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

(D) The radioactive decay law is given by $A = A_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$.
Given: Initial activity $A_0 = 64 \times 10^{-5} \, Ci$,final activity $A = 5 \times 10^{-6} \, Ci$,and half-life $T_{1/2} = 3 \, days$.
Substituting the values into the formula:
$5 \times 10^{-6} = 64 \times 10^{-5} \left( \frac{1}{2} \right)^{t/3}$
Divide 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:
$7 = \frac{t}{3} \Rightarrow t = 21 \, days$.

(C) The fraction of the sample that has decayed is $\frac{3}{4}$.
Therefore,the fraction of the sample that remains undecayed is $N/N_0 = 1 - 3/4 = 1/4$.
Using the radioactive decay formula $N/N_0 = (1/2)^n$,where $n$ is the number of half-lives:
$1/4 = (1/2)^n$
$(1/2)^2 = (1/2)^n$
This gives $n = 2$.
The total time $t$ is given by $t = n \times T_{1/2}$.
Substituting the values,$t = 2 \times 3.8 \ days = 7.6 \ days$.

(A) The radioactive decay law is given by $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{8}$,we have $\frac{1}{8} = (\frac{1}{2})^n$.
Since $\frac{1}{8} = (\frac{1}{2})^3$,we find $n = 3$.
The total time $t$ is given by $t = n \times T_{1/2}$.
Substituting the values,$t = 3 \times 3.8 \ days = 11.4 \ days$.

(D) The number of half-lives $n$ is calculated by dividing the total time $t$ by the half-life $T_{1/2}$.
$n = \frac{t}{T_{1/2}} = \frac{72000}{24000} = 3$.
The fraction of the substance remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$.
Substituting $n = 3$, we get $\frac{N}{N_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.

(A) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$,where $t$ is the total time and $T_{1/2}$ is the half-life.
Given $t = 5 \text{ years}$ and $T_{1/2} = 1 \text{ year}$,we have $n = \frac{5}{1} = 5$.
The fraction of the radioactive material remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting the value of $n$,we get $\frac{N}{N_0} = (\frac{1}{2})^5 = \frac{1}{32}$.

(D) Radioactive decay is a nuclear phenomenon that depends solely on the nature of the nucleus. It is independent of external physical conditions such as temperature,pressure,or chemical environment. Therefore,the activity of a radioactive sample cannot be altered by heating or any other physical or chemical method. Thus,both statements $(b)$ and $(c)$ are correct.

(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{10}{5} = 2$.
The fraction of undecayed nuclei remaining is $\frac{N}{N_0} = (\frac{1}{2})^n = (\frac{1}{2})^2 = \frac{1}{4}$.
The fraction of nuclei that have decayed is $1 - \frac{N}{N_0} = 1 - \frac{1}{4} = \frac{3}{4}$.
To express this as a percentage,multiply by $100$: $\frac{3}{4} \times 100 = 75\%$.
Therefore,the probability of decay in $10$ years is $75\%$.

(B) The number of half-lives $n$ is given by the ratio of the total time $t$ to the half-life $T_{1/2}$.
$n = \frac{t}{T_{1/2}} = \frac{19}{3.8} = 5$.
Using the radioactive decay formula,the remaining quantity $N$ is given by $N = N_0 \times (1/2)^n$.
Substituting the given values,$N = 10.38 \times (1/2)^5$.
$N = 10.38 \times \frac{1}{32}$.
$N = 0.324375\, gm$.
Rounding to two decimal places,the remaining quantity is $0.32\, gm$.

(B) The law of radioactive decay states that the remaining fraction of a radioactive substance after $n$ half-lives is given by the formula:
$\frac{N}{N_0} = (\frac{1}{2})^n$
Here,$n$ represents the number of half-lives.
Given that the number of half-lives $n = 5$,we substitute this value into the formula:
$\frac{N}{N_0} = (\frac{1}{2})^5$
Thus,the fraction of the initial substance remaining after five half-lives is $(\frac{1}{2})^5$.

