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

(A) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N_0$ is the initial number of atoms,$N$ is the remaining number of atoms,$t$ is the time elapsed,and $T$ is the half-life period.
Given: $N_0 = 8 \times 10^4$,$N = 1 \times 10^4$,and $T = 3 \ years$.
Substituting the values into the formula:
$1 \times 10^4 = 8 \times 10^4 \left( \frac{1}{2} \right)^{t/3}$
$\frac{1}{8} = \left( \frac{1}{2} \right)^{t/3}$
Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$,we have:
$\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t/3}$
Comparing the exponents:
$3 = \frac{t}{3}$
$t = 9 \ years$.

(D) The formula for the remaining fraction of a radioactive sample is given by $\frac{N}{N_0} = (1/2)^n$,where $n$ is the number of half-lives.
Given,half-life $T_{1/2} = 1600$ years and total time $t = 6400$ years.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6400}{1600} = 4$.
Therefore,the fraction remaining is $\frac{N}{N_0} = (1/2)^4 = \frac{1}{16}$.

(C) The relationship between half-life $(T_{1/2})$ and average life $(\tau)$ is given by $\tau = \frac{T_{1/2}}{\ln(2)}$.
Given $T_{1/2} = 1600$ years.
Using $\ln(2) \approx 0.693$, we get $\tau = \frac{1600}{0.693}$.
Calculating this value, $\tau \approx 2308.8$ years.
Rounding to the nearest provided option, the average life is approximately $2319$ years.

(D) The fraction of radioactive atoms remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Given that the number of half-lives $n = 5$,the fraction remaining is $\frac{N}{N_0} = (\frac{1}{2})^5 = \frac{1}{32}$.
To find the percentage,we multiply the fraction by $100$:
Percentage $= \frac{1}{32} \times 100 = 3.125\%$.
Therefore,the correct option is $D$.

(C) The formula for the number of radioactive atoms remaining after time $t$ is given by $N_t = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N_0$ is the initial number of atoms,$t$ is the elapsed time,and $T$ is the half-life.
Given:
Initial number of atoms $N_0 = 50,000$
Elapsed time $t = 10 \text{ days}$
Half-life $T = 5 \text{ days}$
Substituting these values into the formula:
$N_t = 50,000 \times \left( \frac{1}{2} \right)^{10/5}$
$N_t = 50,000 \times \left( \frac{1}{2} \right)^2$
$N_t = 50,000 \times \frac{1}{4}$
$N_t = 12,500$
Therefore,the number of atoms left after $10$ days is $12,500$.

(C) The law of radioactive decay is given by $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $T$ is the half-life.
Given that the radioactivity drops to $1/64$ of its initial value in $t = 30 \, s$,we have $\frac{N}{N_0} = \frac{1}{64}$.
Substituting this into the formula: $\frac{1}{64} = \left( \frac{1}{2} \right)^{30/T}$.
Since $64 = 2^6$,we can write $\left( \frac{1}{2} \right)^6 = \left( \frac{1}{2} \right)^{30/T}$.
Equating the exponents: $6 = \frac{30}{T}$.
Therefore,$T = \frac{30}{6} = 5 \, s$.

(C) The decay constant $\lambda$ is a characteristic property of a radioactive isotope.
According to the law of radioactive decay,the rate of decay is proportional to the number of radioactive nuclei present at that instant,given by $\frac{dN}{dt} = -\lambda N$.
Here,$\lambda$ represents the probability of decay per unit time for a single nucleus.
This probability is a constant value for a specific radioactive substance and does not change with time or the age of the sample.
Therefore,$\lambda$ is independent of the age of the atoms.

(A) The average life $T$ (also denoted as $\tau$) of a radioactive substance is defined as the reciprocal of the decay constant $\lambda$.
Mathematically,$T = \frac{1}{\lambda}$.
Rearranging this equation,we get $T\lambda = 1$.
Therefore,the correct relation is $T\lambda = 1$.

