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

(B) Let $N_0 = 10^{20}$ be the initial number of radioactive atoms.
Let $\lambda$ be the decay constant.
The number of atoms remaining after time $t$ is $N(t) = N_0 e^{-\lambda t}$.
The number of $\alpha$-particles emitted in the $n$-th year is the difference in the number of atoms at the start and end of that year: $\Delta N_n = N(n-1) - N(n) = N_0 e^{-\lambda(n-1)} - N_0 e^{-\lambda n} = N_0 e^{-\lambda(n-1)}(1 - e^{-\lambda})$.
The ratio of particles emitted in the third year to the second year is given by:
$\frac{\Delta N_3}{\Delta N_2} = \frac{N_0 e^{-2\lambda}(1 - e^{-\lambda})}{N_0 e^{-\lambda}(1 - e^{-\lambda})} = e^{-\lambda} = 0.3$.
The number of $\alpha$-particles emitted in the first year is $\Delta N_1 = N(0) - N(1) = N_0(1 - e^{-\lambda})$.
Substituting the values: $\Delta N_1 = 10^{20}(1 - 0.3) = 10^{20}(0.7) = 7 \times 10^{19}$.

(C) When a radioactive substance decays through multiple processes simultaneously,the total decay constant is the sum of the individual decay constants.
Thus,the effective decay constant is $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
The half-life $T$ is related to the decay constant $\lambda$ by $T = \frac{\ln 2}{\lambda}$,which implies $\lambda = \frac{\ln 2}{T}$.
Substituting this into the effective decay constant equation:
$\frac{\ln 2}{T_{eff}} = \frac{\ln 2}{T_{\alpha}} + \frac{\ln 2}{T_{\beta}}$.
Dividing by $\ln 2$ on both sides,we get $\frac{1}{T_{eff}} = \frac{1}{T_{\alpha}} + \frac{1}{T_{\beta}}$.
Solving for $T_{eff}$,we obtain $T_{eff} = \frac{T_{\alpha}T_{\beta}}{T_{\alpha} + T_{\beta}}$.

(D) The radioactive decay law states that the activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$,where $n = t/T_{1/2}$ is the number of half-lives.
Given $A_0 = 1600 \ counts \ s^{-1}$ at $t = 0$ and $A = 100 \ counts \ s^{-1}$ at $t = 8 \ s$.
Substituting these values: $100 = 1600 \times (1/2)^{8/T_{1/2}}$.
$1/16 = (1/2)^{8/T_{1/2}} \Rightarrow (1/2)^4 = (1/2)^{8/T_{1/2}}$.
Equating the exponents: $4 = 8/T_{1/2} \Rightarrow T_{1/2} = 2 \ s$.
Now,for $t = 6 \ s$,the number of half-lives $n = 6/2 = 3$.
The activity at $t = 6 \ s$ is $A = A_0 \times (1/2)^n = 1600 \times (1/2)^3$.
$A = 1600 \times (1/8) = 200 \ counts \ s^{-1}$.

(C) The activity of a radioactive sample is given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.
For sample $A$: $R_A = \lambda_A N_A = 10\, mCi$ ... $(1)$
For sample $B$: $R_B = \lambda_B N_B = 20\, mCi$ ... $(2)$
Given $N_A = 2 N_B$ ... $(3)$
Dividing $(1)$ by $(2)$: $\frac{\lambda_A N_A}{\lambda_B N_B} = \frac{10}{20} = \frac{1}{2}$.
Substituting $N_A = 2 N_B$: $\frac{\lambda_A (2 N_B)}{\lambda_B N_B} = \frac{1}{2} \implies 2 \frac{\lambda_A}{\lambda_B} = \frac{1}{2} \implies \frac{\lambda_A}{\lambda_B} = \frac{1}{4}$.
Since half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,we have $\frac{T_{1/2, A}}{T_{1/2, B}} = \frac{\lambda_B}{\lambda_A} = 4$.
This means $T_{1/2, A} = 4 \times T_{1/2, B}$.
Checking the options,for option $B$: $T_{1/2, A} = 10$ days and $T_{1/2, B} = 2.5$ days (not listed). Wait,checking the ratio $T_{1/2, A} / T_{1/2, B} = 4$:
Option $B$: $10/40 = 1/4$ (Incorrect).
Option $C$: $20/5 = 4$ (Correct).
Thus,the correct choice is $C$.