(D) The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the equation $N(t) = N_0 e^{-\lambda t}$.
At $t = T_{1/2}$,$N(t) = N_0 / 2$.
Substituting this,we get $N_0 / 2 = N_0 e^{-\lambda T_{1/2}}$,which simplifies to $1/2 = e^{-\lambda T_{1/2}}$.
Taking the natural logarithm on both sides,$\ln(1/2) = -\lambda T_{1/2}$,or $-\ln 2 = -\lambda T_{1/2}$.
Thus,$T_{1/2} = \frac{\ln 2}{\lambda}$.
Since $\ln 2 = 2.303 \log_{10} 2$,the expression becomes $T_{1/2} = \frac{2.303 \log_{10} 2}{\lambda}$.

(D) The fraction of radioactive material remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$.
Given that the time period is $n = 2$ half-lives.
Substituting the value of $n$,we get $\frac{N}{N_0} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
This represents the fraction of the material that remains undecayed.
The fraction of the material that has disintegrated is given by $1 - \frac{N}{N_0}$.
Therefore,the disintegrated fraction $= 1 - \frac{1}{4} = \frac{3}{4}$.

(C) The number of half-lives $n$ is calculated as $n = \frac{t}{T_{1/2}} = \frac{10 \, days}{2.5 \, days} = 4$.
The activity $A$ after time $t$ is given by the formula $A = A_0 \left( \frac{1}{2} \right)^n$,where $A_0$ is the initial activity.
Substituting the given values: $A = 1.6 \times \left( \frac{1}{2} \right)^4$.
$A = 1.6 \times \frac{1}{16} = 0.1 \, curie$.
Therefore,the activity after $10 \, days$ will be $0.1 \, curie$.

(B) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.
The mean life $\tau$ is defined as $\tau = 1/\lambda$.
At time $t = \tau$,the amount of substance remaining is $N = N_0 e^{-\lambda(1/\lambda)} = N_0 e^{-1} = N_0/e$.
Since $e \approx 2.718$,we have $N \approx N_0 / 2.718 \approx 0.368 N_0 \approx N_0 / 3$.
The amount of substance that has disintegrated is $N_{dis} = N_0 - N = N_0 - N_0/e = N_0(1 - 1/e)$.
$N_{dis} \approx N_0(1 - 0.368) = 0.632 N_0 \approx 2/3 N_0$.
Thus,about $2/3$ of the substance disintegrates.

(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.
If $3/4$ of the nuclei have decayed,the fraction remaining is $1 - 3/4 = 1/4$.
Thus,$N/N_0 = 1/4 = (1/2)^2$.
Comparing this with $(1/2)^n$,we get $n = 2$.
Since $n = t / T_{1/2}$,we have $2 = (3/4) / T_{1/2}$.
Therefore,$T_{1/2} = (3/4) / 2 = 3/8 \, s$.

(C) The number of radioactive nuclei remaining after time $t$ is given by $N = N_0 e^{-\lambda t}$.
The mean life (average life) of a radioactive element is defined as $T_{avg} = \frac{1}{\lambda}$.
Substituting $t = \frac{1}{\lambda}$ into the decay equation,we get the number of nuclei remaining:
$N = N_0 e^{-\lambda (1/\lambda)} = N_0 e^{-1} = \frac{N_0}{e}$.
The fraction of nuclei remaining is $\frac{N}{N_0} = \frac{1}{e}$.
The fraction that disintegrates is given by $1 - \frac{N}{N_0}$.
Therefore,the disintegrated fraction $= 1 - \frac{1}{e} = \frac{e - 1}{e}$.

(A) The half-life of a radioactive substance is the time required for half of the radioactive nuclei in a sample to decay.
To determine the half-life,one must measure the activity of the sample over a period of time.
The activity of a radioactive source is defined as the rate of decay,which is the number of disintegrations per unit time.
$A$ $Geiger-Muller$ counter is a device specifically designed to detect and measure ionizing radiation by counting the number of particles or photons emitted during radioactive decay.
By recording the count rate at different time intervals,the decay constant and subsequently the half-life can be calculated.
Therefore,the $Geiger-Muller$ counter is the correct instrument for this measurement.

(A) The radioactive decay law is given by $\frac{N}{N_0} = \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{16}$,we can write $\frac{1}{16} = \left( \frac{1}{2} \right)^4$.
Comparing the two expressions,we get $n = 4$.
The number of half-lives $n$ is related to the total time $t$ and half-life $T_{1/2}$ by $n = \frac{t}{T_{1/2}}$.
Given $t = 40$ days and $n = 4$,we have $4 = \frac{40}{T_{1/2}}$.
Therefore,$T_{1/2} = \frac{40}{4} = 10$ days.

