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

(B) The intensity of radiation $I$ decreases according to the law of radioactive decay: $I = I_0 \left( \frac{1}{2} \right)^n$, where $n$ is the number of half-lives.
Given that the initial intensity $I_0 = 64 I_{\text{safe}}$, we need the intensity to reach $I_{\text{safe}}$.
Thus, $\frac{I_{\text{safe}}}{64 I_{\text{safe}}} = \left( \frac{1}{2} \right)^n$.
$\frac{1}{64} = \left( \frac{1}{2} \right)^n$.
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 $t$ is given by $t = n \times T_{1/2}$, where $T_{1/2} = 2 \text{ hours}$.
$t = 6 \times 2 = 12 \text{ hours}$.

(A) The rate of decay is given by $\frac{dN}{dt} = \lambda N$.
First,calculate the decay constant $\lambda$ in $s^{-1}$:
$\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1620 \times 365 \times 24 \times 3600} \ s^{-1}$.
Next,calculate the number of atoms $N$ in $1 \ g$ $(10^{-3} \ kg)$ of radium:
$N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{10^{-3} \ kg}{226 \ kg/kmol} \times 6.02 \times 10^{26} \ atoms/kmol = \frac{6.02}{226} \times 10^{23} \ atoms$.
Now,calculate the activity $\frac{dN}{dt}$:
$\frac{dN}{dt} = \left( \frac{0.693}{1620 \times 3.1536 \times 10^7} \right) \times \left( \frac{6.02}{226} \times 10^{23} \right) \approx 3.61 \times 10^{10} \ atoms/s$.

(A) When a radioactive material decays by two simultaneous processes,the effective decay constant $\lambda$ is given by $\lambda = \lambda_1 + \lambda_2$.
Since the half-life $T$ is related to the decay constant by $T = \frac{\ln 2}{\lambda}$,the effective half-life $T$ is given by $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$.
Given $T_1 = 1620$ years and $T_2 = 810$ years,the effective half-life is:
$T = \frac{T_1 T_2}{T_1 + T_2} = \frac{1620 \times 810}{1620 + 810} = \frac{1620 \times 810}{2430} = 540$ years.
We want to find the time $t$ after which $\frac{1}{4}$ of the material remains.
The law of radioactive decay states that $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}$.
Setting $N(t) = \frac{1}{4} N_0$,we get $\frac{1}{4} = \left(\frac{1}{2}\right)^{t/T}$,which implies $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/T}$.
Thus,$\frac{t}{T} = 2$,so $t = 2T = 2 \times 540 = 1080$ years.

(B) The intensity of radiation is proportional to the number of radioactive nuclei $N$ present in the sample.
Let $N_0$ be the initial number of nuclei and $N$ be the number of nuclei after $n$ half-lives.
The safe limit corresponds to $N = \frac{N_0}{128}$.
Using the law of radioactive decay, the remaining fraction is given by $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting the values, we get $\frac{1}{128} = (\frac{1}{2})^n$.
Since $128 = 2^7$, we have $(\frac{1}{2})^7 = (\frac{1}{2})^n$, which implies $n = 7$.
Since the half-life $T_{1/2} = 1$ hour, the total time $t = n \times T_{1/2} = 7 \times 1 = 7$ hours.

(C) Given that the half-life of $X$ is equal to the mean life of $Y$:
$({T_{1/2}})_X = ({\tau})_Y$
We know that ${T_{1/2}} = \frac{0.693}{\lambda}$ and ${\tau} = \frac{1}{\lambda}$.
Substituting these,we get:
$\frac{0.693}{\lambda_X} = \frac{1}{\lambda_Y}$
$\Rightarrow \lambda_X = 0.693 \lambda_Y$
Since $0.693 < 1$,it follows that $\lambda_X < \lambda_Y$.
The rate of decay is given by $R = \lambda N$.
Initially,the number of atoms $N$ is the same for both.
Since $\lambda_Y > \lambda_X$,the decay rate $R_Y = \lambda_Y N$ will be greater than $R_X = \lambda_X N$.
Therefore,$Y$ will decay at a faster rate than $X$.