(C) The activity of a radioactive substance at time $t$ is given by $R = R_0 e^{-\lambda t}$.
Given that at $t = 0$,$R_{A,0} = R_{B,0} = R_0$.
The ratio of activities at time $t$ is $\frac{R_B}{R_A} = \frac{R_0 e^{-\lambda_B t}}{R_0 e^{-\lambda_A t}} = e^{-(\lambda_B - \lambda_A)t}$.
According to the problem,this ratio is $e^{-3t}$.
Comparing the exponents,we get $\lambda_B - \lambda_A = 3$.
The decay constant $\lambda$ is related to half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$.
Given $T_{A,1/2} = \ln 2$,so $\lambda_A = \frac{\ln 2}{\ln 2} = 1$.
Substituting $\lambda_A$ into the equation: $\lambda_B - 1 = 3 \Rightarrow \lambda_B = 4$.
Since $\lambda_B = \frac{\ln 2}{T_{B,1/2}}$,we have $4 = \frac{\ln 2}{T_{B,1/2}}$.
Therefore,$T_{B,1/2} = \frac{\ln 2}{4}$.

(A) The radioactive decay follows the law $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life.
Given $N(0) = 1600$ and $N(8) = 100$.
$100 = 1600 \left( \frac{1}{2} \right)^{8/T_{1/2}}$
$\frac{1}{16} = \left( \frac{1}{2} \right)^{8/T_{1/2}}$
$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{8/T_{1/2}}$
$4 = \frac{8}{T_{1/2}} \implies T_{1/2} = 2 \, s$.
Now, at $t = 6 \, s$, the number of half-lives elapsed is $n = \frac{6}{2} = 3$.
The count rate is $N(6) = 1600 \times \left( \frac{1}{2} \right)^3 = \frac{1600}{8} = 200 \, \text{counts per second}$.

(C) Let $N_0$ be the initial number of nuclei for both materials.
For material $A$,the number of nuclei at time $t$ is $N_A = N_0 e^{-10\lambda t}$.
For material $B$,the number of nuclei at time $t$ is $N_B = N_0 e^{-\lambda t}$.
We are given that the ratio $\frac{N_A}{N_B} = \frac{1}{e}$.
Substituting the expressions,we get $\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.
This simplifies to $e^{-10\lambda t + \lambda t} = e^{-1}$,which means $e^{-9\lambda t} = e^{-1}$.
Equating the exponents,we have $-9\lambda t = -1$.
Therefore,$t = \frac{1}{9\lambda}$.

(D) Let $N_0$ be the initial number of nuclei for both substances $A$ and $B$.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For substance $A$: $N_A(t) = N_0 e^{-5\lambda t}$.
For substance $B$: $N_B(t) = N_0 e^{-\lambda t}$.
The ratio of the number of nuclei is $\frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-4\lambda t}$.
We are given that this ratio is $(1/e)^2 = e^{-2}$.
So,$e^{-4\lambda t} = e^{-2}$.
Equating the exponents: $-4\lambda t = -2$.
Solving for $t$: $t = \frac{-2}{-4\lambda} = \frac{1}{2\lambda}$.

(A) Let the initial number of nuclei be $N_0$ for both $A$ and $B$.
For nucleus $A$,half-life $T_{1/2, A} = 10 \, min$. After $t = 60 \, min$,the number of half-lives $n_A = 60/10 = 6$.
The number of undecayed nuclei $N_A = N_0 / 2^6 = N_0 / 64$.
The number of decayed nuclei $D_A = N_0 - N_A = N_0(1 - 1/64) = 63N_0 / 64$.
For nucleus $B$,half-life $T_{1/2, B} = 20 \, min$. After $t = 60 \, min$,the number of half-lives $n_B = 60/20 = 3$.
The number of undecayed nuclei $N_B = N_0 / 2^3 = N_0 / 8$.
The number of decayed nuclei $D_B = N_0 - N_B = N_0(1 - 1/8) = 7N_0 / 8$.
The ratio of decayed nuclei $D_A / D_B = (63N_0 / 64) / (7N_0 / 8) = (63/64) \times (8/7) = 9/8$.

(D) The activity $R$ of a radioactive sample is given by $R = N \lambda$,where $N$ is the number of radioactive nuclei present 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 radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$ or $N(t) = N_0 (1/2)^{t/T_{1/2}}$.
Given that $10\%$ decays in $5\, days$,the amount remaining is $90\%$ of the initial amount $N_0$.
So,$\frac{N(5)}{N_0} = 0.90$.
We can write $0.90 = (1/2)^{5/T_{1/2}}$.
We need to find the amount remaining after $20\, days$,which is $N(20) = N_0 (1/2)^{20/T_{1/2}}$.
This can be rewritten as $N(20) = N_0 [(1/2)^{5/T_{1/2}}]^4$.
Substituting the value of $(1/2)^{5/T_{1/2}} = 0.90$,we get:
$N(20) = N_0 (0.90)^4$.
Calculating $(0.90)^4 = 0.9 \times 0.9 \times 0.9 \times 0.9 = 0.6561$.
Therefore,the percentage remaining is $0.6561 \times 100\% = 65.61\%$.
Rounding to the nearest integer,the value is approximately $65\%$.