(A) The amount of radioactive substance remaining after $n$ half-lives is given by $N = N_0 \left( \frac{1}{2} \right)^n$.
If $99\%$ of the element decays,the amount remaining is $N = N_0 - 0.99 N_0 = 0.01 N_0$.
Substituting this into the equation: $0.01 N_0 = N_0 \left( \frac{1}{2} \right)^n$.
$\frac{1}{100} = \left( \frac{1}{2} \right)^n$.
Taking the reciprocal: $100 = 2^n$.
We know that $2^6 = 64$ and $2^7 = 128$.
Since $64 < 100 < 128$,the value of $n$ must lie between $6$ and $7$ half-lives.

(D) The formula for radioactive decay is given by $N = N_0 \left( \frac{1}{2} \right)^n$, where $n = \frac{t}{T_{1/2}}$.
Given:
Initial amount $N_0 = 1 \, mg$
Total time $t = 8.1 \, \text{days}$
Half-life $T_{1/2} = 2.7 \, \text{days}$
First, calculate the number of half-lives $n$:
$n = \frac{8.1}{2.7} = 3$
Now, substitute the values into the decay formula:
$N = 1 \times \left( \frac{1}{2} \right)^3$
$N = \frac{1}{8} = 0.125 \, mg$
Thus, the amount left after $8.1 \, \text{days}$ is $0.125 \, mg$.

(C) The mean life $(\tau)$ of a radioactive element is defined as the reciprocal of its decay constant $(\lambda)$.
Given,$\lambda = 1.5 \times 10^{-9} \text{ s}^{-1}$.
The formula for mean life is $\tau = \frac{1}{\lambda}$.
Substituting the value of $\lambda$:
$\tau = \frac{1}{1.5 \times 10^{-9}} \text{ s}$.
$\tau = \frac{1}{1.5} \times 10^9 \text{ s}$.
$\tau = 0.6666... \times 10^9 \text{ s}$.
$\tau = 6.67 \times 10^8 \text{ s}$.
Therefore,the mean life is $6.67 \times 10^8 \text{ s}$.

(C) The amount of radioactive material remaining after time $t$ is given by the formula $N = N_0 \left( \frac{1}{2} \right)^{t/T}$, where $N_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life.
Given: $N_0 = 10 \, g$, $t = 20 \, \text{years}$, $T = 15 \, \text{years}$.
Substituting the values: $N = 10 \times (2)^{-20/15} = 10 \times (2)^{-4/3} = 10 \times (2)^{-1.333}$.
Calculating the value: $2^{1.333} \approx 2.5198$.
So, $N = 10 / 2.5198 \approx 3.968 \, g$.
The amount of disintegrated (decayed) material is $N_{\text{decayed}} = N_0 - N$.
$N_{\text{decayed}} = 10 - 3.968 = 6.032 \, g$.
Rounding to the nearest provided option, the disintegrated material is $6.04 \, g$.

(B) The number of radioactive nuclei remaining undecayed after time $t$ is given by $N = N_0 e^{-\lambda t}$.
The number of nuclei that have decayed after time $t$ is $N_d = N_0 - N = N_0(1 - e^{-\lambda t})$.
One mean life is defined as $\tau = \frac{1}{\lambda}$.
Substituting $t = \tau = \frac{1}{\lambda}$ into the decay equation:
$N_d = N_0(1 - e^{-\lambda \times \frac{1}{\lambda}}) = N_0(1 - e^{-1}) = N_0(1 - \frac{1}{e})$.
Using the value $e \approx 2.718$,we get $\frac{1}{e} \approx 0.368$.
Therefore,$N_d = N_0(1 - 0.368) = N_0(0.632) = 63.2\% \text{ of } N_0$.
Thus,approximately $63\%$ of the initial nuclei will decay during one mean life.

(D) The $S.I.$ unit of radioactivity is the becquerel, denoted by the symbol $Bq$.
One becquerel is defined as the activity of a quantity of radioactive material in which one nucleus decays per second.
Mathematically, $1 \ Bq = 1 \ \text{disintegration per second} \ (dps)$.
Other units like curie $(Ci)$ and rutherford $(Rd)$ are also used, where $1 \ Ci = 3.7 \times 10^{10} \ Bq$ and $1 \ Rd = 10^6 \ Bq$.