(C) The fraction of a radioactive substance remaining after time $t$ is given by the formula:
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T}$ is the number of half-lives.
Given $t = \frac{T}{2}$,the number of half-lives is $n = \frac{T/2}{T} = \frac{1}{2}$.
Substituting this value into the formula:
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^{1/2} = \frac{1}{\sqrt{2}}$.

(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 $(\tau)$ is defined as the reciprocal of the decay constant.
It is given by the formula: $\tau = \frac{1}{\lambda}$.
Therefore,the half-life and mean life are $\frac{log_e 2}{\lambda}$ and $\frac{1}{\lambda}$ respectively.

(A) The law of radioactive decay is given by $N(t) = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N(t)$ is the remaining mass,$N_0$ is the initial mass,$t$ is the time elapsed,and $T$ is the half-life.
Given that $\frac{N(t)}{N_0} = \frac{1}{32}$ and $t = 60\, days$.
We can write $\frac{1}{32} = \left( \frac{1}{2} \right)^5$.
Therefore,$\left( \frac{1}{2} \right)^5 = \left( \frac{1}{2} \right)^{60/T}$.
Equating the exponents,we get $5 = \frac{60}{T}$.
Solving for $T$,we find $T = \frac{60}{5} = 12\, days$.

(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T$ is the half-life.
Given that $(7/8)^{th}$ of the sample decays,the remaining amount $N$ is $N_0 - \frac{7}{8}N_0 = \frac{1}{8}N_0$.
Substituting the values into the formula: $\frac{1}{8}N_0 = N_0 \left( \frac{1}{2} \right)^{t/5}$.
This simplifies to $\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t/5}$.
Comparing the exponents,we get $3 = t/5$,which implies $t = 15 \ days$.

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

(D) The relationship between the half-life $(T_{1/2})$ and the decay constant $(\lambda)$ is given by the formula: $T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.6931}{\lambda}$.
Given the decay constant $\lambda = 4.28 \times 10^{-4} \text{ year}^{-1}$.
Substituting the value into the formula:
$T_{1/2} = \frac{0.6931}{4.28 \times 10^{-4}} \text{ years}$.
$T_{1/2} = \frac{0.6931}{4.28} \times 10^{4} \text{ years}$.
$T_{1/2} \approx 0.161939 \times 10^{4} \text{ years}$.
$T_{1/2} \approx 1619.39 \text{ years}$.
Rounding to the nearest whole number,we get $1620 \text{ years}$.

(A) The initial amount of radioactive material is $N_0 = 16\, g$.
The half-life of the material is $T_{1/2} = 2\, days$.
The total time elapsed is $t = 32\, days$.
The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{32}{2} = 16$.
The amount of radioactive material remaining $N$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values, we get $N = 16 \times (\frac{1}{2})^{16} = 2^4 \times (\frac{1}{2})^{16} = (\frac{1}{2})^{12}$.
Since $(\frac{1}{2})^{10} = \frac{1}{1024} \approx 10^{-3} = 1\, mg$, then $(\frac{1}{2})^{12} = \frac{1}{4096}\, g \approx 0.244\, mg$.
Therefore, the remaining amount is less than $1\, mg$.

(C) The half-life of the radio-isotope is $T_{1/2} = 5$ years.
The total time elapsed is $t = 15$ years.
The number of half-lives $n$ is given by $n = t / T_{1/2} = 15 / 5 = 3$.
The fraction of atoms remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.
Substituting the values,$N/N_0 = (1/2)^3 = 1/8$.
The fraction of atoms that decay is given by $1 - N/N_0$.
Therefore,the decayed fraction $= 1 - 1/8 = 7/8$.

(D) The law of radioactive decay is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the number of nuclei remaining at time $t$,$N_0$ is the initial number of nuclei,and $\lambda$ is the decay constant.
For all nuclei to disintegrate,we require $N(t) = 0$.
Setting $N_0 e^{-\lambda t} = 0$,we find that this condition is only satisfied as $t \to \infty$.
Since the decay process is exponential,the number of nuclei approaches zero asymptotically but never reaches zero in a finite amount of time.
Therefore,the time taken for all nuclei to disintegrate is theoretically infinite,making the answer uncertain or undefined in a practical sense.