(B) Let $N_0$ be the initial amount of the radioactive substance.
The amount remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.
At time $t_1$,$\frac{1}{4}$ has decayed,so the remaining amount is $N(t_1) = N_0 - \frac{1}{4}N_0 = \frac{3}{4}N_0$.
At time $t_2$,$\frac{3}{4}$ has decayed,so the remaining amount is $N(t_2) = N_0 - \frac{3}{4}N_0 = \frac{1}{4}N_0$.
Using the decay law: $\frac{N(t_2)}{N(t_1)} = \frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} = e^{-\lambda(t_2 - t_1)}$.
Substituting the values: $\frac{1/4 N_0}{3/4 N_0} = e^{-\lambda(t_2 - t_1)} \Rightarrow \frac{1}{3} = e^{-\lambda(t_2 - t_1)}$.
Taking the natural logarithm on both sides: $\ln(1/3) = -\lambda(t_2 - t_1) \Rightarrow \ln 3 = \lambda(t_2 - t_1)$.
Substituting $\lambda = \frac{\ln 2}{20}$: $\ln 3 = \frac{\ln 2}{20} (t_2 - t_1)$.
Therefore,$t_2 - t_1 = 20 \frac{\ln 3}{\ln 2} \, \text{min}$.

(D) The formula for radioactive decay is given by $N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$.
Given:
Initial number of atoms $N_0 = 6.4 \times 10^{10}$.
Half-life $T_{1/2} = 4 \text{ days}$.
Total time $t = 12 \text{ days}$.
Substituting the values into the formula:
$N = 6.4 \times 10^{10} \times \left(\frac{1}{2}\right)^{\frac{12}{4}}$
$N = 6.4 \times 10^{10} \times \left(\frac{1}{2}\right)^3$
$N = 6.4 \times 10^{10} \times \frac{1}{8}$
$N = 0.8 \times 10^{10}$ atoms.

(B) The activity of a radioactive substance at time $t$ is given by $A = A_0 e^{-\lambda t}$, where $A_0$ is the initial activity.
Let $A_1 = 6000\, \text{dps}$ at $t_1 = 280\, \text{days}$.
Let $A_2 = 3000\, \text{dps}$ at $t_2 = 280 + 140 = 420\, \text{days}$.
Since the activity drops from $6000\, \text{dps}$ to $3000\, \text{dps}$ in $140\, \text{days}$, the half-life $T_{1/2}$ is $140\, \text{days}$.
Using the relation $A = A_0 (1/2)^{t/T_{1/2}}$, we find the initial activity $A_0$ at $t = 0$ using the state at $t_1 = 280\, \text{days}$:
$6000 = A_0 (1/2)^{280/140}$
$6000 = A_0 (1/2)^2$
$6000 = A_0 / 4$
$A_0 = 6000 \times 4 = 24000\, \text{dps}$.

(C) Let $N_0$ be the initial number of nuclei for both $X$ and $Y$.
Given half-lives: $T_{1/2, X} = 1\, h$ and $T_{1/2, Y} = 2\, h$.
After time $t = 2\, h$,the number of nuclei remaining is given by $N = N_0 (1/2)^{t/T_{1/2}}$.
For $X$: $N_X = N_0 (1/2)^{2/1} = N_0 / 4$.
For $Y$: $N_Y = N_0 (1/2)^{2/2} = N_0 / 2$.
Activity $A = \lambda N$,where $\lambda = \ln(2) / T_{1/2}$.
Ratio of activities: $\frac{A_X}{A_Y} = \frac{\lambda_X N_X}{\lambda_Y N_Y} = \frac{(\ln 2 / T_{1/2, X}) N_X}{(\ln 2 / T_{1/2, Y}) N_Y} = \frac{T_{1/2, Y}}{T_{1/2, X}} \times \frac{N_X}{N_Y}$.
Substituting the values: $\frac{A_X}{A_Y} = \frac{2}{1} \times \frac{N_0/4}{N_0/2} = 2 \times \frac{1}{2} = 1$.
Thus,the ratio is $1 : 1$.

(D) Radioactive decay is a nuclear process that depends solely on the nature of the nucleus itself.
It is independent of external physical factors such as temperature,pressure,or chemical environment.
Since the rate of disintegration is determined by the decay constant $\lambda$ and the number of radioactive nuclei $N$ (given by $R = \lambda N$),and $\lambda$ is a constant characteristic of the specific radioactive isotope,it cannot be altered by external conditions.
Therefore,it is not possible to increase the rate of disintegration of a fixed quantity of a radioactive element by any external means.

(D) Let $t$ be the time after which the amounts of $A_1$ and $A_2$ become equal.
Using the radioactive decay law,the amount $N$ at time $t$ is given by $N = N_0 (1/2)^{t/T_{1/2}}$.
For $A_1$: $N_1 = 40 \times (1/2)^{t/20}$.
For $A_2$: $N_2 = 160 \times (1/2)^{t/10}$.
Setting $N_1 = N_2$,we get:
$40 \times (1/2)^{t/20} = 160 \times (1/2)^{t/10}$.
Dividing both sides by $40$:
$(1/2)^{t/20} = 4 \times (1/2)^{t/10}$.
Rearranging the terms:
$(1/2)^{t/20} / (1/2)^{t/10} = 4$.
$(1/2)^{(t/20 - t/10)} = 4$.
$(1/2)^{-t/20} = 4$.
$2^{t/20} = 2^2$.
Comparing the exponents,$t/20 = 2$,which gives $t = 40 \, s$.