(B) The relation between the remaining amount $N$ and initial amount $N_0$ after $n$ half-lives is given by:
$N = N_0 \left( \frac{1}{2} \right)^n$
Given $N_0 = 16 \, gm$ and $N = 1 \, gm$:
$1 = 16 \left( \frac{1}{2} \right)^n$
$\frac{1}{16} = \left( \frac{1}{2} \right)^n$
$\left( \frac{1}{2} \right)^4 = \left( \frac{1}{2} \right)^n$
So,the number of half-lives $n = 4$.
We know that $n = \frac{t}{T_{1/2}}$,where $t$ is the total time and $T_{1/2}$ is the half-life.
$4 = \frac{120}{T_{1/2}}$
$T_{1/2} = \frac{120}{4} = 30 \, days$.

(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 $T_{1/2} = 10 \text{ years}$ and $N = \frac{1}{4} N_0$.
Substituting these values: $\frac{1}{4} N_0 = N_0 \left( \frac{1}{2} \right)^{t/10}$.
$\frac{1}{4} = \left( \frac{1}{2} \right)^{t/10}$.
Since $\frac{1}{4} = \left( \frac{1}{2} \right)^2$,we have $\left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/10}$.
Comparing the exponents: $2 = \frac{t}{10}$.
Therefore,$t = 20 \text{ years}$.

(A) The formula for the remaining amount of a radioactive substance is given by $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Given the initial mass is $N_0$,the half-life period $T_{1/2} = 5 \text{ years}$,and the time elapsed $t = 15 \text{ years}$.
Substituting these values into the formula:
$N = N_0 \left( \frac{1}{2} \right)^{\frac{15}{5}}$
$N = N_0 \left( \frac{1}{2} \right)^3$
$N = N_0 \left( \frac{1}{8} \right) = \frac{N_0}{8}$.
Thus,the amount of substance left after $15 \text{ years}$ is $\frac{N_0}{8}$.

(B) 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 that the activity becomes $\frac{1}{64}$ of its original value in $t = 60 \ s$,we have:
$\frac{A}{A_0} = \frac{1}{64} = \left( \frac{1}{2} \right)^{\frac{60}{T_{1/2}}}$
Since $\frac{1}{64} = \left( \frac{1}{2} \right)^6$,we can equate the exponents:
$6 = \frac{60}{T_{1/2}}$
$T_{1/2} = \frac{60}{6} = 10 \ s$.
Therefore,the half-life period is $10 \ s$.

(C) The activity of a radioactive sample at any time $t$ is given by the law of radioactive decay: $A = A_0 e^{-\lambda t}$.
Since the mean life $T = 1/\lambda$,we can write the equation as $A = A_0 e^{-t/T}$.
At time $t_1$,the activity is $A_1 = A_0 e^{-t_1/T}$. From this,we find $A_0 = A_1 e^{t_1/T}$.
At time $t_2$,the activity is $A_2 = A_0 e^{-t_2/T}$.
Substituting the expression for $A_0$ into the equation for $A_2$,we get:
$A_2 = (A_1 e^{t_1/T}) e^{-t_2/T} = A_1 e^{(t_1 - t_2)/T}$.
Thus,the correct relation is $A_2 = A_1 e^{(t_1 - t_2)/T}$.

(A) The law of radioactive decay states that the activity $A$ at time $t$ is given by $A = A_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given that the activity decays to $\frac{1}{16}$ of its initial value,we have $\frac{A}{A_0} = \frac{1}{16}$.
Since $\frac{1}{16} = (\frac{1}{2})^4$,we find that $n = 4$.
The total time taken $t$ is given by $t = n \times T_{1/2}$,where $T_{1/2} = 100 \mu s$.
Therefore,$t = 4 \times 100 \mu s = 400 \mu s$.

(D) The law of radioactive decay is given by $\frac{N}{N_0} = (\frac{1}{2})^{t/T}$,where $N$ is the remaining amount,$N_0$ is the initial amount,$t$ is the time elapsed,and $T$ is the half-life.
Given,$\frac{N}{N_0} = \frac{1}{16}$ and $T = 48 \text{ hours}$.
Substituting the values: $\frac{1}{16} = (\frac{1}{2})^{t/48}$.
Since $\frac{1}{16} = (\frac{1}{2})^4$,we have $(\frac{1}{2})^4 = (\frac{1}{2})^{t/48}$.
Equating the exponents: $4 = \frac{t}{48}$.
Therefore,$t = 4 \times 48 = 192 \text{ hours}$.