(C) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.
Given: $N_0 = 16 \, g$,$N = 1 \, g$,and $T_{1/2} = 140 \, days$.
Substituting the values: $\frac{1}{16} = \left( \frac{1}{2} \right)^{t/140}$.
Since $\frac{1}{16} = \left( \frac{1}{2} \right)^4$,we have $\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{t/140}$.
Equating the exponents: $4 = \frac{t}{140}$.
Therefore,$t = 4 \times 140 = 560 \, days$.

(B) The rate of decay $R$ is proportional to the number of radioactive nuclei $N$,i.e.,$R = \frac{dN}{dt} \propto N$.
Given the initial rate $R_1 = 200$ particles/s and final rate $R_2 = 25$ particles/s after time $t = 3$ hours.
The ratio of rates is given by $\frac{R_2}{R_1} = \frac{N_2}{N_1} = \left( \frac{1}{2} \right)^n$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.
Substituting the values: $\frac{25}{200} = \frac{1}{8} = \left( \frac{1}{2} \right)^3$.
Comparing the exponents,we get $n = 3$.
Since $n = \frac{t}{T_{1/2}}$,we have $3 = \frac{3 \text{ hours}}{T_{1/2}}$.
Therefore,$T_{1/2} = 1$ hour.
Since $1$ hour = $60$ minutes,the half-life period is $60$ minutes.

(D) The relationship between the half-life period $(T_{1/2})$ and the decay constant $(\lambda)$ is given by the formula:
$T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda}$.
Given the decay constant $\lambda = 0.01 \ s^{-1}$.
Substituting the value into the formula:
$T_{1/2} = \frac{0.693}{0.01} = 69.3 \ s$.
Therefore,the half-life period is $69.3 \ s$.

(D) Radioactive decay is a spontaneous and random process.
There is no fixed time interval between the decay of two successive nuclei.
While the half-life represents the statistical average time for half of the sample to decay,the decay of an individual nucleus is governed by probability.
Therefore,the next nucleus can decay at any time after the current one.
Thus,the correct option is $D$.

(D) The fraction of the isotope that has decayed is $\frac{7}{8}$.
Therefore,the fraction of the isotope that remains undecayed is $N/N_0 = 1 - \frac{7}{8} = \frac{1}{8}$.
Using the radioactive decay law: $\frac{N}{N_0} = (\frac{1}{2})^{t/T}$,where $T = 15 \, hrs$ is the half-life.
Substituting the values: $\frac{1}{8} = (\frac{1}{2})^{t/15}$.
Since $\frac{1}{8} = (\frac{1}{2})^3$,we have $(\frac{1}{2})^3 = (\frac{1}{2})^{t/15}$.
Equating the exponents: $3 = \frac{t}{15}$.
Solving for $t$: $t = 3 \times 15 = 45 \, hrs$.

(A) The relationship between mean life $(\tau)$ and half-life $(T_{1/2})$ is given by the formula: $\tau = \frac{T_{1/2}}{\ln(2)}$.
Given that $T_{1/2} = 10\, hours$ and $\ln(2) \approx 0.6931$.
Substituting the values: $\tau = \frac{10}{0.6931} \approx 14.427\, hours$.
Rounding to one decimal place,we get $\tau = 14.4\, hours$.