(B) The activity $A(t)$ of a radioactive sample at time $t$ is given by the law of radioactive decay:
$A(t) = -\frac{dN}{dt} = \lambda N = \lambda N_0 e^{-\lambda t}$
where $\lambda$ is the decay constant and $N_0$ is the initial number of radioactive nuclei.
This equation represents an exponential decay function.
As time $t$ increases,the activity $A(t)$ decreases exponentially from its initial value $\lambda N_0$ towards zero.
Looking at the provided graph,curve $B$ represents an exponential decay,which matches the mathematical relationship between activity and time.
Therefore,the correct curve is $B$.

(A) Initial activity $R_0 = 1 \, \mu Ci = 10^{-6} \times 3.7 \times 10^{10} \, dps = 3.7 \times 10^4 \, dps$.
Activity of $1 \, cm^3$ sample at $t = 5 \, hours$ is $r = \frac{296}{60} \, dps \approx 4.933 \, dps$.
The activity of the total blood volume $V$ at time $t$ is given by $R = R_0 e^{-\lambda t}$.
Since the radioactive substance is uniformly mixed,the activity per unit volume is $r = \frac{R}{V}$.
Therefore,$V = \frac{R}{r} = \frac{R_0 e^{-\lambda t}}{r}$.
Substituting the values: $V = \frac{3.7 \times 10^4 \times 0.7927}{4.933} \approx 5945 \, cm^3$.
Since $1000 \, cm^3 = 1 \, L$,the total volume $V \approx 5.94 \, L$.

(D) The activity $A$ at time $t$ is given by $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.
Given that the activity is $3\%$ of the original, we have $0.03 = (1/2)^n$.
Taking the logarithm, $\ln(0.03) = n \ln(0.5)$, which gives $n = \ln(0.03) / \ln(0.5) \approx 5.06$.
Thus, the time elapsed for the tritium decay is $t = n \times T_{1/2} = 5.06 \times 12.5 \text{ years} \approx 63.25 \text{ years}$.
Since the reference bottle was already $7 \text{ years}$ old, the total age of the sample is $63.25 + 7 \approx 70.25 \text{ years}$.
Therefore, the sample was prepared about $70 \text{ years}$ back.

(B) The half-life $(T_{1/2})$ of a radioactive substance is defined as the time required for half of the radioactive nuclei to decay. It is given by the formula: $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\log_e 2}{\lambda}$.
The mean life or average life ($T_{av}$ or $\tau$) is defined as the sum of the lives of all atoms divided by the total number of atoms. It is given by the formula: $T_{av} = \frac{1}{\lambda}$.
Therefore,the half-life and mean life are $\frac{\log_e 2}{\lambda}$ and $\frac{1}{\lambda}$ respectively.

(D) The formula for the remaining fraction of a radioactive substance is given by $\frac{N}{N_{0}} = \left(\frac{1}{2}\right)^{n}$,where $n$ is the number of half-lives.
First,calculate the number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6400}{800} = 8$.
Now,substitute the value of $n$ into the formula:
$\frac{N}{N_{0}} = \left(\frac{1}{2}\right)^{8} = \frac{1}{256}$.
Therefore,the fraction remaining after $6400$ years is $\frac{1}{256}$.

(A) The number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.
Given initial number of nuclei $N_0$ is the same for both materials.
Let $N_1$ and $N_2$ be the number of nuclei of $X_1$ and $X_2$ at time $t$ respectively.
$N_1 = N_0 e^{-5\lambda t}$ and $N_2 = N_0 e^{-\lambda t}$.
The ratio is $\frac{N_1}{N_2} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t}$.
We are given that $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Equating the exponents: $-4\lambda t = -1$.
Therefore,$t = \frac{1}{4\lambda}$.

(B) Let $N_0$ be the initial number of radioactive nuclei.
The fraction of nuclei that decay in time $t$ is $x = \frac{N_0 - N_t}{N_0} = 1 - \frac{N_t}{N_0}$.
From this,the fraction remaining after time $t$ is $\frac{N_t}{N_0} = 1 - x$.
According to the law of radioactive decay,the number of nuclei remaining after time $2t$ is $N_{2t} = N_0 \left( \frac{N_t}{N_0} \right)^2$.
Substituting the value of $\frac{N_t}{N_0}$,we get $N_{2t} = N_0 (1 - x)^2$.
The fraction of nuclei that decay in time $2t$ is given by $\frac{N_0 - N_{2t}}{N_0} = 1 - \frac{N_{2t}}{N_0}$.
Substituting $N_{2t}$,we get $1 - (1 - x)^2 = 1 - (1 - 2x + x^2) = 1 - 1 + 2x - x^2 = 2x - x^2$.