(C) The average life (mean life) $\tau$ is given as $5 \text{ hours}$.
The decay constant $\lambda$ is related to the average life by $\tau = \frac{1}{\lambda}$.
The number of radioactive nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
Substituting $t = \tau = \frac{1}{\lambda}$,we get $N(\tau) = N_0 e^{-\lambda(1/\lambda)} = N_0 e^{-1} = \frac{N_0}{e} \approx \frac{N_0}{2.718} \approx 0.368 N_0$.
The number of nuclei that have decayed is $N_{decayed} = N_0 - N(\tau) = N_0 - 0.368 N_0 = 0.632 N_0$.
This means $63.2\%$ of the nuclei have decayed,which is more than half $(50\%)$.
Therefore,the correct option is $C$.

(D) The mass of a radioactive sample at time $t$ is given by the decay law: $M = M_0 e^{-\lambda t}$.
Here,$M_0 = 10 \, g$ and the time $t$ is given as two mean lives,i.e.,$t = 2 \tau$,where $\tau = \frac{1}{\lambda}$ is the mean life.
Substituting $t = \frac{2}{\lambda}$ into the decay equation:
$M = 10 e^{-\lambda (2/\lambda)} = 10 e^{-2}$.
Using the value $e \approx 2.718$,we have $e^2 \approx 7.389$.
$M = \frac{10}{7.389} \approx 1.35 \, g$.
Therefore,the correct option is $D$.

(B) The disintegration rate $A$ at any time $t$ is given by $A = A_0 e^{-\lambda t}$.
Given,$A_0 = 5000$ disintegrations per minute at $t = 0$.
At $t = 5$ minutes,$A = 1250$ disintegrations per minute.
Substituting these values into the equation: $1250 = 5000 e^{-\lambda (5)}$.
Dividing both sides by $5000$: $\frac{1250}{5000} = e^{-5\lambda}$.
$\frac{1}{4} = e^{-5\lambda}$.
Taking the natural logarithm on both sides: $\ln(1/4) = -5\lambda$.
$-\ln(4) = -5\lambda$.
$\ln(2^2) = 5\lambda$.
$2 \ln 2 = 5\lambda$.
$\lambda = \frac{2}{5} \ln 2 = 0.4 \ln 2 \text{ per minute}$.

(B) The number of atoms remaining at time $t$ is given by the formula $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $N_0 = 8 \times 10^{10}$ and $T_{1/2} = 1 \text{ hour}$.
Number of atoms remaining at $t = 2 \text{ hours}$:
$N_1 = 8 \times 10^{10} \times \left( \frac{1}{2} \right)^{\frac{2}{1}} = 8 \times 10^{10} \times \frac{1}{4} = 2 \times 10^{10}$.
Number of atoms remaining at $t = 4 \text{ hours}$:
$N_2 = 8 \times 10^{10} \times \left( \frac{1}{2} \right)^{\frac{4}{1}} = 8 \times 10^{10} \times \frac{1}{16} = 0.5 \times 10^{10}$.
The number of atoms decayed between $t = 2 \text{ hours}$ and $t = 4 \text{ hours}$ is the difference between the number of atoms present at these two times:
$\Delta N = N_1 - N_2 = (2 - 0.5) \times 10^{10} = 1.5 \times 10^{10}$.

(B) Carbon dating is primarily used to determine the age of organic materials such as fossils.
It relies on the decay of the radioactive isotope Carbon-$14$ $(^{14}C)$,which has a half-life of approximately $5730$ years.
Due to the half-life of $^{14}C$,this method is effective for dating samples up to approximately $45,000$ to $50,000$ years old.
Therefore,the age range for which carbon dating is best suited is of the order of $10^4$ years.
For much older geological samples,other methods like the Potassium-Argon dating method are preferred.

(D) The radioactive decay law is given by $A = A_0 (1/2)^n$,where $n$ is the number of half-lives.
Given initial activity $A_0 = 240$ counts/min and final activity $A = 30$ counts/min.
Substituting the values: $30 = 240 (1/2)^n$.
$(1/2)^n = 30/240 = 1/8$.
$(1/2)^n = (1/2)^3$,which implies $n = 3$.
The total time elapsed is $t = 1$ hour = $60$ minutes.
Since $n = t / T_{1/2}$,we have $3 = 60 / T_{1/2}$.
Therefore,the half-life $T_{1/2} = 60 / 3 = 20$ minutes.