(B) The half-life $T_{1/2}$ is the time taken for a substance to reduce to half its initial mass.
Given that $20 \, g$ reduces to $10 \, g$ in $4 \, minutes$, the half-life $T_{1/2} = 4 \, minutes$.
We use the radioactive decay formula: $M = M_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$, where $M$ is the final mass, $M_0$ is the initial mass, and $t$ is the time elapsed.
Substituting the values $M = 10 \, g$, $M_0 = 80 \, g$, and $T_{1/2} = 4 \, minutes$:
$10 = 80 \left( \frac{1}{2} \right)^{t / 4}$
$\frac{10}{80} = \left( \frac{1}{2} \right)^{t / 4}$
$\frac{1}{8} = \left( \frac{1}{2} \right)^{t / 4}$
Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$, we have:
$\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t / 4}$
Equating the exponents: $3 = \frac{t}{4}$
$t = 12 \, minutes$.

(C) The radioactive decay law is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.
Given: $N_0 = 16\, g$,$N = 1\, g$,and $t = 2\, hours$.
Substituting these values into the formula:
$1 = 16 \left( \frac{1}{2} \right)^{2/T_{1/2}}$
$\frac{1}{16} = \left( \frac{1}{2} \right)^{2/T_{1/2}}$
$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{2/T_{1/2}}$
Equating the exponents:
$4 = \frac{2}{T_{1/2}}$
$T_{1/2} = \frac{2}{4} = \frac{1}{2}\, hour$.

(D) Radioactivity is defined as the rate of disintegration of a radioactive substance.
One Rutherford $(Rd)$ is defined as the quantity of a radioactive substance that undergoes $10^6$ disintegrations per second.
Therefore,$1 \ Rd = 1.0 \times 10^6 \text{ disintegrations/sec}$.
This unit is named after Ernest Rutherford.
Comparing this with the given options,the correct value is $1.0 \times 10^6 \text{ disintegrations/sec}$.

(B) The radioactive decay law is given by the formula: $A = A_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$
Given:
Initial time $t = 2\, hours = 120\, minutes$
Half-life $T_{1/2} = 30\, minutes$
Final count rate $A = 5\, s^{-1}$
Number of half-lives $n = \frac{t}{T_{1/2}} = \frac{120}{30} = 4$
Substituting the values into the formula:
$5 = A_0 \left( \frac{1}{2} \right)^4$
$5 = A_0 \left( \frac{1}{16} \right)$
$A_0 = 5 \times 16 = 80\, s^{-1}$
Therefore,the initial count rate was $80\, s^{-1}$.

(B) The number of atoms remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Given,half-life $T_{1/2} = 60\, minutes = 1\, hour$.
Total time $t = 3\, hours$.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{3}{1} = 3$.
The fraction of atoms remaining is $\frac{N}{N_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
The fraction of atoms that have decayed is $1 - \frac{N}{N_0} = 1 - \frac{1}{8} = \frac{7}{8}$.
In percentage,this is $\frac{7}{8} \times 100 = 87.5\%$.

(A) Radioactive carbon dating is a method used to determine the age of organic materials by measuring the amount of the radioactive isotope $C-14$ present in the sample.
The half-life of $C-14$ is approximately $5730$ years.
Because of this specific half-life,it is the standard isotope used for dating archaeological artifacts that are thousands of years old.
Therefore,the correct option is $A$.

(B) The law of radioactive decay is given by the formula: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T}$, where $N$ is the remaining amount, $N_0$ is the initial amount, $t$ is the elapsed time, and $T$ is the half-life.
Given that $\frac{N}{N_0} = \frac{1}{16}$ and $t = 2 \, hours$.
Substituting these values into the formula:
$\frac{1}{16} = \left( \frac{1}{2} \right)^{2/T}$
Since $\frac{1}{16} = \left( \frac{1}{2} \right)^4$, we have:
$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^{2/T}$
Equating the exponents:
$4 = \frac{2}{T}$
$T = \frac{2}{4} = 0.5 \, hours$.
Converting to minutes: $0.5 \times 60 = 30 \, minutes$.