(C) The decay constant for $\alpha$-emission is $\lambda_{\alpha} = \frac{1}{1620} \text{ year}^{-1}$.
The decay constant for $\beta$-emission is $\lambda_{\beta} = \frac{1}{405} \text{ year}^{-1}$.
The total decay constant is $\lambda = \lambda_{\alpha} + \lambda_{\beta} = \frac{1}{1620} + \frac{1}{405} = \frac{1+4}{1620} = \frac{5}{1620} = \frac{1}{324} \text{ year}^{-1}$.
The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$,where $\frac{N}{N_0} = \frac{1}{4}$.
Taking the natural logarithm on both sides: $\ln\left(\frac{N_0}{N}\right) = \lambda t$.
Substituting the values: $\ln(4) = \left(\frac{1}{324}\right) t$.
$t = 324 \times \ln(4) = 324 \times 2 \times \ln(2) \approx 648 \times 0.6931 = 449.13 \text{ years}$.
Thus,the time required is approximately $449 \text{ years}$.

(D) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{138.6 \times 24 \text{ hours}} = \frac{0.693}{3326.4} \text{ h}^{-1}$.
Alternatively,using the activity formula for a small time interval $\Delta t$:
Activity $A = \lambda N = \frac{0.693}{T_{1/2}} N$.
Given $T_{1/2} = 138.6 \text{ days} = 138.6 \times 24 \text{ hours} = 3326.4 \text{ hours}$.
Number of atoms $N = 10^6$.
Disintegrations in $\Delta t = 24 \text{ hours}$ (which is $1$ day):
$\Delta N = \lambda N \Delta t = \left( \frac{0.693}{138.6 \text{ days}} \right) \times 10^6 \times 1 \text{ day}$.
$\Delta N = \frac{0.693}{138.6} \times 10^6 = 0.005 \times 10^6 = 5000$.
Thus,the number of disintegrations is $5000$.

(B) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} = 0.03465 \ min^{-1}$.
The time $t$ required for a substance to decay is given by $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right)$.
For $33\%$ disintegration,the remaining amount $N = N_0 - 0.33N_0 = 0.67N_0$. Thus,$t_1 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{67} \right) \approx 66.46 \times 0.1739 \approx 11.56 \ min$.
For $67\%$ disintegration,the remaining amount $N = N_0 - 0.67N_0 = 0.33N_0$. Thus,$t_2 = \frac{2.303}{0.03465} \log_{10} \left( \frac{100}{33} \right) \approx 66.46 \times 0.4815 \approx 32.00 \ min$.
The time difference is $\Delta t = t_2 - t_1 = 32.00 - 11.56 = 20.44 \ min \approx 20 \ min$.

(D) Let $N_0$ be the initial number of radioactive atoms for both elements at $t = 0$.
According to the law of radioactive decay,the number of atoms remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For element $X_1$ with decay constant $\lambda_1 = 10\lambda$,the number of atoms is $N_1 = N_0 e^{-10\lambda t}$.
For element $X_2$ with decay constant $\lambda_2 = \lambda$,the number of atoms is $N_2 = N_0 e^{-\lambda t}$.
We are given that the ratio $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Substituting the expressions for $N_1$ and $N_2$:
$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$
$e^{-10\lambda t + \lambda t} = e^{-1}$
$e^{-9\lambda t} = e^{-1}$
Comparing the exponents:
$-9\lambda t = -1$
$t = \frac{1}{9\lambda}$.

(D) The activity of a radioactive sample follows the law $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.
Given that the activity reduces from $6000 \, dps$ to $3000 \, dps$ in $140$ days,the half-life $T_{1/2}$ is $140$ days.
After $280$ days,the number of half-lives elapsed is $n = 280 / 140 = 2$.
Let $A_{280}$ be the activity after $280$ days,which is $6000 \, dps$.
Using the formula $A_{280} = A_{initial} \times (1/2)^n$,we get $6000 = A_{initial} \times (1/2)^2$.
$6000 = A_{initial} \times (1/4)$.
$A_{initial} = 6000 \times 4 = 24000 \, dps$.