(B) The number of atoms $N$ in a mass $m$ is given by $N = \frac{m}{M_w} \times N_A$.
Radioactivity $R$ is defined as the rate of decay,given by $R = \lambda N$.
Substituting the value of $N$,we get $R = \lambda \left( \frac{m}{M_w} \times N_A \right) = \left( \frac{N_A}{M_w} m \right) \lambda$.

(B) The activity of a radioactive sample follows the law $A = A_0 \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
Given,initial activity $A_0 = 8 \text{ counts}$ and final activity $A = 1 \text{ count}$.
Substituting the values: $1 = 8 \left( \frac{1}{2} \right)^n$.
This simplifies to $\left( \frac{1}{2} \right)^n = \frac{1}{8}$.
Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$,we have $n = 3$.
The total time $t = 3 \text{ hours}$.
The number of half-lives is given by $n = \frac{t}{T_{1/2}}$,where $T_{1/2}$ is the half-life.
Therefore,$3 = \frac{3}{T_{1/2}}$,which gives $T_{1/2} = 1 \text{ hour}$.

(B) Let $N_0$ be the initial number of nuclei. The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
At $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}{5} = \frac{4}{5}$. Thus,$N(t_1) = N_0 \frac{4}{5} = N_0 e^{-\lambda t_1}$.
At $t_2$,the decay fraction is $\frac{4}{5}$,so the remaining fraction is $1 - \frac{4}{5} = \frac{1}{5}$. Thus,$N(t_2) = N_0 \frac{1}{5} = N_0 e^{-\lambda t_2}$.
Dividing the two equations: $\frac{N(t_1)}{N(t_2)} = \frac{4/5}{1/5} = 4 = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = e^{\lambda(t_2 - t_1)}$.
Taking the natural logarithm on both sides: $\ln 4 = \lambda(t_2 - t_1)$.
Since the mean life $\tau = \frac{1}{\lambda}$,we have $\lambda = \frac{1}{\tau}$.
Substituting this into the equation: $\ln 4 = \frac{t_2 - t_1}{\tau}$.
Therefore,$\tau = \frac{t_2 - t_1}{\ln 4}$.

(B) The half-life of the substance is $T_{1/2} = 20 \, min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.
Let $N_0$ be the initial amount of the substance.
For $20\%$ decay,the amount remaining is $N_1 = N_0 - 0.20 N_0 = 0.80 N_0$. The time taken is $t_1$,where $0.80 N_0 = N_0 e^{-\lambda t_1}$.
For $80\%$ decay,the amount remaining is $N_2 = N_0 - 0.80 N_0 = 0.20 N_0$. The time taken is $t_2$,where $0.20 N_0 = N_0 e^{-\lambda t_2}$.
Dividing the two equations:
$\frac{0.80 N_0}{0.20 N_0} = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}$
$4 = e^{\lambda(t_2 - t_1)}$
Taking the natural logarithm on both sides:
$\ln 4 = \lambda(t_2 - t_1)$
$2 \ln 2 = \left( \frac{\ln 2}{20} \right) (t_2 - t_1)$
$2 = \frac{t_2 - t_1}{20}$
$t_2 - t_1 = 40 \, min$.

(C) The radioactive decay law is given by $N(t) = N_0(1/2)^{t/T_{1/2}}$,where $N(t)$ is the amount remaining at time $t$,$N_0$ is the initial amount,and $T_{1/2} = 10$ days.
If $90\%$ of the element disintegrates,the amount remaining is $N(t) = 10\%$ of $N_0 = 0.1 N_0$.
Substituting this into the decay equation: $0.1 N_0 = N_0(1/2)^{t/10}$.
$0.1 = (1/2)^{t/10}$.
Taking the logarithm on both sides: $\log_{10}(0.1) = (t/10) \log_{10}(0.5)$.
$-1 = (t/10) \times (-0.3010)$.
$t/10 = 1 / 0.3010 \approx 3.322$.
$t \approx 33.22$ days.
Rounding to the nearest integer,the time taken is $33$ days.

