Rate of decay and Half-life Questions in English

(B) The initial amount $N_0 = 16 \ g$.
The amount decayed is $15 \ g$,so the remaining amount $N_t = 16 - 15 = 1 \ g$.
The formula for radioactive decay is $N_t = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
$1 = 16 \times (1/2)^n$
$(1/2)^n = 1/16 = (1/2)^4$.
Therefore,$n = 4$.
The total time $t = n \times t_{1/2} = 4 \times 140 \ days = 560 \ days$.

(B) The number of half-lives $n$ is calculated as $n = \frac{t}{t_{1/2}} = \frac{20 \ s}{4 \ s} = 5$.
The fraction of the substance remaining is given by $\frac{N_t}{N_o} = (\frac{1}{2})^n = (\frac{1}{2})^5 = \frac{1}{32}$.
The fraction decayed is $1 - \frac{N_t}{N_o} = 1 - \frac{1}{32} = \frac{31}{32}$.
The percentage decayed is $\frac{31}{32} \times 100 = 96.875 \% \approx 96.87 \%$.

(A) The formula for the amount remaining is $N_t = N_o (1/2)^n$,where $n = T / t_{1/2}$.
Given: $t_{1/2} = 22 \ years$,$T = 11 \ years$,$N_o = 2 \ g$.
Calculate the number of half-lives: $n = 11 / 22 = 0.5$.
Substitute the values into the formula: $N_t = 2 \times (1/2)^{0.5}$.
$N_t = 2 \times (1 / \sqrt{2}) = 2 / 1.414 = 1.414 \ g$.

(C) The radioactive decay follows the first-order kinetics equation: $t = \frac{2.303}{\lambda} \log \frac{N_0}{N_t}$.
Here,$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5000} \ min^{-1}$.
Given $N_0 = 15 \ counts/min/gm$ and $N_t = 5.0 \ counts/min/gm$.
Substituting the values: $t = \frac{2.303 \times 5000}{0.693} \times \log \frac{15}{5}$.
$t = \frac{2.303 \times 5000}{0.693} \times \log 3$.
$t = \frac{2.303 \times 5000}{0.693} \times 0.4771 \approx 7927 \ years$.
Thus,the age of the specimen is approximately $7.92 \times 10^3 \ years$.

(B) The radioactive disintegration reactions follow first-order kinetics.
This is because the rate of disintegration is directly proportional to the concentration of the radioactive material raised to the power of $1$,i.e.,$\text{Rate} = k[N]$,where $[N]$ is the number of radioactive nuclei present.

(B) The radioactive disintegration of $_{92}U^{235}$ is a first-order reaction.
Radioactive decay follows first-order kinetics because the rate of disintegration depends only on the concentration of the radioactive material present at that time.

(C) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{12 \ days}{3 \ days} = 4$.
The remaining amount of substance $(N)$ is given by the formula: $N = N_0 \left( \frac{1}{2} \right)^n$.
Substituting the value of $n$: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Therefore,the fraction of the substance remaining is $1/16$.

(A) The half-life period is $T_{1/2} = 4 \ hours$.
The total time elapsed is $t = 24 \ hours$.
The number of half-lives $n$ is calculated as $n = t / T_{1/2} = 24 / 4 = 6$.
The remaining mass $N$ is given by the formula $N = N_0 / 2^n$,where $N_0 = 200 \ g$.
Substituting the values,$N = 200 / 2^6 = 200 / 64 = 3.125 \ g$.

(D) For a first-order reaction,the amount remaining after $40 \text{ min}$ is $100\% - 87.5\% = 12.5\%$.
Using the formula $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$:
$k = \frac{2.303}{40} \log \frac{100}{12.5} = \frac{2.303}{40} \log 8 = \frac{2.303 \times 3 \times 0.3010}{40} = \frac{0.693 \times 3}{40} \text{ min}^{-1}$.
The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693 \times 40}{0.693 \times 3} = \frac{40}{3} \text{ min} = 13 \text{ min } 20 \text{ sec}$.

