated as $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2.6}{26 - 0} = -\frac{2.6}{26} = -0.1$.
Thus,the equation of the line is $\ln A = -0.1t + 2.6$.
Taking the exponential of both sides,we get $A = e^{-0.1t + 2.6} = e^{2.6} \cdot e^{-0.1t}$.
Given $\ln 12 = 2.6$,it follows that $e^{2.6} = 12$.
Substituting this value,we get $A = 12 e^{-0.1t}$.

(B) Let the decay constants for the two modes be $\lambda_{\alpha}$ and $\lambda_{\beta}$ respectively.
Given that the probability of $\alpha$-decay is $66.6\%$ $(2/3)$ and $\beta$-decay is $33.3\%$ $(1/3)$,the ratio of decay constants is $\frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{2/3}{1/3} = 2$,so $\lambda_{\beta} = \frac{\lambda_{\alpha}}{2}$.
The effective decay constant $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
The effective half-life $T_{eff} = \frac{\ln 2}{\lambda_{eff}} = 60 \text{ years}$.
Substituting $\lambda_{\beta} = \frac{\lambda_{\alpha}}{2}$ into the equation: $\lambda_{eff} = \lambda_{\alpha} + \frac{\lambda_{\alpha}}{2} = \frac{3\lambda_{\alpha}}{2}$.
Thus,$\frac{\ln 2}{\frac{3\lambda_{\alpha}}{2}} = 60 \Rightarrow \frac{\ln 2}{\lambda_{\alpha}} = 60 \times \frac{3}{2} = 90 \text{ years}$.
The half-life for $\alpha$-decay only is $T_{\alpha} = \frac{\ln 2}{\lambda_{\alpha}} = 90 \text{ years}$.

(C) Let $N_0$ be the initial number of atoms of radioactive nuclide $X$. At any time $t$,the number of atoms of $X$ remaining is given by $N_X(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln(2)}{\tau}$.
Since $X$ decays into $Y$,the number of atoms of $Y$ at time $t$ is $N_Y(t) = N_0 - N_X(t) = N_0(1 - e^{-\lambda t})$.
At the intersection point $T$,$N_X(T) = N_Y(T)$.
Therefore,$N_0 e^{-\lambda T} = N_0(1 - e^{-\lambda T})$.
Dividing by $N_0$,we get $e^{-\lambda T} = 1 - e^{-\lambda T}$,which implies $2e^{-\lambda T} = 1$,or $e^{-\lambda T} = \frac{1}{2}$.
Taking the natural logarithm on both sides,$-\lambda T = \ln(1/2) = -\ln(2)$.
Thus,$\lambda T = \ln(2)$.
Substituting $\lambda = \frac{\ln(2)}{\tau}$,we get $\left(\frac{\ln(2)}{\tau}\right) T = \ln(2)$.
Solving for $T$,we find $T = \tau$.

(C) The rate of disintegration $R$ is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms remaining at time $t$.
Given $T_{1/2, A} = 2 \ hr$ and $T_{1/2, B} = 4 \ hr$. The decay constants are $\lambda_A = \frac{\ln 2}{2}$ and $\lambda_B = \frac{\ln 2}{4}$.
Initially,$N_A(0) = N_B(0) = N_0$.
At $t = 2 \ hr$,the number of atoms remaining is $N(t) = N_0 e^{-\lambda t} = N_0 (1/2)^{t/T_{1/2}}$.
For source $A$: $N_A(2) = N_0 (1/2)^{2/2} = N_0/2$.
For source $B$: $N_B(2) = N_0 (1/2)^{2/4} = N_0/\sqrt{2}$.
The rates of disintegration are $R_A = \lambda_A N_A$ and $R_B = \lambda_B N_B$.
Ratio $\frac{R_A}{R_B} = \frac{\lambda_A N_A}{\lambda_B N_B} = \frac{(\ln 2 / 2) \times (N_0 / 2)}{(\ln 2 / 4) \times (N_0 / \sqrt{2})} = \frac{1/4}{1/(4\sqrt{2})} = \sqrt{2}$.
Thus,the ratio is $\sqrt{2} : 1$.