(B) Given,half-life $T_{1/2} = 20 \ min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20} \ min^{-1}$.
Let $N_0$ be the initial amount of radioactive substance.
At time $t_1$,$20\%$ has decayed,so the remaining amount is $N(t_1) = 0.8 N_0$. Using $N(t) = N_0 e^{-\lambda t}$,we have $0.8 N_0 = N_0 e^{-\lambda t_1} \implies e^{-\lambda t_1} = 0.8$.
At time $t_2$,$80\%$ has decayed,so the remaining amount is $N(t_2) = 0.2 N_0$. Thus,$0.2 N_0 = N_0 e^{-\lambda t_2} \implies e^{-\lambda t_2} = 0.2$.
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 the 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) Let $N_0$ be the initial number of radioactive nuclei.
At time $t$,the fraction of nuclei that have disintegrated is $x$.
Therefore,the fraction of nuclei remaining is $N(t)/N_0 = 1 - x$.
According to the law of radioactive decay,$N(t) = N_0(1/2)^{t/T}$.
So,$1 - x = (1/2)^{t/T}$.
Now,we need to find the fraction decayed in time $t/2$,which is $1 - N(t/2)/N_0$.
$N(t/2)/N_0 = (1/2)^{(t/2)/T} = [(1/2)^{t/T}]^{1/2}$.
Substituting $1 - x = (1/2)^{t/T}$,we get $N(t/2)/N_0 = \sqrt{1 - x}$.
The fraction that decays in time $t/2$ is $1 - N(t/2)/N_0 = 1 - \sqrt{1 - x}$.

(A) Let $N_x$ be the amount of radioactive isotope $x$ and $N_y$ be the amount of stable element $y$ present in the rock.
The ratio is given as $N_x : N_y = 1 : 7$.
The initial amount of the radioactive isotope $N_0$ is the sum of the remaining isotope and the decayed product:
$N_0 = N_x + N_y = 1 + 7 = 8$.
We know that $N_x = N_0 \times (1/2)^n$, where $n$ is the number of half-lives.
$1 = 8 \times (1/2)^n \Rightarrow (1/2)^n = 1/8 = (1/2)^3$.
Thus, $n = 3$.
The age of the rock $t$ is given by $t = n \times T_{1/2}$, where $T_{1/2} = 50$ years.
$t = 3 \times 50 = 150$ $years$.

(B) The activity $R$ of a radioactive sample at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.
Rearranging the formula,we get: $\frac{R_0}{R} = e^{\lambda t}$.
Taking the natural logarithm on both sides: $\lambda t = \log_e \left( \frac{R_0}{R} \right)$.
Therefore,the decay constant $\lambda$ is: $\lambda = \frac{1}{t} \log_e \left( \frac{R_0}{R} \right)$.
Given values are: $R_0 = 5000 \text{ disintegrations/min}$,$R = 1250 \text{ disintegrations/min}$,and $t = 5 \text{ min}$.
Substituting these values into the equation:
$\lambda = \frac{1}{5} \log_e \left( \frac{5000}{1250} \right) = \frac{1}{5} \log_e (4)$.
Since $4 = 2^2$,we have:
$\lambda = \frac{1}{5} \log_e (2^2) = \frac{2}{5} \log_e 2$.
Calculating the fraction: $\frac{2}{5} = 0.4$.
Thus,the decay constant is $\lambda = 0.4 \log_e 2 \text{ min}^{-1}$.

(B) The half-life of the radioactive substance is $T_{1/2} = 5\, years$.
The total time elapsed is $t = 10\, years$.
The number of half-lives passed is $n = t / T_{1/2} = 10 / 5 = 2$.
The fraction of undecayed nuclei remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.
Substituting $n = 2$, we get $N/N_0 = (1/2)^2 = 1/4$.
This means $1/4$ of the original nuclei remain undecayed.
The fraction of nuclei that have decayed is $1 - N/N_0 = 1 - 1/4 = 3/4$.
To express this as a percentage, we multiply by $100$: $(3/4) \times 100\% = 75\%$.
Therefore, the probability of decay for a nucleus in $10\, years$ is $75\%$.

(B) The intensity of radiation $I$ follows the law $I = I_0 \left( \frac{1}{2} \right)^n$, where $n$ is the number of half-lives.
Given that the initial intensity $I_0 = 64 \times I_{\text{safe}}$, we want to find the time when $I = I_{\text{safe}}$.
Substituting the values: $I_{\text{safe}} = 64 \times I_{\text{safe}} \times \left( \frac{1}{2} \right)^n$.
This simplifies to $\left( \frac{1}{2} \right)^n = \frac{1}{64}$.
Since $64 = 2^6$, we have $\left( \frac{1}{2} \right)^n = \left( \frac{1}{2} \right)^6$, which implies $n = 6$.
The total time $t = n \times T_{1/2}$, where $T_{1/2} = 2\, hours$.
Therefore, $t = 6 \times 2\, hours = 12\, hours$.

(B) The formula for radioactive decay is given by $M = M_{0} \left(\frac{1}{2}\right)^{n}$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Given: $M = 1 \; g$,$M_{0} = 256 \; g$,and $T_{1/2} = 12.5 \; h$.
Substituting the values: $1 = 256 \left(\frac{1}{2}\right)^{n}$.
This simplifies to $\frac{1}{256} = \left(\frac{1}{2}\right)^{n}$.
Since $256 = 2^{8}$,we have $\left(\frac{1}{2}\right)^{8} = \left(\frac{1}{2}\right)^{n}$.
Therefore,$n = 8$.
Using $n = \frac{t}{T_{1/2}}$,we get $t = n \times T_{1/2} = 8 \times 12.5 \; h = 100 \; h$.