(B) The rate of disintegration is given by $\frac{dN}{dt} = \lambda N_0 = 10^{17} \, s^{-1}$.
The half-life $T_{1/2} = 1445 \, \text{years}$.
Convert the half-life into seconds: $T_{1/2} = 1445 \times 365 \times 24 \times 3600 \approx 4.55 \times 10^{10} \, s$.
The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{4.55 \times 10^{10}} \approx 1.52 \times 10^{-11} \, s^{-1}$.
Using the relation $\frac{dN}{dt} = \lambda N_0$,we find the original number of particles $N_0$:
$N_0 = \frac{dN/dt}{\lambda} = \frac{10^{17}}{1.52 \times 10^{-11}} \approx 6.58 \times 10^{27} \approx 6.6 \times 10^{27}$.

(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{1.5 \times 10^9}{4.5 \times 10^9} = \frac{1}{3}$.
The remaining number of $U^{238}$ nuclei $N_U$ is given by $N_U = N_0 \left( \frac{1}{2} \right)^n = N_0 \left( \frac{1}{2} \right)^{1/3}$.
Given $2^{1/3} = 1.26$,we have $\left( \frac{1}{2} \right)^{1/3} = \frac{1}{1.26} \approx 0.7937$.
Thus,$N_U = N_0 \times 0.7937$.
The number of $Pb$ nuclei formed is $N_{Pb} = N_0 - N_U = N_0(1 - 0.7937) = N_0(0.2063)$.
Alternatively,using the ratio: $\frac{N_U}{N_0} = \frac{1}{1.26}$.
Since $N_0 = N_U + N_{Pb}$,we have $\frac{N_U}{N_U + N_{Pb}} = \frac{1}{1.26}$.
$1.26 N_U = N_U + N_{Pb} \implies N_{Pb} = 0.26 N_U$.
Therefore,the ratio $\frac{N_{Pb}}{N_U} = 0.26$.

(A) The activity $A$ is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$.
Given $T_{1/2} = 138.6 \text{ days} = 138.6 \times 24 \times 3600 \text{ seconds} = 1.1975 \times 10^7 \text{ s}$.
Given activity $A = 2000 \text{ disintegrations/sec}$.
Substituting the values into $N = \frac{A}{\lambda} = \frac{A \times T_{1/2}}{\ln 2}$:
$N = \frac{2000 \times 138.6 \times 24 \times 3600}{0.693} = \frac{2000 \times 1.1975 \times 10^7}{0.693} \approx 3.45 \times 10^{10}$.

(A) The rate of change of the number of nuclei $N$ is given by the differential equation: $\frac{dN}{dt} = \alpha - \lambda N$.
At the maximum number of nuclei,the rate of change is zero,i.e.,$\frac{dN}{dt} = 0$.
Setting the equation to zero: $0 = \alpha - \lambda N$.
Solving for $N$,we get $N_{max} = \frac{\alpha}{\lambda}$.
Note that if the initial number of nuclei $N_0$ is already greater than $\frac{\alpha}{\lambda}$,the number of nuclei will decrease until it reaches the equilibrium value $\frac{\alpha}{\lambda}$. If $N_0 < \frac{\alpha}{\lambda}$,it will increase until it reaches $\frac{\alpha}{\lambda}$.

(C) The half-life of the substance is $T_{1/2} = 12 \, \text{hrs}$.
The total time elapsed is $t = 1 \, \text{week} = 7 \, \text{days} = 7 \times 24 \, \text{hrs} = 168 \, \text{hrs}$.
The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{168}{12} = 14$.
The activity $A$ after $n$ half-lives is given by $A = A_0 \left( \frac{1}{2} \right)^n$, where $A_0 = 1 \, \text{curie}$.
$A = 1 \times \left( \frac{1}{2} \right)^{14} \, \text{curie} = \frac{1}{16384} \, \text{curie}$.
Calculating the value: $A \approx 0.000061 \, \text{curie} = 61 \times 10^{-6} \, \text{curie} = 61 \, \mu\text{Ci}$.
Thus, the activity left is approximately $60 \, \mu\text{Ci}$.