(D) The activity of a radioactive sample at time $t$ is given by the formula: $A = A_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$.
Given: $A_0 = 64 \times 10^{-5} \text{ Ci}$,$A = 5 \times 10^{-6} \text{ Ci}$,and $T_{1/2} = 3 \text{ days}$.
Substituting the values: $5 \times 10^{-6} = 64 \times 10^{-5} \left( \frac{1}{2} \right)^{t / 3}$.
Dividing both sides by $64 \times 10^{-5}$: $\frac{5 \times 10^{-6}}{64 \times 10^{-5}} = \left( \frac{1}{2} \right)^{t / 3}$.
$\frac{0.5}{64} = \left( \frac{1}{2} \right)^{t / 3} \Rightarrow \frac{1}{128} = \left( \frac{1}{2} \right)^{t / 3}$.
Since $\frac{1}{128} = \left( \frac{1}{2} \right)^7$,we have $\left( \frac{1}{2} \right)^7 = \left( \frac{1}{2} \right)^{t / 3}$.
Equating the exponents: $7 = \frac{t}{3} \Rightarrow t = 21 \text{ days}$.

(C) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
For substance $A$: $T_{1/2, A} = 20 \text{ min}$,$t = 80 \text{ min}$. Number of half-lives $n_A = \frac{80}{20} = 4$.
Remaining nuclei $N_A = N_0 \left( \frac{1}{2} \right)^4 = \frac{N_0}{16}$.
For substance $B$: $T_{1/2, B} = 40 \text{ min}$,$t = 80 \text{ min}$. Number of half-lives $n_B = \frac{80}{40} = 2$.
Remaining nuclei $N_B = N_0 \left( \frac{1}{2} \right)^2 = \frac{N_0}{4}$.
The ratio of remaining nuclei is $\frac{N_A}{N_B} = \frac{N_0/16}{N_0/4} = \frac{4}{16} = \frac{1}{4}$.

(A) Let $N_0$ be the initial number of nuclei for both materials.
The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For $X_1$,$N_1 = N_0 e^{-6\lambda t}$.
For $X_2$,$N_2 = N_0 e^{-4\lambda t}$.
The ratio is given as $\frac{N_1}{N_2} = \frac{1}{e}$.
Substituting the expressions: $\frac{N_0 e^{-6\lambda t}}{N_0 e^{-4\lambda t}} = e^{-6\lambda t + 4\lambda t} = e^{-2\lambda t}$.
Equating to $\frac{1}{e} = e^{-1}$,we get $e^{-2\lambda t} = e^{-1}$.
Comparing the exponents: $-2\lambda t = -1$.
Therefore,$t = \frac{1}{2\lambda} \, s$.

(C) The number of active nuclei present in the original sample is $N_0 = 4 \times 10^{16}$.
The half-life of the element is $T_{1/2} = 10$ days.
The number of half-lives in $30$ days is $n = \frac{30}{10} = 3$.
The number of active nuclei remaining after $30$ days is $N = \frac{N_0}{2^n} = \frac{4 \times 10^{16}}{2^3} = \frac{4 \times 10^{16}}{8} = 0.5 \times 10^{16}$.
The number of decayed nuclei is given by $N_{decayed} = N_0 - N$.
$N_{decayed} = 4 \times 10^{16} - 0.5 \times 10^{16} = 3.5 \times 10^{16}$.

(B) The activity of a radioactive sample at time $t$ is given by $A(t) = A_0 e^{-\lambda t}$.
Given that at $t = 1 \text{ hour}$,$A(1) = \frac{A_0}{\sqrt{3}}$.
Substituting this into the formula: $\frac{A_0}{\sqrt{3}} = A_0 e^{-\lambda(1)}$,which implies $e^{-\lambda} = \frac{1}{\sqrt{3}}$.
We need to find the activity after $3$ hours more,which means at a total time $t = 1 + 3 = 4 \text{ hours}$.
$A(4) = A_0 e^{-\lambda(4)} = A_0 (e^{-\lambda})^4$.
Substituting $e^{-\lambda} = \frac{1}{\sqrt{3}}$ into the equation:
$A(4) = A_0 \left(\frac{1}{\sqrt{3}}\right)^4 = A_0 \left(\frac{1}{3^2}\right) = \frac{A_0}{9}$.

(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
At $t = 0$,$N(0) = N_0 = 1000$.
At $t = 2 \ s$,$N(2) = 1000 e^{-2\lambda} = 900$.
This implies $e^{-2\lambda} = \frac{900}{1000} = 0.9$.
At $t = 4 \ s$,the number of nuclei is $N(4) = N_0 e^{-4\lambda} = N_0 (e^{-2\lambda})^2$.
Substituting the values,we get $N(4) = 1000 \times (0.9)^2$.
$N(4) = 1000 \times 0.81 = 810$.
Therefore,the number of nuclei at $t = 4 \ s$ is $810$.