(B) The rate of disintegration is given by $R = \lambda N$,where $\lambda = \frac{\ln 2}{T_{1/2}}$.
For nucleus $A$: $T_{1/2, A} = 1 \, min$,so $\lambda_A = \frac{\ln 2}{1} = \ln 2 \, min^{-1}$.
For nucleus $B$: $T_{1/2, B} = 2 \, min$,so $\lambda_B = \frac{\ln 2}{2} \, min^{-1}$.
The number of nuclei at time $t$ is $N_A(t) = 4N_0 e^{-\lambda_A t}$ and $N_B(t) = N_0 e^{-\lambda_B t}$.
The rates of disintegration are $R_A = \lambda_A N_A = \lambda_A (4N_0 e^{-\lambda_A t})$ and $R_B = \lambda_B N_B = \lambda_B (N_0 e^{-\lambda_B t})$.
Setting $R_A = R_B$:
$4 \lambda_A N_0 e^{-\lambda_A t} = \lambda_B N_0 e^{-\lambda_B t}$
$4 (\ln 2) e^{-(\ln 2) t} = \frac{\ln 2}{2} e^{-(\frac{\ln 2}{2}) t}$
$8 = \frac{e^{-(\frac{\ln 2}{2}) t}}{e^{-(\ln 2) t}} = e^{(\ln 2 - \frac{\ln 2}{2}) t} = e^{(\frac{\ln 2}{2}) t}$
Taking the natural logarithm on both sides:
$\ln 8 = \frac{\ln 2}{2} t$
$3 \ln 2 = \frac{\ln 2}{2} t$
$t = 6 \, min$.

(B) The number of active nuclei at any time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
The mean life is $\tau = \frac{1}{\lambda}$. At $t = \tau$,the number of remaining nuclei is $N_1 = N_0 e^{-\lambda(1/\lambda)} = \frac{N_0}{e}$.
The half-life is $T_{1/2} = \frac{\ln 2}{\lambda}$. At $t = 2T_{1/2}$,the number of remaining nuclei is $N_2 = N_0 (\frac{1}{2})^2 = \frac{N_0}{4}$.
The number of nuclei that decay between these two times is the difference in the number of remaining nuclei: $\Delta N = N_1 - N_2 = \frac{N_0}{e} - \frac{N_0}{4}$.

(A) From the graph, at $t = 0 \, hr$, the initial decay rate $R_0 = 200 \, \text{decays/sec}$.
At $t = 2 \, hr$, the decay rate $R = 100 \, \text{decays/sec}$.
Since the decay rate becomes half of its initial value in $2 \, hr$, the half-life $T_{1/2} = 2 \, hr$.
The decay rate at any time $t$ is given by $R = R_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$.
For $t = 8 \, hr$, $R = 200 \left( \frac{1}{2} \right)^{8/2} = 200 \left( \frac{1}{2} \right)^4 = \frac{200}{16} = \frac{25}{2} \, \text{decays/sec}$.

(C) Let $A_0$ be the initial activity of the radioactive sample.
The activity at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$,where $\lambda$ is the decay constant.
At time $t_1$,the activity is $A_1 = A_0 e^{-\lambda t_1}$.
At time $t_2$,the activity is $A_2 = A_0 e^{-\lambda t_2}$.
Dividing the two equations: $\frac{A_2}{A_1} = \frac{A_0 e^{-\lambda t_2}}{A_0 e^{-\lambda t_1}} = e^{-\lambda(t_2 - t_1)}$.
Since the mean life $T = \frac{1}{\lambda}$,we can substitute $\lambda = \frac{1}{T}$.
Thus,$\frac{A_2}{A_1} = e^{-(t_2 - t_1)/T} = e^{(t_1 - t_2)/T}$.
Therefore,$A_2 = A_1 e^{(t_1 - t_2)/T}$.

(D) Given that $\frac{7}{8}$ of the sample of ${}^{66}Cu$ decays in $15 \ minutes$.
The fraction of the sample remaining undecayed is $N = 1 - \frac{7}{8} = \frac{1}{8}$.
We know that the remaining amount $N$ after $n$ half-lives is given by $N = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,the number of half-lives $n$ is $3$.
The relationship between time $t$,number of half-lives $n$,and half-life $T$ is $n = \frac{t}{T}$.
Substituting the values,$3 = \frac{15}{T}$.
Therefore,the half-life $T = \frac{15}{3} = 5 \ minutes$.