(D) Let $N_0$ be the initial number of atoms of each species. The number of radioactive nuclei of the first species at time $t$ is $N_1(t) = N_0 e^{-t/\tau}$.
The number of radioactive nuclei of the second species at time $t$ is $N_2(t) = N_0 e^{-t/(5\tau)}$.
The total number of radioactive nuclei is $N(t) = N_1(t) + N_2(t) = N_0(e^{-t/\tau} + e^{-t/(5\tau)})$.
At $t = 0$,$N(0) = 2N_0$. As $t \to \infty$,$N(t) \to 0$.
Since both $e^{-t/\tau}$ and $e^{-t/(5\tau)}$ are monotonically decreasing functions of time,their sum $N(t)$ must also be a monotonically decreasing function of time.
Therefore,the total number of radioactive nuclei will continuously decrease with time,which is represented by the graph in option $(d)$.

(C) According to the law of radioactive decay,the number of undecayed nuclei $N$ at time $t$ is given by $N = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei and $\lambda$ is the decay constant.
The rate of decay is given by the derivative of $N$ with respect to time $t$:
$\frac{dN}{dt} = -\lambda N_0 e^{-\lambda t}$
The magnitude of the rate of decay is $|\frac{dN}{dt}| = \lambda N_0 e^{-\lambda t}$.
This expression shows that the rate of decay decreases exponentially with time $t$. Among the given options,Figure $C$ represents an exponential decay curve,which correctly depicts the relationship between the rate of decay and time.

(D) The rate of radioactive disintegration $R$ is given by the law of radioactive decay as $R = -\frac{dN}{dt} = \lambda N$,where $\lambda$ is the decay constant.
From this relation,we can write the ratio of the rate of disintegration to the number of atoms as:
$\frac{R}{N} = \lambda$
Since the decay constant $\lambda$ is a characteristic property of a radioactive substance and remains constant over time,the ratio $R/N$ is independent of time.
Therefore,the graph representing the variation of $R/N$ with time $t$ will be a horizontal straight line parallel to the time axis.

(B) $1$. To find the half-life,observe the graph. The initial count rate at $t = 0$ is $100$. The half-life is the time taken for the count rate to become half of its initial value,i.e.,$50$. From the graph,at a count rate of $50$,the time is $3\,h$. Thus,the half-life $T_{1/2} = 3\,h$.
$2$. To find the total counts in the first half-life period ($0$ to $3\,h$),we calculate the area under the curve between $t = 0$ and $t = 3\,h$.
$3$. Each small square on the graph represents a count rate of $10$ units on the y-axis and $1\,h$ on the x-axis. However,the question specifies $10\,g$ of material. The area under the curve represents the total number of counts. By counting the number of small squares under the curve from $t = 0$ to $t = 3\,h$,we find approximately $24$ squares.
$4$. Each square represents $10 \times 60 = 600$ counts (assuming the rate is per minute,so $10 \text{ counts/min} \times 60 \text{ min} = 600 \text{ counts/hour}$).
$5$. Total counts $\approx 24 \times 600 = 14400$.

(B) The number of undecayed atoms at time $t$ is given by $N = N_0 e^{-\lambda t}$.
The number of decayed atoms is $N_d = N_0 - N = N_0(1 - e^{-\lambda t})$.
The fraction of radioactive material that has decayed is $f = \frac{N_d}{N_0} = \frac{N_0(1 - e^{-\lambda t})}{N_0} = 1 - e^{-\lambda t}$.
At $t = 0$,$f = 1 - e^0 = 0$.
As $t \to \infty$,$f \to 1 - 0 = 1$.
The function $f(t) = 1 - e^{-\lambda t}$ represents an exponential growth curve that starts from $0$ and approaches $1$ asymptotically. This corresponds to curve $B$ in the graph.

(B) The number of undecayed atoms at time $t$ is given by $N = N_0 e^{-\lambda t}$.
The activity of the radioactive substance is given by $A = \lambda N$.
From this relation, we can see that $N = \frac{A}{\lambda}$.
Since $\lambda$ (decay constant) is a constant for a given radioactive substance, $N$ is directly proportional to $A$ $(N \propto A)$.
Therefore, the plot of the number of undecayed atoms $(N)$ versus activity $(A)$ is a straight line passing through the origin.