(C) The rate of decay of a radioactive substance is given by the law of radioactive decay: $\frac{dN}{dt} = -\lambda N$.
Here,the magnitude of the rate of emission of alpha particles is given as $n = |\frac{dN}{dt}|$.
Therefore,$n = \lambda N$.
From this,the decay constant $\lambda$ is calculated as $\lambda = \frac{n}{N}$.
The half-life $T_{1/2}$ of a radioactive element is related to the decay constant by the formula $T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda}$.
Substituting the value of $\lambda$ into the half-life formula,we get $T_{1/2} = \frac{0.693}{n/N} = \frac{0.693 N}{n} \, s$.

(C) The radioactive decay law is given by $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T_{1/2}$ is the half-life.
Given $A_0 = 1600$ counts/sec at $t = 0$ and $A = 100$ counts/sec at $t = 8$ seconds.
Substituting these values: $100 = 1600 \left( \frac{1}{2} \right)^{\frac{8}{T_{1/2}}}$.
$\frac{100}{1600} = \left( \frac{1}{2} \right)^{\frac{8}{T_{1/2}}} \implies \frac{1}{16} = \left( \frac{1}{2} \right)^{\frac{8}{T_{1/2}}}$.
Since $\frac{1}{16} = (\frac{1}{2})^4$,we have $\frac{8}{T_{1/2}} = 4$,which gives $T_{1/2} = 2$ seconds.
Now,to find the counting rate at $t = 6$ seconds:
$A = 1600 \left( \frac{1}{2} \right)^{\frac{6}{2}} = 1600 \left( \frac{1}{2} \right)^3$.
$A = 1600 \times \frac{1}{8} = 200$ counts/sec.

(B) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
By definition,at half-life $T_{1/2}$,the number of radioactive nuclei remaining is half of the initial amount,i.e.,$N(T_{1/2}) = \frac{N_0}{2}$.
Substituting this into the decay law: $\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}$.
$\frac{1}{2} = e^{-\lambda T_{1/2}}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -\lambda T_{1/2}$.
$-\ln 2 = -\lambda T_{1/2}$.
Therefore,$T_{1/2} = \frac{\ln 2}{\lambda}$.

(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$.
By definition,at half-life $t = T$,the number of nuclei remaining is $N = \frac{N_0}{2}$.
Substituting these values into the equation:
$\frac{N_0}{2} = N_0 e^{-\lambda T}$
$\frac{1}{2} = e^{-\lambda T}$
$2 = e^{\lambda T}$
Taking the natural logarithm $(\ln)$ on both sides:
$\ln(2) = \lambda T$
Since $\ln(2) \approx 0.693$,we get:
$\lambda T = 0.693$.

(A) The number of half-lives $n$ in a total time $t$ is given by $n = \frac{t}{T_{1/2}}$.
Given $t = 20\, min$ and $T_{1/2} = 5\, min$,we have $n = \frac{20}{5} = 4$.
The fraction of the substance remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.
Substituting $n = 4$,we get $N/N_0 = (1/2)^4 = 1/16$.
The fraction of the substance that has decayed is $1 - N/N_0 = 1 - 1/16 = 15/16$.
To express this as a percentage,we calculate $(15/16) \times 100\% = 93.75\%$.

(D) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$,where $t = 12\, days$ and $T_{1/2} = 4\, days$.
Thus,$n = \frac{12}{4} = 3$.
The number of atoms remaining $N$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$,where $N_0 = 6.4 \times 10^{10}$.
Substituting the values,we get $N = 6.4 \times 10^{10} \times (\frac{1}{2})^3$.
$N = 6.4 \times 10^{10} \times \frac{1}{8} = 0.8 \times 10^{10}$ atoms.

(A) The decay constant $(\lambda)$ is a characteristic property of the radioactive nucleus itself.
It depends only on the nature of the nucleus and is independent of the chemical state or the physical environment of the atom.
Since radium bromide is a chemical compound containing radium atoms,the radium nuclei within the compound remain unchanged.
Therefore,the decay constant of radium bromide is the same as that of pure radium,which is $\lambda$.