(B) The radioactive decay formula is given by $M = M_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Here, the initial mass $M_0 = 100 \, g$, the remaining mass $M = 25 \, g$, and the half-life $T_{1/2} = 1600 \, years$.
Substituting these values into the formula:
$25 = 100 \left( \frac{1}{2} \right)^{\frac{t}{1600}}$
$\frac{25}{100} = \left( \frac{1}{2} \right)^{\frac{t}{1600}}$
$\frac{1}{4} = \left( \frac{1}{2} \right)^{\frac{t}{1600}}$
Since $\frac{1}{4} = \left( \frac{1}{2} \right)^2$, we have:
$\left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{\frac{t}{1600}}$
Equating the exponents:
$2 = \frac{t}{1600}$
$t = 2 \times 1600 = 3200 \, years$.
Thus, the correct option is $B$.

(C) The activity of a radioactive substance is given by $R = R_0 e^{-\lambda t}$.
Given that after $t = 9$ years,the activity becomes $R = \frac{R_0}{3}$.
Substituting these values: $\frac{R_0}{3} = R_0 e^{-\lambda \times 9}$,which implies $e^{-9\lambda} = \frac{1}{3}$ ... $(i)$.
After a further $9$ years,the total time elapsed is $t = 9 + 9 = 18$ years.
The new activity $R'$ is given by $R' = R_0 e^{-\lambda \times 18}$.
This can be written as $R' = R_0 (e^{-9\lambda})^2$.
Substituting the value from equation $(i)$: $R' = R_0 \left(\frac{1}{3}\right)^2 = \frac{R_0}{9}$.

(C) The radioactive decay law states that the amount remaining after $n$ half-lives is given by $N = N_0 (1/2)^n$.
To reduce the substance to one-fourth of its original amount $(N = N_0/4)$, we have $(1/2)^n = 1/4$, which implies $n = 2$.
Since one half-life $(T_{1/2})$ is $40 \, years$, the total time taken is $t = n \times T_{1/2} = 2 \times 40 = 80 \, years$.
The decay constant $\lambda$ is given by the formula $\lambda = \frac{0.693}{T_{1/2}}$.
Substituting the given value: $\lambda = \frac{0.693}{40} = 0.0173 \, year^{-1}$.

(D) The formula for the remaining mass of a radioactive substance is given by $M = M_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Here, the initial mass $M_0 = 20\, mg$, the total time $t = 36\, days$, and the half-life $T_{1/2} = 3.6\, days$.
The number of half-lives $n = \frac{t}{T_{1/2}} = \frac{36}{3.6} = 10$.
Substituting these values into the formula:
$M = 20 \times \left( \frac{1}{2} \right)^{10}$.
$M = 20 \times \frac{1}{1024}$.
$M = 0.01953125\, mg$.
Rounding to three decimal places, we get $M \approx 0.019\, mg$.

(C) The number of radioactive nuclei remaining after time $t$ is given by the formula: $N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$.
Here,the initial amount is $N_0$,the half-life $T_{1/2} = 10$ days,and the total time $t = 30$ days.
The fraction of the material remaining is $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{\frac{30}{10}}$.
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.
$\frac{N}{N_0} = 0.125$.

(D) The radioactive decay law is given by the formula: $\frac{A}{A_0} = (\frac{1}{2})^{t/T_{1/2}}$, 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 that the activity decreases to $1/8$ of its original value, we have $\frac{A}{A_0} = 1/8$.
The half-life $T_{1/2} = 8$ years.
Substituting these values into the formula: $1/8 = (\frac{1}{2})^{t/8}$.
Since $1/8 = (\frac{1}{2})^3$, we can write: $(\frac{1}{2})^3 = (\frac{1}{2})^{t/8}$.
Comparing the exponents, we get $3 = t/8$.
Therefore, $t = 3 \times 8 = 24$ years.

(B) The formula for the remaining amount of a radioactive substance is given by $N = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
The number of half-lives $n$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{11400}{5700} = 2$.
Substituting this into the formula:
$N = N_0 \times (1/2)^2$
$N = N_0 \times (1/4)$
$N = 0.25 N_0$.
Therefore,the amount left is $0.25$ of the original amount.