(A) The activity of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.
Given that in $t = 3\, \text{days}$, the activity becomes $R = R_0/3$:
$\frac{1}{3} = e^{-\lambda \times 3} = e^{-3\lambda}$ .........$(1)$
We need to find the activity $R'$ after $t = 9\, \text{days}$:
$R' = R_0 e^{-\lambda \times 9} = R_0 (e^{-3\lambda})^3$
Substituting the value from equation $(1)$:
$R' = R_0 \times (1/3)^3$
$R' = R_0 \times (1/27)$
Therefore, the activity becomes $(1/27)$ of the original value.

(D) The decay rate (activity) is given by the formula: $\frac{dN}{dt} = \lambda N$.
First, calculate the decay constant $\lambda$ using the half-life $T_{1/2} = 1.2 \times 10^7 \, s$:
$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.2 \times 10^7} \, s^{-1}$.
Now, substitute the values of $\lambda$ and $N = 4.0 \times 10^{15}$ atoms into the decay rate formula:
$\frac{dN}{dt} = \left( \frac{0.693}{1.2 \times 10^7} \right) \times (4.0 \times 10^{15})$
$\frac{dN}{dt} = \frac{0.693 \times 4.0}{1.2} \times 10^{15-7}$
$\frac{dN}{dt} = \frac{2.772}{1.2} \times 10^8$
$\frac{dN}{dt} = 2.31 \times 10^8 \, \text{atoms/s}$.
Rounding to the appropriate significant figures, the decay rate is $2.3 \times 10^8 \, \text{atoms/s}$.

(C) The number of radioactive nuclei $N$ at any time $t$ is given by the radioactive decay law:
$N(t) = N_0 e^{-\lambda t}$
Where $N_0$ is the initial number of radioactive nuclei at $t = 0$ and $\lambda$ is the decay constant.
This equation represents an exponential decay function,which is of the form $y = a e^{-kx}$.
As $t$ increases,$N$ decreases exponentially from the initial value $N_0$ and approaches zero as $t \to \infty$.
Therefore,the correct graph is the one showing an exponential decay curve starting from a positive value on the $N$-axis,which corresponds to option $C$.

(C) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.
Given that the ratio of $C^{14}$ to $C^{12}$ is one-fourth of the original amount,we have $\frac{N(t)}{N_0} = \frac{1}{4}$.
Substituting the values,we get $\frac{1}{4} = \left(\frac{1}{2}\right)^{t/5700}$.
Since $\frac{1}{4} = \left(\frac{1}{2}\right)^2$,we can equate the exponents: $2 = \frac{t}{5700}$.
Solving for $t$,we get $t = 2 \times 5700 = 11400 \, years$.

(C) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate: $\frac{dN}{dt} = n - \lambda N$.
Rearranging the terms,we get $\frac{dN}{n - \lambda N} = dt$.
Integrating both sides with limits from $t = 0$ $(N = N_0)$ to $t = t$ $(N = N)$:
$\int_{N_0}^{N} \frac{dN}{n - \lambda N} = \int_{0}^{t} dt$.
Using the substitution method,$-\frac{1}{\lambda} [\ln(n - \lambda N)]_{N_0}^{N} = t$.
$-\frac{1}{\lambda} \ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = t$.
$\ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = -\lambda t$.
$\frac{n - \lambda N}{n - \lambda N_0} = e^{-\lambda t}$.
$n - \lambda N = (n - \lambda N_0) e^{-\lambda t}$.
$\lambda N = n - (n - \lambda N_0) e^{-\lambda t}$.
$N = \frac{n}{\lambda} - \left( \frac{n}{\lambda} - N_0 \right) e^{-\lambda t} = \frac{n}{\lambda} + \left( N_0 - \frac{n}{\lambda} \right) e^{-\lambda t}$.

(C) The ratio of ${}^{14}C$ to ${}^{12}C$ in the fossil bone is given as $\frac{N}{N_0} = \frac{1}{16}$.
The law of radioactive decay states that $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Equating the two,we get $\left(\frac{1}{2}\right)^n = \frac{1}{16} = \left(\frac{1}{2}\right)^4$.
Therefore,the number of half-lives $n = 4$.
The age of the fossil $t$ is given by $t = n \times T_{1/2}$,where $T_{1/2} = 5730 \, years$.
$t = 4 \times 5730 = 22920 \, years$.

(D) The activity of a radioactive sample follows the law $N(t) = N_0 e^{-\lambda t}$.
Given that at $t = 5 \, minutes$,the activity is $N = N_0/e$.
Substituting these values into the equation: $N_0/e = N_0 e^{-\lambda(5)}$.
This simplifies to $e^{-1} = e^{-5\lambda}$,which implies $5\lambda = 1$,so $\lambda = 1/5 \, min^{-1}$.
The half-life $T_{1/2}$ is defined as the time required for the activity to reduce to half its initial value,given by $T_{1/2} = \frac{\ln 2}{\lambda}$.
Substituting $\lambda = 1/5$,we get $T_{1/2} = \frac{\ln 2}{1/5} = 5 \ln 2 \, minutes$.
Therefore,the correct option is $D$.