(C) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$.
For substance $A$,$n_A = \frac{80}{20} = 4$.
For substance $B$,$n_B = \frac{80}{40} = 2$.
The number of remaining nuclei is given by $N = N_0 \left( \frac{1}{2} \right)^n$.
Since initial numbers $N_0$ are equal,the ratio is $\frac{N_A}{N_B} = \frac{N_0 (1/2)^{n_A}}{N_0 (1/2)^{n_B}} = \frac{(1/2)^4}{(1/2)^2} = \frac{2^2}{2^4} = \frac{4}{16} = \frac{1}{4}$.
Thus,the ratio is $1 : 4$.

(C) The half-life period $T_{1/2}$ of a radioactive substance is related to the decay constant $\lambda$ by the formula: $T_{1/2} = \frac{0.693}{\lambda}$.
Given the decay constant $\lambda = 1.07 \times 10^{-4} \text{ year}^{-1}$.
Substituting the value into the formula:
$T_{1/2} = \frac{0.693}{1.07 \times 10^{-4}}$
$T_{1/2} = \frac{0.693}{1.07} \times 10^4$
$T_{1/2} \approx 0.6476 \times 10^4 = 6476 \text{ years}$.
Thus,the half-life period is $6476 \text{ years}$.

(D) Radioactive decay is a stochastic (probabilistic) process.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$,where $\lambda$ is the decay constant.
This equation implies that the probability of a nucleus decaying is distributed over time from $t = 0$ to $t = \infty$.
Therefore,it is impossible to predict exactly when a specific nucleus will decay. $A$ given nucleus has a non-zero probability of decaying at any instant $t > 0$.
Thus,option $(D)$ is correct.

(A) The radioactive decay law states that the amount of substance remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$.
Given that the ratio of Carbon-$14$ to Carbon-$12$ is $\frac{1}{4}$ of the original amount,we have $\frac{N(t)}{N_0} = \frac{1}{4}$.
Since $\frac{1}{4} = (1/2)^2$,it implies that the time elapsed $t$ is equal to two half-lives.
$t = 2 \times T_{1/2} = 2 \times 5,800$ years.
Since $1$ century $= 100$ years,$t = 2 \times 58$ centuries.
Therefore,$x = 2 \times 58$.

(A) The age of organic materials like trees is determined using the radio-isotope of Carbon,specifically Carbon-$14$ $(^{14}C)$.
This process is known as Carbon dating or radiocarbon dating.
Living organisms absorb Carbon-$14$ from the atmosphere during their lifetime.
Once the organism dies,it stops taking in Carbon-$14$,and the existing amount begins to decay at a known rate (half-life of approximately $5730$ years).
By measuring the remaining ratio of Carbon-$14$ to stable Carbon-$12$,scientists can calculate the time elapsed since the organism's death.

(C) Statement $(I)$ is true: Radioactive decay follows the law $N(t) = N_0 e^{-\lambda t}$,which is an exponential decay.
Statement $(II)$ is true: By definition,the half-life $(T_{1/2})$ is the time taken for the number of radioactive nuclei to reduce to half of their initial value.
Statement $(III)$ is true: Radioactive dating (e.g.,uranium-lead dating) is a standard method used to determine the age of rocks and the Earth.
Statement $(IV)$ is false: The relationship between half-life $(T_{1/2})$ and mean life $(\tau)$ is $T_{1/2} = \tau \ln(2) \approx 0.693 \tau$. Thus,the half-life is approximately $69.3\%$ of the mean life,not $50\%$.
Therefore,statements $(I), (II),$ and $(III)$ are correct. The correct option is $C$.

(C) The radioactive decay law is given by $N(t) = N_0 (1/2)^{t/T_{1/2}}$,where $T_{1/2} = 10 \, days$.
We want to find $t$ such that $N(t) = N_0 / 10$.
Substituting the values: $N_0 / 10 = N_0 (1/2)^{t/10}$.
$1/10 = (1/2)^{t/10}$.
Taking the logarithm on both sides: $\log_{10}(1/10) = (t/10) \log_{10}(1/2)$.
$-1 = (t/10) \times (-0.3010)$.
$t = 10 / 0.3010 \approx 33.22 \, days$.
Therefore,the time is approximately $33 \, days$.