(B) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$,where $T$ is the half-life.
Given that the substance reduces to $25\%$ of its initial value,we have $N(t) = 0.25 N_0 = \frac{1}{4} N_0$.
Substituting this into the equation: $\frac{1}{4} N_0 = N_0 \left(\frac{1}{2}\right)^{16/T}$.
This simplifies to $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{16/T}$.
Equating the exponents,we get $2 = \frac{16}{T}$.
Solving for $T$,we find $T = \frac{16}{2} = 8\, days$.

(C) The activity of a radioactive sample follows the law $A = A_0 (1/2)^{t/T_{1/2}}$,where $A$ is the final activity,$A_0$ is the initial activity,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.
Given $A_0 = 16000 \ counts \ min^{-1}$,$A = 2000 \ counts \ min^{-1}$,and $t = 12 \ h$.
Substituting these values into the equation: $2000 = 16000 \times (1/2)^{12/T_{1/2}}$.
Dividing both sides by $16000$: $2000/16000 = (1/2)^{12/T_{1/2}}$.
$1/8 = (1/2)^{12/T_{1/2}}$.
Since $1/8 = (1/2)^3$,we have $(1/2)^3 = (1/2)^{12/T_{1/2}}$.
Equating the exponents: $3 = 12/T_{1/2}$.
Solving for $T_{1/2}$: $T_{1/2} = 12/3 = 4 \ h$.

(C) The number of nuclei remaining after time $t$ is given by $N = N_0 (1/2)^n$,where $n = t / T_{1/2}$ is the number of half-lives.
For substance $A$: $T_{1/2, A} = 20 \ min$. The number of half-lives in $80 \ min$ is $n_A = 80 / 20 = 4$.
Remaining nuclei $N_A = N_0 (1/2)^4 = N_0 / 16$.
For substance $B$: $T_{1/2, B} = 40 \ min$. The number of half-lives in $80 \ min$ is $n_B = 80 / 40 = 2$.
Remaining nuclei $N_B = N_0 (1/2)^2 = N_0 / 4$.
The ratio of remaining nuclei is $N_A / N_B = (N_0 / 16) / (N_0 / 4) = 4 / 16 = 1 / 4$.

(A) The activity $A$ of a radioactive sample is given by $A = |dN/dt| = \lambda N$.
Given that the activity decreases by $4\%$ in $1 \ h$,we have the fractional change in activity: $\frac{\Delta A}{A} = 0.04$ over $\Delta t = 1 \ h$.
Since $A = A_0 e^{-\lambda t}$,for small $\lambda t$,we can approximate $A = A_0(1 - \lambda t)$,which gives $\frac{A_0 - A}{A_0} = \lambda t$.
Substituting the given values: $0.04 = \lambda \times (1 \ h)$.
Thus,the decay constant $\lambda = 0.04 \ h^{-1}$.
The mean life time $T$ is the reciprocal of the decay constant: $T = \frac{1}{\lambda}$.
$T = \frac{1}{0.04} \ h = 25 \ h$.

(D) The activity of a radioactive sample at time $t$ is given by $A(t) = A_0 (1/2)^n$,where $n = t / T_{1/2}$ is the number of half-lives elapsed.
For nuclide $X$ at $t = 48$ years:
Number of half-lives $n_X = 48 / 24 = 2$.
Activity $A_X = A_0 (1/2)^2 = A_0 / 4$.
For nuclide $Y$ at $t = 48$ years:
Number of half-lives $n_Y = 48 / 16 = 3$.
Activity $A_Y = A_0 (1/2)^3 = A_0 / 8$.
The total activity of the mixture is the sum of the individual activities:
$A_{total} = A_X + A_Y = A_0 / 4 + A_0 / 8 = (2 A_0 + A_0) / 8 = 3 A_0 / 8$.

(B) Let $N_{0}$ be the initial number of radioactive nuclei.
After $33 \%$ decay,the number of undecayed nuclei remaining is $N_{1} = N_{0} - 0.33 N_{0} = 0.67 N_{0}$.
After $67 \%$ decay,the number of undecayed nuclei remaining is $N_{2} = N_{0} - 0.67 N_{0} = 0.33 N_{0}$.
We observe that $N_{2} \approx \frac{N_{1}}{2}$,since $0.33 N_{0} \approx \frac{0.67 N_{0}}{2}$.
By the definition of half-life,the time required for the number of undecayed nuclei to reduce to half of its initial value is exactly one half-life.
Therefore,the time taken for the substance to decay from $33 \%$ to $67 \%$ is equal to the half-life of the substance,which is $20 \, minutes$.