(A) The formula for the remaining mass of a radioactive substance is $N_t = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
First,calculate the number of half-lives $(n)$: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{18 \ h}{3 \ h} = 6$.
Now,substitute the values into the formula: $N_t = 256 \ g \times (1/2)^6$.
$N_t = 256 \times \frac{1}{64}$.
$N_t = 4.0 \ g$.

(B) Radioactive decay is a first-order reaction.
This is because the rate of the reaction depends on the concentration of the radioactive element.

(C) The radioactive decay follows first-order kinetics.
Given:
Final activity $(A_t)$ = $0.01 \mu Ci$
Time $(t)$ = $24 \ hours$
Half-life $(t_{1/2})$ = $6 \ hours$
Number of half-lives $(n)$ = $t / t_{1/2} = 24 / 6 = 4$.
The relation between initial activity $(A_0)$ and final activity $(A_t)$ is given by:
$A_t = A_0 \times (1/2)^n$
$0.01 = A_0 \times (1/2)^4$
$0.01 = A_0 \times (1/16)$
$A_0 = 0.01 \times 16 = 0.16 \mu Ci$.
Therefore,the maximum initial activity that can be administered is $0.16 \mu Ci$.

(C) The relationship between half-life $(t_{1/2})$ and decay constant $(k)$ for a first-order process is given by:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 2.25 \times 10^{-4} \text{ year}^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{2.25 \times 10^{-4}} = \frac{0.693}{2.25} \times 10^{4} = 0.308 \times 10^{4} = 3080 \text{ years}$.

(B) The formula for the remaining amount after $n$ half-lives is $N = \frac{N_0}{2^n}$.
First,calculate the number of half-lives $(n)$: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{49.2}{12.3} = 4$.
Now,calculate the remaining amount: $N = \frac{32}{2^4} = \frac{32}{16} = 2 \ mg$.

(A) The half-life $(t_{1/2})$ is defined as the time required for $50\%$ of a radioactive sample to disintegrate.
Given that carbon-$14$ disintegrates $50\%$ in $5770$ years,the half-life is $5770$ years.
Therefore,the correct option is $A$.

(B) The half-life $(t_{1/2})$ of the radioactive isotope $^{14}C$ is approximately $5730$ $years$.
Therefore,the correct option is $(B)$.

(D) The formula for radioactive decay is $N_t = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $N_0 = 200 \ mg$,$N_t = 25 \ mg$,and $t_{1/2} = 5760 \ years$.
Substituting the values: $25 = 200 \times (\frac{1}{2})^n$.
$(\frac{1}{2})^n = \frac{25}{200} = \frac{1}{8} = (\frac{1}{2})^3$.
Therefore,$n = 3$.
The total time required is $t = n \times t_{1/2} = 3 \times 5760 = 17280 \ years$.

(A) The half-life $(t_{1/2})$ of a radioactive element is related to the decay constant $(\lambda)$ by the formula:
$t_{1/2} = \frac{0.693}{\lambda}$
Given $\lambda = 3 \times 10^{-6} \ min^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{3 \times 10^{-6} \ min^{-1}} = 0.231 \times 10^6 \ min = 2.31 \times 10^5 \ min$.

(A) The half-life period $(t_{1/2})$ of the radioactive decay is given as $6.93 \ min$.
For a first-order reaction,the rate constant $(k)$ is calculated using the formula: $k = \frac{0.693}{t_{1/2}}$.
Substituting the given value: $k = \frac{0.693}{6.93} = 0.10 \ \min^{-1}$.

(B) The initial amount $N_0 = 1 \ g$. Given the molar mass $M = 1 \ g/mol$,the initial number of atoms is $N_0 = (1 \ g / 1 \ g/mol) \times 6.022 \times 10^{23} \ \text{atoms/mol} = 6.022 \times 10^{23} \ \text{atoms}$.
Using the radioactive decay formula $N = N_0 \times (1/2)^{(t/T_{1/2})}$,where $t = 4 \ hrs$ and $T_{1/2} = 10 \ hrs$.
$N = 6.022 \times 10^{23} \times (0.5)^{(4/10)} = 6.022 \times 10^{23} \times (0.5)^{0.4}$.
Since $(0.5)^{0.4} \approx 0.7578$,we get $N \approx 6.022 \times 10^{23} \times 0.7578 \approx 4.56 \times 10^{23} \ \text{atoms}$.