(D) The law of radioactive decay states that the number of radioactive nuclei $N$ at any time $t$ is given by the equation: $N = N_0 e^{-\lambda t}$,where $N_0$ is the initial number of nuclei and $\lambda$ is the decay constant.
This equation represents an exponential decay function.
As time $t$ increases,the value of $N$ decreases exponentially,approaching zero as $t$ approaches infinity.
Therefore,the graph of $N$ versus $t$ is an exponential decay curve,which corresponds to option $(D)$.

(B) The activity $A$ of a radioactive sample is given by the formula $A = -\frac{dN}{dt} = \lambda N = \lambda N_0 e^{-\lambda t}$.
Here,$N_0$ is the initial number of nuclei,$\lambda$ is the decay constant,and $t$ is time.
This equation shows that the activity $A$ decreases exponentially with time $t$.
Graphically,an exponential decay starting from a positive value at $t=0$ and approaching zero as $t \to \infty$ is represented by curve $B$.

(D) The radioactive decay law for activity $A$ is given by $A = A_0 e^{-\lambda t}$,where $A_0 = \lambda N_0$ is the initial activity.
Taking the natural logarithm on both sides:
$\ln A = \ln(A_0 e^{-\lambda t})$
$\ln A = \ln A_0 + \ln(e^{-\lambda t})$
$\ln A = \ln A_0 - \lambda t$
This equation is of the form $y = mx + c$,where $y = \ln A$,$x = t$,slope $m = -\lambda$,and intercept $c = \ln A_0$.
Since the slope is negative $(-\lambda)$ and the intercept is positive $(\ln A_0)$,the graph of $\log A$ versus $t$ is a straight line with a negative slope. Looking at the provided figure,the line labeled $D$ represents this linear decrease.

(B) The activity $A$ of a radioactive sample is defined as the rate of decay of the radioactive nuclei,which is given by the relation:
$A = \left| \frac{dN}{dt} \right| = \lambda N$
where $\lambda$ is the decay constant and $N$ is the number of active atoms present at that instant.
This equation is of the form $y = mx$,where $y = A$,$x = N$,and the slope $m = \lambda$.
Since $\lambda$ is a constant for a given radioactive substance,the relationship between $A$ and $N$ is a straight line passing through the origin.
Therefore,the correct graph is a straight line passing through the origin,which corresponds to option $B$.

(D) When a radioactive substance decays by two different processes simultaneously,the total decay constant $\lambda_{eff}$ is the sum of the individual decay constants: $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
We know that the half-life $T$ is related to the decay constant $\lambda$ by the formula $T = \frac{\ln 2}{\lambda}$,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 both sides by $\ln 2$,we get: $\frac{1}{T_{eff}} = \frac{1}{T_{\alpha}} + \frac{1}{T_{\beta}}$.
Solving for $T_{eff}$: $\frac{1}{T_{eff}} = \frac{T_{\beta} + T_{\alpha}}{T_{\alpha} T_{\beta}}$.
Therefore,$T_{eff} = \frac{T_{\alpha} T_{\beta}}{T_{\alpha} + T_{\beta}}$.

(B) The relationship between half-life $T_{1/2}$ and decay constant $\lambda$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$.
Given $T_{1/2} = 2.2 \times 10^9 \; s$,we find $\lambda = \frac{0.693}{2.2 \times 10^9} \approx 3.15 \times 10^{-10} \; s^{-1}$.
The rate of disintegration $R$ is related to the number of radioactive atoms $N$ by the formula $R = \lambda N$.
Given $R = 10^{10} \; s^{-1}$,we calculate $N = \frac{R}{\lambda} = \frac{10^{10}}{3.15 \times 10^{-10}} \approx 3.17 \times 10^{19}$ atoms.

(A) curie $(Ci)$ is a unit of radioactivity. It is defined as the quantity of a radioactive substance that undergoes $3.7 \times 10^{10}$ disintegrations per second,which is approximately the activity of $1 \ g$ of radium-$226$.