(A) The half-life of the sample is $T_{1/2} = 15$ years.
If $96.875\%$ of the sample decays,the remaining amount $N$ is $100\% - 96.875\% = 3.125\%$ of the initial amount $N_0$.
We can write $N = N_0 \times \frac{3.125}{100} = N_0 \times \frac{3125}{100000} = N_0 \times \frac{1}{32}$.
Since $\frac{1}{32} = (\frac{1}{2})^5$,the number of half-lives $n$ is $5$.
The total time taken is $t = n \times T_{1/2} = 5 \times 15 = 75$ years.

(D) Let $N_0$ be the initial number of nuclei for both materials.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For material $x_1$,$N_1(t) = N_0 e^{-(10\lambda)t}$.
For material $x_2$,$N_2(t) = N_0 e^{-\lambda t}$.
Given the ratio $\frac{N_1}{N_2} = \frac{1}{e}$,we have:
$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$
$e^{-10\lambda t + \lambda t} = e^{-1}$
$e^{-9\lambda t} = e^{-1}$
Comparing the exponents: $-9\lambda t = -1$
Therefore,$t = \frac{1}{9\lambda}$.

(D) Let $N_0$ be the initial number of radioactive nuclei.
At time $t_1$,the amount decayed is $\frac{1}{3}N_0$,so the remaining amount is $N(t_1) = N_0 - \frac{1}{3}N_0 = \frac{2}{3}N_0$.
At time $t_2$,the amount decayed is $\frac{2}{3}N_0$,so the remaining amount is $N(t_2) = N_0 - \frac{2}{3}N_0 = \frac{1}{3}N_0$.
We observe that $N(t_2) = \frac{1}{2} N(t_1)$.
Since the number of active nuclei is halved in the interval $(t_2 - t_1)$,this time interval must be equal to the half-life $T_{1/2}$.
Therefore,$t_2 - t_1 = T_{1/2} = 50 \, \text{days}$.

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.
The amount of stable element $Y$ formed is $N_0 - N$.
Given the ratio $\frac{N}{N_0 - N} = \frac{1}{15}$.
This implies $\frac{N_0 - N}{N} = 15$,so $\frac{N_0}{N} - 1 = 15$,which gives $\frac{N_0}{N} = 16$.
We know the radioactive decay law: $N = N_0 (\frac{1}{2})^{t/T_{1/2}}$,where $T_{1/2} = 50$ years.
Thus,$\frac{N_0}{N} = 2^{t/T_{1/2}} = 16$.
Since $16 = 2^4$,we have $\frac{t}{T_{1/2}} = 4$.
Therefore,$t = 4 \times 50 = 200$ years.

(A) The decay constant for $\alpha$-emission is $\lambda_1 = \frac{1}{\tau_1} = \frac{1}{30} \, \text{year}^{-1}$.
The decay constant for $\beta$-emission is $\lambda_2 = \frac{1}{\tau_2} = \frac{1}{60} \, \text{year}^{-1}$.
Since the sample decays by both processes simultaneously,the effective decay constant is $\lambda_{eff} = \lambda_1 + \lambda_2 = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20} \, \text{year}^{-1}$.
The effective mean life is $\tau_{eff} = \frac{1}{\lambda_{eff}} = 20 \, \text{years}$.
The half-life is $T_{1/2} = \tau_{eff} \ln(2) = 20 \times 0.693 = 13.86 \, \text{years}$.
For the sample to reduce to one-fourth of its initial amount,the time $t$ required is given by $N(t) = N_0 (1/2)^n$,where $n$ is the number of half-lives.
Since $(1/2)^2 = 1/4$,we need $n = 2$ half-lives.
Therefore,$t = 2 \times T_{1/2} = 2 \times 13.86 \approx 27.72 \, \text{years}$,which is approximately $28 \, \text{years}$.

(B) The number of radioactive 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 $N_0 = 1.414 \times 10^6$,$N = 10^6$,and $t = 10 \text{ min}$.
Substituting the values: $10^6 = 1.414 \times 10^6 \times \left( \frac{1}{2} \right)^n$.
$\frac{1}{1.414} = \left( \frac{1}{2} \right)^n$.
Since $1.414 \approx \sqrt{2}$,we have $\frac{1}{\sqrt{2}} = \left( \frac{1}{2} \right)^n$.
$(2)^{-1/2} = (2)^{-n}$.
Therefore,$n = 1/2$.
Since $n = \frac{t}{T_{1/2}}$,we have $1/2 = \frac{10}{T_{1/2}}$.
$T_{1/2} = 20 \text{ min}$.