(D) . The half-life $t_{1/2}$ of a radioactive element is given as $N$ years.
Radioactive decay follows first-order kinetics,where the amount of substance remaining is given by $N(t) = N_0 e^{-\lambda t}$.
For the substance to decay completely,$N(t)$ must be $0$,which implies $e^{-\lambda t} = 0$.
This condition is only satisfied as $t \to \infty$.
Therefore,the time required for complete decay is infinity.

(B) The radioactive decay follows the relation: $\frac{N}{N_o} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_o} = \frac{1}{8}$,we have $(\frac{1}{2})^n = \frac{1}{8} = (\frac{1}{2})^3$.
Therefore,$n = 3$.
The total time $t$ is given by $t = n \times t_{1/2}$.
Given $t_{1/2} = 20 \ \text{minutes}$,so $t = 3 \times 20 \ \text{minutes} = 60 \ \text{minutes}$.

(B) The number of half-lives $n$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{45 \ min}{15 \ min} = 3$.
The remaining amount of radioactive material $N$ is given by the formula $N = N_o \times (\frac{1}{2})^n$.
Assuming the initial amount $N_o = 100 \%$,the remaining amount is $N = 100 \times (\frac{1}{2})^3 = \frac{100}{8} = 12.5 \%$.
Thus,$12.5 \%$ of the radioactivity remains.

(C) The radioactive decay formula is given by $N = N_o (1/2)^n$,where $n$ is the number of half-lives.
Given that after $3160 \ years$,the remaining amount $N = (1/4)N_o$,so $N/N_o = 1/4$.
Substituting this into the formula: $1/4 = (1/2)^n$.
Since $1/4 = (1/2)^2$,we get $n = 2$.
The number of half-lives $n$ is related to total time $t$ and half-life $t_{1/2}$ by $n = t / t_{1/2}$.
Therefore,$2 = 3160 / t_{1/2}$.
$t_{1/2} = 3160 / 2 = 1580 \ years$.

(B) At radioactive equilibrium,the rate of decay of $X$ is equal to the rate of decay of $Y$.
Thus,$\lambda_X N_X = \lambda_Y N_Y$.
This implies $\frac{N_X}{N_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{t_{1/2}(X)}{t_{1/2}(Y)}$.
Given $\frac{N_X}{N_Y} = 1 : 2 \times 10^{-6}$ and $t_{1/2}(Y) = 4.9 \times 10^{-4} \ days$.
Therefore,$t_{1/2}(X) = \frac{N_X}{N_Y} \times t_{1/2}(Y) = \frac{1}{2 \times 10^{-6}} \times 4.9 \times 10^{-4} \ days$.
$t_{1/2}(X) = 0.5 \times 10^6 \times 4.9 \times 10^{-4} \ days = 245 \ days$.

(D) Given: Half-life $(t_{1/2})$ = $5 \, \text{years}$,Total time $(t)$ = $15 \, \text{years}$,Initial amount $(N_0)$ = $64 \, \text{g}$.
Number of half-lives $(n)$ = $\frac{t}{t_{1/2}} = \frac{15}{5} = 3$.
The amount remaining $(N)$ is given by the formula: $N = \frac{N_0}{2^n}$.
Substituting the values: $N = \frac{64}{2^3} = \frac{64}{8} = 8 \, \text{g}$.
Therefore,the correct option is $(D)$.

(A) Given: Half-life $(T_{1/2})$ = $1600 \ years$,Total time $(T)$ = $6400 \ years$.
The number of half-lives $(n)$ is calculated as: $n = \frac{T}{T_{1/2}} = \frac{6400}{1600} = 4$.
The remaining fraction of the substance is given by the formula: $N = N_o \times (1/2)^n$.
Assuming initial amount $N_o = 1$,the remaining amount $N = 1 \times (1/2)^4 = 1/16$.
Therefore,the mass left after $6400 \ years$ is $1/16$ of the initial mass.