(D) The activity $A$ of a radioactive sample at time $t$ is given by the formula $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A_0$ is the initial activity and $T_{1/2}$ is the half-life.
Given: $T_{1/2} = 3 \ hours$ and $t = 9 \ hours$.
The number of half-lives elapsed is $n = \frac{t}{T_{1/2}} = \frac{9}{3} = 3$.
Substituting these values into the formula:
$\frac{A}{A_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
Thus,the activity becomes $1/8$ of its initial value.

(B) The activity of a radioactive sample at any time $t$ is given by $A = A_0 (1/2)^n$, where $n$ is the number of half-lives.
Given, half-life $T_{1/2} = 1$ month.
We need to find the activity $2$ months earlier, so $n = 2$ half-lives.
Let $A_{initial}$ be the activity $2$ months earlier and $A_{final} = 2 \, \mu Ci$ be the activity on $1-8-1991$.
Since $A_{final} = A_{initial} \times (1/2)^n$, we have $2 = A_{initial} \times (1/2)^2$.
$2 = A_{initial} \times (1/4)$.
$A_{initial} = 2 \times 4 = 8 \, \mu Ci$.
Therefore, the activity two months earlier was $8 \, \mu Ci$.

(A) The number of half-lives in two days ($48$ hours) for substance $1$ and $2$ are $n_1 = \frac{48}{12} = 4$ and $n_2 = \frac{48}{16} = 3$.
Using the radioactive decay formula $N = N_0 \left( \frac{1}{2} \right)^n$,the ratio of the remaining amounts is:
$\frac{N_1}{N_2} = \frac{(N_0)_1}{(N_0)_2} \times \frac{(1/2)^{n_1}}{(1/2)^{n_2}}$
Substituting the given values:
$\frac{N_1}{N_2} = \frac{2}{1} \times \frac{(1/2)^4}{(1/2)^3} = 2 \times \frac{1}{2} = 1$.
Thus,the ratio is $1 : 1$.

(C) The relationship between the decay constant $\lambda$ and the half-life $T_{1/2}$ is given by the formula: $\lambda = \frac{0.693}{T_{1/2}}$.
Given that $T_{1/2} = 2.3 \, days$.
Substituting the value into the formula: $\lambda = \frac{0.693}{2.3} = 0.3 \, day^{-1}$.
Therefore,the correct option is $C$.

(D) The number of half-lives $n$ in a time $t$ is given by $n = \frac{t}{T_{1/2}}$.
Given $t = 150\, \text{years}$ and $T_{1/2} = 75\, \text{years}$, we have $n = \frac{150}{75} = 2$.
The fraction of atoms remaining after $n$ half-lives is given by $N/N_0 = (1/2)^n$.
Substituting $n = 2$, we get $N/N_0 = (1/2)^2 = 1/4 = 0.25$.
The fraction of atoms that have decayed is $1 - N/N_0 = 1 - 0.25 = 0.75$.
To express this as a percentage, we multiply by $100$, which gives $0.75 \times 100 = 75\%$.

(B) The radioactive decay law is given by $A = A_0 e^{-\lambda t}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $\lambda$ is the decay constant.
Given: $A_0 = 9750$ counts/min,$A = 975$ counts/min,and $t = 5$ minutes.
Substituting the values: $975 = 9750 e^{-\lambda \times 5}$.
Dividing both sides by $9750$: $0.1 = e^{-5\lambda}$,which implies $e^{5\lambda} = 10$.
Taking the natural logarithm on both sides: $5\lambda = \ln(10)$.
Since $\ln(10) \approx 2.3026$,we have $5\lambda = 2.3026$.
Therefore,$\lambda = \frac{2.3026}{5} = 0.46052 \approx 0.461$ per minute.