(B) The fraction of the sample remaining after time $t$ is $N/N_0 = 1 - 7/8 = 1/8$.
Using the radioactive decay law $N = N_0(1/2)^n$,we have $1/8 = (1/2)^n$,which implies $(1/2)^3 = (1/2)^n$,so $n = 3$.
Since $n = t/T_{1/2}$,the half-life is $T_{1/2} = t/3$.
Now,for $15/16$ of the sample to disintegrate,the remaining fraction is $N/N_0 = 1 - 15/16 = 1/16$.
Using $N/N_0 = (1/2)^n$,we have $1/16 = (1/2)^n$,which implies $(1/2)^4 = (1/2)^n$,so $n = 4$.
The total time required is $t' = n \times T_{1/2} = 4 \times (t/3) = \frac{4}{3}t$.

(C) The activity of a radioactive sample is given by $A = |dN/dt| = N\lambda$. Thus,option $A$ is correct.
By definition,the decay constant $\lambda$ represents the probability of decay per unit time for a single nucleus. Thus,option $B$ is correct.
The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For $t = 1/\lambda$,the number of nuclei remaining is $N(1/\lambda) = N e^{-\lambda(1/\lambda)} = N e^{-1} = N/e$.
Option $C$ states that the number of nuclei remaining is $N[1 - 1/e]$,which represents the number of nuclei that have decayed,not the number remaining. Therefore,option $C$ is incorrect.
The half-life $T_{1/2}$ is given by $\ln(2)/\lambda$. Thus,option $D$ is correct.

(C) The radioactive decay law states that the rate of decay is proportional to the number of radioactive nuclei present at that time,given by $\frac{dN}{dt} = -\lambda N$.
Here,$\lambda$ is the decay constant,which represents the probability of decay per unit time for a single radioactive nucleus.
Radioactive decay is a spontaneous and random process,meaning the probability of decay for a nucleus does not depend on how long the nucleus has existed.
Therefore,$\lambda$ is a constant characteristic of the radioactive isotope and is independent of the age of the atom.

(B) The specific activity of a radioactive substance is defined as the activity per unit mass.
By definition,the unit $1 \, Curie$ $(1 \, Ci)$ was originally defined as the activity of $1 \, gram$ of Radium-$226$.
Therefore,the specific activity of radium is approximately $1 \, Ci/g$.

(A) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
At half-life $t = \tau_{1/2}$,the number of nuclei remaining is $N = N_0 / 2$.
Substituting these values: $\frac{N_0}{2} = N_0 e^{-\lambda \tau_{1/2}}$.
$1/2 = e^{-\lambda \tau_{1/2}}$,which implies $2 = e^{\lambda \tau_{1/2}}$.
Taking the natural logarithm on both sides: $\ln 2 = \lambda \tau_{1/2}$.
Therefore,the relationship is $\tau_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$.

(A) Specific activity is defined as the activity per unit mass of the radioactive substance.
Given,specific activity $A_s = 1 \, Ci/g$.
The mass of the sample $m = 1 \, \mu g = 10^{-6} \, g$.
The activity $A$ of the sample is given by the product of specific activity and the mass of the sample:
$A = A_s \times m$
$A = 1 \, Ci/g \times 10^{-6} \, g$
$A = 10^{-6} \, Ci$
Since $1 \, \mu Ci = 10^{-6} \, Ci$,the activity is $1 \, \mu Ci$.

(D) The radioactive decay law is given by the formula: $\frac{N}{N_0} = (1/2)^n$, where $n$ is the number of half-lives.
Number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6400}{1600} = 4$.
Substituting the value of $n$ into the formula:
$\frac{N}{N_0} = (1/2)^4 = \frac{1}{16}$.
Thus, the fraction of the radium sample remaining after $6400 \, \text{years}$ is $1/16$.