(A) For the same radioactive substance,the decay constant $\lambda$ is identical for all lumps.
The activity $A$ of a radioactive sample is given by $A = \lambda N$,where $N$ is the number of radioactive nuclei present at time $t$.
Since $\lambda$ is constant,the activity is directly proportional to the number of nuclei: $A \propto N$.
At any time $t$,the number of nuclei in each lump is $N(t) = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei.
Therefore,the ratio of activities at any time $t$ is:
$A_1 : A_2 : A_3 = (\lambda N_{0,1} e^{-\lambda t}) : (\lambda N_{0,2} e^{-\lambda t}) : (\lambda N_{0,3} e^{-\lambda t})$
$A_1 : A_2 : A_3 = N_{0,1} : N_{0,2} : N_{0,3}$
Since the initial ratio of activities was $1 : 2 : 3$,the ratio of the initial number of nuclei $N_{0,1} : N_{0,2} : N_{0,3}$ must also be $1 : 2 : 3$.
Thus,the ratio of their activities at any future date remains $1 : 2 : 3$.

(B) The half-life of $X$ is $T_{1/2, X} = \frac{0.693}{\lambda_X}$.
The mean life of $Y$ is $\tau_Y = \frac{1}{\lambda_Y}$.
Given $T_{1/2, X} = \tau_Y$,we have $\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$,which implies $\frac{\lambda_X}{\lambda_Y} = 0.693$.
The decay rate $R$ is given by $R = \lambda N$.
Given initial number of atoms $N_X = N_Y = N_0$,the initial decay rates are $R_X = \lambda_X N_0$ and $R_Y = \lambda_Y N_0$.
Taking the ratio,$\frac{R_X}{R_Y} = \frac{\lambda_X}{\lambda_Y} = 0.693$.
Since $0.693 < 1$,it follows that $R_X < R_Y$,meaning $Y$ decays faster than $X$ initially.

(D) The activity of a radioactive sample is given by $R = N \lambda$,where $N$ is the number of radioactive nuclei and $\lambda$ is the decay constant.
The decay constant $\lambda$ is related to the half-life $T$ by $\lambda = \frac{\ln 2}{T}$.
At time $T_1$,the activity is $R_1 = N_1 \lambda$,so $N_1 = \frac{R_1}{\lambda} = \frac{R_1 T}{\ln 2}$.
At time $T_2$,the activity is $R_2 = N_2 \lambda$,so $N_2 = \frac{R_2}{\lambda} = \frac{R_2 T}{\ln 2}$.
The number of atoms that have disintegrated in the time interval $(T_2 - T_1)$ is the difference in the number of nuclei present at these times: $\Delta N = N_1 - N_2$.
Substituting the expressions for $N_1$ and $N_2$:
$\Delta N = \frac{R_1 T}{\ln 2} - \frac{R_2 T}{\ln 2} = \frac{T}{\ln 2} (R_1 - R_2)$.
Since $T$ and $\ln 2$ are constants,the number of disintegrated atoms is proportional to $(R_1 - R_2)$.

(B) The number of radioactive nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{1}{\tau_{mean}}$.
For the first species with mean-life $\tau_1 = \tau$,the decay constant is $\lambda_1 = \frac{1}{\tau}$.
The number of nuclei remaining at $t = 5\tau$ is $N_1 = N_0 e^{-\lambda_1 t} = N_0 e^{-(1/\tau)(5\tau)} = N_0 e^{-5} = \frac{N_0}{e^5}$.
For the second species with mean-life $\tau_2 = 5\tau$,the decay constant is $\lambda_2 = \frac{1}{5\tau}$.
The number of nuclei remaining at $t = 5\tau$ is $N_2 = N_0 e^{-\lambda_2 t} = N_0 e^{-(1/5\tau)(5\tau)} = N_0 e^{-1} = \frac{N_0}{e}$.
The total number of radioactive nuclei is $N_{total} = N_1 + N_2 = \frac{N_0}{e^5} + \frac{N_0}{e}$.
Taking $N_0$ as a common factor,$N_{total} = N_0 \left( \frac{1 + e^4}{e^5} \right)$.

(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.
Given at $t = 0$,$N = N_0$ and at $t = 5$ minutes,$N = N_0/e$.
Substituting these values: $\frac{N_0}{e} = N_0 e^{-\lambda(5)}$.
This simplifies to $e^{-1} = e^{-5\lambda}$,which gives $5\lambda = 1$,so $\lambda = 1/5 \text{ min}^{-1}$.
The half-life $T_{1/2}$ is the time when the activity reduces to half of its initial value,i.e.,$N = N_0/2$.
Using $N = N_0 e^{-\lambda t}$,we have $\frac{N_0}{2} = N_0 e^{-\lambda t} \Rightarrow 2 = e^{\lambda t}$.
Taking the natural logarithm on both sides: $\ln 2 = \lambda t$.
Substituting $\lambda = 1/5$: $\ln 2 = \frac{t}{5} \Rightarrow t = 5 \ln 2$ minutes.