(A) The decay of carbon follows first-order kinetics.
Given: Initial activity $(r_o)$ = $15.2 \ min^{-1} g^{-1}$,Final activity $(r)$ = $7.6 \ min^{-1} g^{-1}$,and $t_{1/2} = 5760$ years.
Since $r = \frac{r_o}{2}$,the artifact has undergone exactly one half-life.
Therefore,the age of the artifact $(t)$ = $1 \times t_{1/2} = 5760$ years.
Alternatively,using the formula $t = \frac{t_{1/2}}{0.693} \ln(\frac{r_o}{r}) = \frac{5760}{0.693} \ln(\frac{15.2}{7.6}) = \frac{5760}{0.693} \ln(2) = 5760$ years.

(B) Let the initial number of uranium atoms be $N_0$.
Since the rock contains an equal number of uranium and lead atoms,the number of uranium atoms remaining $(N)$ is half of the initial number,i.e.,$N = \frac{N_0}{2}$.
This implies that one half-life has passed.
The age of the rock $(t)$ is equal to the half-life $(t_{1/2})$ of uranium.
Therefore,$t = 4.5 \times 10^9$ years.

(A) The radioactive decay follows first-order kinetics,where the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$.
Given that the initial activity $A_0 = 10A$,where $A$ is the permissible safe activity level.
The decay constant $\lambda$ is related to the half-life $t_{1/2}$ by $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{30 \text{ days}}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{\lambda} \log_{10} \left( \frac{A_0}{A} \right)$.
Substituting the values: $t = \frac{2.303}{(0.693 / 30)} \log_{10} (10)$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.434} \approx 3.32$ (or more precisely $\frac{2.303}{0.693} \approx 3.32$ and $3.32 \times 30 \approx 100$),we get $t = 30 \times \frac{2.303}{0.693} \times 1 = 30 \times 3.325 \approx 100 \text{ days}$.

(C) The half-life $(t_{1/2})$ of $Gd-153$ is $242$ days.
The total time $(t)$ is $2$ years,which is $2 \times 365 = 730$ days.
The number of half-lives $(n)$ is calculated as $n = \frac{t}{t_{1/2}} = \frac{730}{242} \approx 3.016$.
Since $n \approx 3$,the remaining amount of the substance is given by the formula: $\text{Remaining percentage} = \frac{100}{2^n} = \frac{100}{2^3} = \frac{100}{8} = 12.5\%$.
The value closest to $12.5\%$ among the given options is $12.0\%$.

(C) The formula for radioactive decay is $N_t = N_0 \left(\frac{1}{2}\right)^n$,where $n$ is the number of half-lives.
Given $N_0 = 200 \text{ mg}$,$N_t = 25 \text{ mg}$,and $t_{1/2} = 4650 \text{ years}$.
Substituting the values: $25 = 200 \times \left(\frac{1}{2}\right)^n$.
$\frac{25}{200} = \left(\frac{1}{2}\right)^n \Rightarrow \frac{1}{8} = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{8} = \left(\frac{1}{2}\right)^3$,we get $n = 3$.
Total time $T = n \times t_{1/2} = 3 \times 4650 \text{ years} = 13950 \text{ years}$.

(D) The radioactive decay follows first-order kinetics.
The intensity of radiation is proportional to the amount of the radioactive substance present.
Let the initial intensity be $I_0 = 64 \times I_{safe}$ and the final intensity be $I = I_{safe}$.
Using the relation $I = I_0 \times (1/2)^n$,where $n$ is the number of half-lives:
$I_{safe} = 64 \times I_{safe} \times (1/2)^n$
$1/64 = (1/2)^n$
$(1/2)^6 = (1/2)^n$
So,$n = 6$.
The total time $t$ is given by $t = n \times t_{1/2}$.
Given $t_{1/2} = 2 \ hr$,we have $t = 6 \times 2 = 12 \ hr$.