(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Given, $T_{1/2} = 10 \, days$ and $\frac{N}{N_0} = \frac{1}{10}$.
Substituting these values: $\frac{1}{10} = \left( \frac{1}{2} \right)^{\frac{t}{10}}$.
Taking the natural logarithm on both sides: $\ln(10^{-1}) = \frac{t}{10} \ln(2^{-1})$.
$- \ln(10) = - \frac{t}{10} \ln(2)$.
$t = 10 \times \frac{\ln(10)}{\ln(2)} = 10 \times \frac{2.303}{0.693} \approx 10 \times 3.32$.
$t \approx 33.2 \, days$.
Therefore, the time required is approximately $33 \, days$.

(A) The number of undecayed nuclei at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For sample $A_1$,$N_1 = N_0 e^{-(10\lambda_0)t}$.
For sample $A_2$,$N_2 = N_0 e^{-\lambda_0 t}$.
The ratio of undecayed nuclei is $N_1/N_2 = \frac{N_0 e^{-10\lambda_0 t}}{N_0 e^{-\lambda_0 t}} = e^{-9\lambda_0 t}$.
Given $t = 1/(9\lambda_0)$,substitute this value into the ratio:
$N_1/N_2 = e^{-9\lambda_0 (1/9\lambda_0)} = e^{-1} = 1/e$.

(A) Given: Half-life $T_{1/2} = 30 \ min$.
Total time $t = 2 \ \text{hours} = 120 \ min$.
Number of half-lives $n = \frac{t}{T_{1/2}} = \frac{120}{30} = 4$.
The relationship between initial activity $A_0$ and final activity $A$ is given by $A = \frac{A_0}{2^n}$.
Given final activity $A = 5 \ \text{disintegrations/sec}$.
Substituting the values: $5 = \frac{A_0}{2^4} = \frac{A_0}{16}$.
Therefore, $A_0 = 5 \times 16 = 80 \ \text{disintegrations/sec}$.

(D) The half-life $T_{1/2} = 5$ years.
The total time $t = 10$ years.
The number of half-lives $n = t / T_{1/2} = 10 / 5 = 2$.
The fraction of the substance remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.
$N/N_0 = (1/2)^2 = 1/4$.
The fraction of the substance that has decayed is $1 - N/N_0$.
Decayed fraction $= 1 - 1/4 = 3/4 = 0.75$.

(A) The decay process is $X \rightarrow Y$.
Given half-life $T_{1/2} = 3$ days.
Total time elapsed $t = 6$ days.
The number of half-lives $n = t / T_{1/2} = 6 / 3 = 2$.
The remaining mass of $X$ after $n$ half-lives is given by $N = N_0(1/2)^n$.
$N = 10 \times (1/2)^2 = 10 / 4 = 2.5 \, g$.
Since the total mass is conserved,the mass of $Y$ formed is $M_Y = N_0 - N = 10 - 2.5 = 7.5 \, g$.
Thus,the mass of $X$ is $2.5 \, g$ and the mass of $Y$ is $7.5 \, g$.

(A) The activity $R$ is given by the formula $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.
Number of nuclei $N = \frac{m}{M} \times N_A$,where $m = 1 \, mg = 10^{-3} \, g$,$M = 64 \, g/mol$,and $N_A = 6.022 \times 10^{23} \, atoms/mol$.
$N = \frac{10^{-3}}{64} \times 6.022 \times 10^{23} \approx 9.41 \times 10^{18} \, atoms$.
Now,$R = (1.6 \times 10^{-5} \, s^{-1}) \times (9.41 \times 10^{18}) = 1.5056 \times 10^{14} \, Bq$.
To convert to Curies $(Ci)$,divide by $3.7 \times 10^{10} \, Bq/Ci$:
$R = \frac{1.5056 \times 10^{14}}{3.7 \times 10^{10}} \approx 4069 \, Ci$.
Rounding to the nearest provided option,the answer is $4054 \, Ci$.

(B) Given: Half-life of $X$ $(T_{1/2, X})$ = Mean life of $Y$ $(\tau_Y)$.
We know that $T_{1/2, X} = \frac{\ln 2}{\lambda_X} \approx 0.693 \tau_X$.
Also, $\tau_Y = \frac{1}{\lambda_Y}$.
Since $T_{1/2, X} = \tau_Y$, we have $0.693 \tau_X = \tau_Y$, which implies $\tau_X > \tau_Y$.
Since the decay constant $\lambda = \frac{1}{\tau}$, it follows that $\lambda_Y > \lambda_X$.
The rate of decay is given by $R = \lambda N$. Since the initial number of atoms $N_0$ is the same for both, the element with the larger decay constant will decay faster.
Therefore, $Y$ decays faster than $X$.