(C) Let $N_0$ be the initial number of atoms of $X$. At any time $T$,the number of atoms of $X$ remaining is $N_X(T) = N_0 e^{-\lambda T}$.
Since $X$ decays to $Y$,the number of atoms of $Y$ formed at time $T$ is $N_Y(T) = N_0(1 - e^{-\lambda T})$.
At the intersection point,$N_X(T) = N_Y(T)$.
Therefore,$N_0 e^{-\lambda T} = N_0(1 - e^{-\lambda T})$.
$e^{-\lambda T} = 1 - e^{-\lambda T} \implies 2e^{-\lambda T} = 1 \implies e^{-\lambda T} = \frac{1}{2}$.
Taking the natural logarithm on both sides,$-\lambda T = \ln(1/2) = -\ln(2)$.
So,$T = \frac{\ln(2)}{\lambda}$.
Since the half-life $t = \frac{\ln(2)}{\lambda}$,we get $T = t$.

(A) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.
At $t = 0$, activity is $N_0$.
At $t = 5\, \text{minutes}$, activity is $N(5) = N_0/e$.
Substituting these values: $N_0/e = N_0 e^{-\lambda(5)}$.
This simplifies to $e^{-1} = e^{-5\lambda}$, which gives $5\lambda = 1$, so $\lambda = 1/5\, \text{min}^{-1}$.
The half-life $T_{1/2}$ is the time when activity becomes $N_0/2$.
$N_0/2 = N_0 e^{-\lambda T_{1/2}} \Rightarrow 1/2 = e^{-\lambda T_{1/2}}$.
Taking natural logarithm on both sides: $\ln(1/2) = -\lambda T_{1/2} \Rightarrow -\ln 2 = -\lambda T_{1/2}$.
$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\ln 2}{1/5} = 5 \ln 2\, \text{minutes}$.

(A) Given,half-life $T_{1/2} = 20 \ min$.
Let $N_0$ be the initial amount of the substance.
At $33\%$ decay,the remaining amount is $N_1 = N_0(1 - 0.33) = 0.67 N_0$.
At $67\%$ decay,the remaining amount is $N_2 = N_0(1 - 0.67) = 0.33 N_0$.
Using the radioactive decay law,$N = N_0(1/2)^{t/T_{1/2}}$,we have:
For $N_1$: $0.67 N_0 = N_0(1/2)^{t_1/20} \Rightarrow 0.67 = (1/2)^{t_1/20} \dots (1)$
For $N_2$: $0.33 N_0 = N_0(1/2)^{t_2/20} \Rightarrow 0.33 = (1/2)^{t_2/20} \dots (2)$
Dividing equation $(1)$ by $(2)$:
$\frac{0.67}{0.33} = \frac{(1/2)^{t_1/20}}{(1/2)^{t_2/20}} = (1/2)^{(t_1 - t_2)/20}$
Since $0.67/0.33 \approx 2$,we have $2^1 = 2^{(t_2 - t_1)/20}$.
Therefore,$(t_2 - t_1)/20 = 1$,which gives $t_2 - t_1 = 20 \ min$.

(A) The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$.
For substance $A$: $N_A = 10 \left( \frac{1}{2} \right)^{t/1}$.
For substance $B$: $N_B = 1 \left( \frac{1}{2} \right)^{t/2}$.
Given that the remaining quantities are equal, $N_A = N_B$.
$10 \left( \frac{1}{2} \right)^t = 1 \left( \frac{1}{2} \right)^{t/2}$.
$10 = \frac{(1/2)^{t/2}}{(1/2)^t} = (1/2)^{-t/2} = 2^{t/2}$.
Taking the logarithm on both sides: $\log_{10}(10) = \frac{t}{2} \log_{10}(2)$.
$1 = \frac{t}{2} \times 0.3010$.
$t = \frac{2}{0.3010} \approx 6.64 \, years$. (Rounding to the nearest provided option, $6.62$).