(D) The mean life $(T)$ is given by $T = 1/\lambda = 100 \, \text{s}$.
The half-life $(T_{1/2})$ is related to the decay constant $(\lambda)$ by the formula $T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda}$.
Substituting $\lambda = 1/T$,we get $T_{1/2} = 0.693 \times T$.
Given $T = 100 \, \text{s}$,the half-life in seconds is $T_{1/2} = 0.693 \times 100 = 69.3 \, \text{s}$.
To convert the half-life into minutes,divide by $60$:
$T_{1/2} = \frac{69.3}{60} \, \text{min} = 1.155 \, \text{min}$.

(C) Radioactivity is a natural phenomenon in which particles are emitted from nuclei as a result of nuclear instability.
It is a spontaneous nuclear process and does not depend on external factors such as temperature,pressure,or chemical environment.
Because the nucleus experiences an intense conflict between the strong nuclear force and the electrostatic repulsive force,many nuclear isotopes are unstable and emit radiation to achieve a more stable state.

(B) 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 the half-life $T_{1/2} = 77 \, days$.
Substituting the value into the formula: $\lambda = \frac{0.693}{77}$.
Calculating the division: $\lambda = 0.009 \, day^{-1}$.
Expressing in scientific notation: $\lambda = 9 \times 10^{-3} \, day^{-1}$.
Therefore,the correct option is $B$.

(C) The law of radioactive decay is given by the formula $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $N$ is the number of nuclei remaining,$N_0$ is the initial number of nuclei,$t$ is the time elapsed,and $T_{1/2}$ is the half-life.
Given that the time elapsed is half of a half-life,we have $t = \frac{1}{2} T_{1/2}$.
Substituting this into the decay formula:
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^{(T_{1/2}/2) / T_{1/2}}$
$\frac{N}{N_0} = \left( \frac{1}{2} \right)^{1/2}$
$\frac{N}{N_0} = \frac{1}{\sqrt{2}}$
Therefore,the fraction of nuclei remaining is $1/\sqrt{2}$.

(D) Radioactive decay is a spontaneous and statistical process governed by the laws of probability.
For any given radioactive nucleus,the probability of decay per unit time (decay constant,$\lambda$) is a characteristic property of that specific nuclide.
This probability does not depend on the age of the nucleus or the history of its formation.
Therefore,both nuclei,being of the same radioactive nuclide,have the same probability of decay regardless of whether they were created $5$ billion years ago or $5$ minutes ago.
Thus,the probability of decay is independent of the time of creation.

(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{3.8 \ days}$.
Substituting the values into the decay equation: $\frac{1}{20} = e^{-\lambda t}$,which implies $20 = e^{\lambda t}$.
Taking the natural logarithm on both sides: $\ln 20 = \lambda t$.
Converting to base $10$ logarithm: $2.303 \log_{10} 20 = \lambda t$.
Given $\log_{10} 20 = \log_{10} (2 \times 10) = 0.3010 + 1 = 1.3010$.
Substituting $\lambda = \frac{0.6931}{3.8}$ and $\log_{10} e = 0.4343$,we use the relation $\ln x = 2.303 \log_{10} x$ or $\lambda = \frac{0.6931}{T_{1/2}}$.
$t = \frac{\ln 20}{\lambda} = \frac{2.303 \times 1.3010 \times 3.8}{0.6931} \approx 16.43 \ days$.
Rounding to the nearest provided option,$t \approx 16.5 \ days$.

(C) Let $N_0$ be the initial amount of radioactive isotope $X$.
After some time $t$,the amount of $X$ remaining is $N_X$ and the amount of $Y$ formed is $N_Y$.
Given the ratio $N_X : N_Y = 1 : 7$,the total amount of substance is $N_X + N_Y = N_0$.
Thus,$N_X = \frac{1}{1+7} N_0 = \frac{1}{8} N_0$.
Using the radioactive decay law,$N_X = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives.
$\frac{1}{8} N_0 = N_0 (\frac{1}{2})^n \implies (\frac{1}{2})^3 = (\frac{1}{2})^n \implies n = 3$.
The age of the rock $t = n \times T_{1/2} = 3 \times 1.37 \times 10^9$ years $= 4.11 \times 10^9$ years.