(D) The decay constant for $\alpha$-decay is $\lambda_1 = \frac{\ln 2}{T_1}$.
The decay constant for $\beta$-decay is $\lambda_2 = \frac{\ln 2}{T_2}$.
When a radioactive nucleus can decay by two different processes,the total decay constant is the sum of the individual decay constants: $\lambda_{eq} = \lambda_1 + \lambda_2$.
Since $\lambda_{eq} = \frac{\ln 2}{T_{eq}}$,we have $\frac{\ln 2}{T_{eq}} = \frac{\ln 2}{T_1} + \frac{\ln 2}{T_2}$.
Dividing by $\ln 2$,we get $\frac{1}{T_{eq}} = \frac{1}{T_1} + \frac{1}{T_2} = \frac{T_1 + T_2}{T_1 T_2}$.
Therefore,the effective half-life is $T_{eq} = \frac{T_1 T_2}{T_1 + T_2}$.

(C) The radioactive decay law is given by $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $T_{1/2}$ is the half-life.
Given that $3/4$ of the sample decays,the remaining amount is $N(t) = N_0 - \frac{3}{4} N_0 = \frac{1}{4} N_0$.
Substituting this into the decay equation:
$\frac{1}{4} N_0 = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$
$\left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/T_{1/2}}$
Comparing the exponents,we get $2 = \frac{t}{T_{1/2}}$.
Given $t = 3/4 \, s$,we have $2 = \frac{3/4}{T_{1/2}}$.
Therefore,$T_{1/2} = \frac{3/4}{2} = 3/8 \, s$.

(C) Given: Half-life $T_{1/2} = 30$ days,Total time $t = 90$ days.
Number of half-lives $n = \frac{t}{T_{1/2}} = \frac{90}{30} = 3$.
The remaining amount of the radioactive substance is given by $N = N_0 \left(\frac{1}{2}\right)^n$.
Substituting the values: $N = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$.
The amount that has decayed is $N_{decayed} = N_0 - N = N_0 - \frac{N_0}{8} = \frac{7N_0}{8}$.
The percentage of decay is $\frac{N_{decayed}}{N_0} \times 100 = \frac{7}{8} \times 100 = 87.5\%$.

(A) The law of radioactive decay is given by the formula: $M = M_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$
Given:
Initial mass = $M_0$
Half-life $(t_{1/2})$ = $5 \text{ years}$
Total time $(t)$ = $15 \text{ years}$
Number of half-lives $(n)$ = $\frac{t}{t_{1/2}} = \frac{15}{5} = 3$
Substituting the values into the formula:
$M = M_0 \left( \frac{1}{2} \right)^3$
$M = M_0 \left( \frac{1}{8} \right)$
$M = \frac{M_0}{8}$

(A) The number of nuclei remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Given that $7/8$ of the nuclei have decayed,the remaining fraction is $1 - 7/8 = 1/8$.
So,$\frac{N}{N_0} = \frac{1}{8}$.
Substituting this into the formula: $\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.
Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$,we have $\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{15}{T_{1/2}}}$.
Equating the exponents: $3 = \frac{15}{T_{1/2}}$.
Therefore,$T_{1/2} = \frac{15}{3} = 5 \ min$.

(B) The formula for the remaining mass of a radioactive substance is given by $N = \frac{N_0}{2^n}$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Given: Initial mass $N_0 = 10.38 \, g$, half-life $T_{1/2} = 3.8 \, days$, and total time $t = 19 \, days$.
First, calculate the number of half-lives: $n = \frac{19}{3.8} = 5$.
Now, substitute the values into the formula: $N = \frac{10.38}{2^5}$.
$N = \frac{10.38}{32} = 0.324375 \, g$.
Rounding to the nearest provided option, the remaining mass is $0.32 \, g$.