(C) The number of days from $January \, 1^{st}$ to $January \, 24^{th}$ is $23 \, days$.
The number of half-lives $n$ is calculated as $n = \frac{23}{8.04} \approx 2.86$.
Since $n < 3$,the sample has not yet undergone three full half-lives.
After $3$ half-lives,the activity would be $A = A_0 \times (1/2)^3 = 600 \times (1/8) = 75 \, Bq$.
Because the elapsed time $(23 \, days)$ is less than $3$ half-lives $(3 \times 8.04 = 24.12 \, days)$,the remaining activity must be greater than $75 \, Bq$.

(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
At $t = 0$,$N_0 = 1000$.
At $t = 2 \, s$,$N(2) = 900$.
So,$900 = 1000 e^{-\lambda(2)}$,which gives $e^{-2\lambda} = 0.9$.
At $t = 4 \, s$,the number of nuclei is $N(4) = N_0 e^{-\lambda(4)} = N_0 (e^{-2\lambda})^2$.
Substituting the values: $N(4) = 1000 \times (0.9)^2 = 1000 \times 0.81 = 810$.
Therefore,the number of nuclei at $t = 4 \, s$ is $810$.

(B) Given: Half-life $T_{1/2} = 20 \ min$. The decay constant $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.
Let $N_0$ be the initial amount of the substance.
At $20\%$ decay,the remaining amount is $N_1 = N_0 - 0.20 N_0 = 0.80 N_0$. The time taken is $t_1$.
Using the decay law $N_1 = N_0 e^{-\lambda t_1}$,we have $0.80 N_0 = N_0 e^{-\lambda t_1}$,so $e^{-\lambda t_1} = 0.8$.
At $80\%$ decay,the remaining amount is $N_2 = N_0 - 0.80 N_0 = 0.20 N_0$. The time taken is $t_2$.
Using the decay law $N_2 = N_0 e^{-\lambda t_2}$,we have $0.20 N_0 = N_0 e^{-\lambda t_2}$,so $e^{-\lambda t_2} = 0.2$.
Dividing the two equations: $\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = \frac{0.8}{0.2} = 4$.
$e^{\lambda(t_2 - t_1)} = 4$.
Taking natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln 4 = 2 \ln 2$.
Substituting $\lambda = \frac{\ln 2}{20}$:
$\frac{\ln 2}{20} (t_2 - t_1) = 2 \ln 2$.
$t_2 - t_1 = 2 \times 20 = 40 \ min$.

(D) Initial activity $A_0 = 0.8\,\mu Ci = 0.8 \times 3.7 \times 10^4\,dps = 29600\,dps$.
Activity at time $t = 10\,hrs$ is given by $A_t = A_0 e^{-\lambda t}$.
Substituting the given values: $A_t = 29600 \times 0.84 = 24864\,dps$.
Activity in $1\,cm^3$ of blood at $t = 10\,hrs$ is $n = 300\,decays/min = 300/60 = 5\,dps$.
Let $V$ be the total volume of blood in $cm^3$. The total activity $A_t$ is distributed in volume $V$,so $A_t = V \times n$.
$V = A_t / n = 24864 / 5 = 4972.8\,cm^3$.
Since $1000\,cm^3 = 1\,litre$,$V = 4972.8 / 1000 \approx 5\,litres$.

(D) Given: $N_1 = 2N_2$.
Activity of a radioactive substance is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.
Half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,which implies $\lambda = \frac{\ln 2}{T_{1/2}}$.
For sample $S_1$: $A_1 = \lambda_1 N_1 = \frac{\ln 2}{T_1} N_1 = 5\,\mu Ci$ ...... $(i)$
For sample $S_2$: $A_2 = \lambda_2 N_2 = \frac{\ln 2}{T_2} N_2 = 10\,\mu Ci$ ...... $(ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{A_2}{A_1} = \frac{\lambda_2 N_2}{\lambda_1 N_1} = \frac{T_1}{T_2} \times \frac{N_2}{N_1} = \frac{10}{5} = 2$
Since $N_1 = 2N_2$,we have $\frac{N_2}{N_1} = \frac{1}{2}$.
Substituting this into the ratio:
$\frac{T_1}{T_2} \times \frac{1}{2} = 2 \Rightarrow \frac{T_1}{T_2} = 4 \Rightarrow T_1 = 4T_2$.
This means the half-life of $S_1$ is four times that of $S_2$. Looking at the options,if $T_2 = 5$ years,then $T_1 = 20$ years. Thus,the half-lives are $20$ years and $5$ years,respectively.

(B) The initial number of nuclei $N_0$ is given by $N_0 = \frac{m}{M} \times N_A = \frac{1}{24} \times 6.023 \times 10^{23} \approx 2.51 \times 10^{22}$.
The number of nuclei decayed in time $t$ is $N_{\beta} = N_0(1 - e^{-\lambda t})$.
Given $t = 7.5 \, hrs$ and $T_{1/2} = 15 \, hrs$,the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$.
Thus,$N_{\beta} = N_0(1 - e^{-\frac{\ln 2}{15} \times 7.5}) = N_0(1 - e^{-\frac{\ln 2}{2}}) = N_0(1 - 2^{-1/2})$.
Using $2^{-1/2} \approx 0.707$,we get $N_{\beta} = 2.51 \times 10^{22} \times (1 - 0.707) = 2.51 \times 10^{22} \times 0.293 \approx 7.35 \times 10^{21}$.
Rounding to the nearest option,$N_{\beta} \approx 7.5 \times 10^{21}$.

(B) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate:
$\frac{dN}{dt} = P - \lambda N$
Given $P = 100$ and $\lambda = 0.5$,we have $\frac{dN}{dt} = 100 - 0.5N$.
Rearranging and integrating from $t=0$ $(N=0)$ to $t$ $(N=50)$:
$\int_0^N \frac{dN}{100 - 0.5N} = \int_0^t dt$
$-\frac{1}{0.5} [\ln(100 - 0.5N)]_0^N = t$
$-2 [\ln(100 - 0.5N) - \ln(100)] = t$
$\ln \left( \frac{100 - 0.5N}{100} \right) = -0.5t$
$1 - \frac{0.5N}{100} = e^{-0.5t}$
$N = \frac{100}{0.5} (1 - e^{-0.5t}) = 200(1 - e^{-0.5t})$.
Setting $N = 50$:
$50 = 200(1 - e^{-0.5t})$
$0.25 = 1 - e^{-0.5t}$
$e^{-0.5t} = 0.75 = \frac{3}{4}$
$-0.5t = \ln(3/4) = -\ln(4/3)$
$t = \frac{\ln(4/3)}{0.5} = 2 \ln \left( \frac{4}{3} \right) s$.

(B) Given,for $^{14}C$:
Initial activity $A_{0} = 16$ disintegrations $\text{min}^{-1} \text{g}^{-1}$.
Final activity $A = 12$ disintegrations $\text{min}^{-1} \text{g}^{-1}$.
Half-life $t_{1/2} = 5760$ years.
The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \text{ year}^{-1}$.
Using the radioactive decay formula $A = A_{0} e^{-\lambda t}$,we have $\frac{A_{0}}{A} = e^{\lambda t}$.
Taking the natural logarithm on both sides: $\ln\left(\frac{A_{0}}{A}\right) = \lambda t$.
Therefore,$t = \frac{1}{\lambda} \ln\left(\frac{A_{0}}{A}\right) = \frac{t_{1/2}}{0.693} \times 2.303 \log_{10}\left(\frac{A_{0}}{A}\right)$.
Substituting the values: $t = \frac{5760}{0.693} \times 2.303 \times \log_{10}\left(\frac{16}{12}\right)$.
$t = \frac{5760 \times 2.303}{0.693} \times \log_{10}(1.333)$.
$t \approx 19142.8 \times 0.1249 \approx 2391$ years.

(C) Given: Initial activity $A_0 = 20$ decays/min.
Final activity $A = 2$ decays/min.
Half-life $T_{1/2} = 5730$ years.
We know that the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$,where $\lambda = \frac{0.693}{T_{1/2}}$.
Rearranging for $t$,we get $t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A}\right) = \frac{T_{1/2}}{0.693} \times 2.303 \times \log_{10}\left(\frac{A_0}{A}\right)$.
Substituting the values: $t = \frac{5730}{0.693} \times 2.303 \times \log_{10}\left(\frac{20}{2}\right)$.
Since $\log_{10}(10) = 1$,we have $t = \frac{5730 \times 2.303}{0.693} \approx 5730 \times 3.322$.
$t \approx 19039$ years.

(D) Given: $(T_{1/2})_A = (\tau)_B$, where $\tau$ is the mean life.
We know that $T_{1/2} = \frac{0.693}{\lambda}$ and $\tau = \frac{1}{\lambda}$.
Therefore, $\frac{0.693}{\lambda_A} = \frac{1}{\lambda_B}$.
This implies $\lambda_A = 0.693 \lambda_B$, which means $\lambda_A < \lambda_B$.
The rate of decay is given by $R = \lambda N$.
Initially, the number of atoms $N$ is the same for both elements.
Since $\lambda_B > \lambda_A$, the decay rate $R_B = \lambda_B N$ will be greater than $R_A = \lambda_A N$.
Thus, $B$ will decay at a faster rate than $A$.