(B) Radioactive decay follows first-order kinetics. The rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \ min^{-1}$.
For a first-order reaction,the time $t$ required for decay from $x_1\%$ to $x_2\%$ is given by $t = \frac{2.303}{K} \log \frac{100 - x_1}{100 - x_2}$.
Here,$x_1 = 33$ and $x_2 = 67$,so the remaining amounts are $100 - 33 = 67$ and $100 - 67 = 33$.
Substituting the values: $t = \frac{2.303}{(0.693/20)} \log \frac{67}{33}$.
Since $0.693 \approx 2.303 \times \log_{10}(2)$,we have $t = 20 \times \frac{\log(67/33)}{\log(2)} \approx 20 \times \frac{0.307}{0.301} \approx 20 \ min$.

(C) Given half-life $t_{1/2} = 3 \ days$.
Total time elapsed $T = 12 \ days$.
Number of half-lives $(n) = \frac{T}{t_{1/2}} = \frac{12}{3} = 4$.
The relationship between initial mass $(N_0)$ and remaining mass $(N)$ is given by $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $3 = N_0 \times (\frac{1}{2})^4$.
$3 = N_0 \times \frac{1}{16}$.
$N_0 = 3 \times 16 = 48 \ g$.

(C) For a radioactive decay (first-order kinetics),the time taken for $50\%$ decay is $t_{50\%} = 1 \times t_{1/2} = 100 \, \text{min}$.
For $87.5\%$ decay,the remaining amount is $100\% - 87.5\% = 12.5\%$.
Since $12.5\% = (1/2)^3$,this corresponds to $3$ half-lives.
Thus,$t_{87.5\%} = 3 \times t_{1/2} = 3 \times 100 = 300 \, \text{min}$.
The time interval between these two stages is $\Delta t = t_{87.5\%} - t_{50\%} = 300 \, \text{min} - 100 \, \text{min} = 200 \, \text{min}$.

(B) Radioactive decay follows first-order kinetics.
The decay constant $(\lambda)$ is given by $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ year^{-1}$.
Given,the initial amount $[R]_0 = 100$ and the remaining amount $[R] = 80$.
Using the first-order integrated rate equation:
$t = \frac{2.303}{\lambda} \log \frac{[R]_0}{[R]}$
Substituting the values:
$t = \frac{2.303 \times 5730}{0.693} \times \log \left( \frac{100}{80} \right)$
$t = \frac{2.303 \times 5730}{0.693} \times \log(1.25)$
$t = \frac{2.303 \times 5730}{0.693} \times 0.0969$
$t \approx 1845 \ years$.

(C) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{a}{a - x}$.
Given that the sample decreases by $90\%$, the remaining amount $(a - x)$ is $10\%$ of the initial amount $a$, so $(a - x) = 0.1a$.
Substituting the values: $k = \frac{2.303}{10} \log \frac{a}{0.1a} = \frac{2.303}{10} \log 10 = \frac{2.303}{10} \times 1 = 0.2303 \ \text{year}^{-1}$.
The half-life $t_{1/2}$ is calculated as: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.2303} \approx 3 \ \text{years}$.

(A) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{16 \ h}{4 \ h} = 4$.
Remaining mass $(N_t)$ is given by the formula $N_t = N_0 \times (\frac{1}{2})^n$,where $N_0 = 200 \ g$.
$N_t = 200 \times (\frac{1}{2})^4 = 200 \times \frac{1}{16} = 12.5 \ g$.
The disintegrated amount is the initial mass minus the remaining mass: $\text{Disintegrated amount} = N_0 - N_t = 200 \ g - 12.5 \ g = 187.5 \ g$.

(A) Radioactive decay follows first-order kinetics. The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ year^{-1}$.
Using the first-order integrated rate equation: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Here,$[A]_0 = 100$ and $[A]_t = 80$.
Substituting the values: $t = \frac{2.303}{k} \log \left( \frac{100}{80} \right) = \frac{2.303 \times 5730}{0.693} \log(1.25)$.
$t = \frac{2.303 \times 5730}{0.693} \times 0.0969 \approx 1845 \ years$.

(C) Radioactive decay follows first-order kinetics.
Let $N_0$ be the initial number of uranium atoms.
At time $t$,the number of uranium atoms remaining is $N_t$ and the number of lead atoms formed is $N_0 - N_t$.
Given that the number of uranium atoms equals the number of lead atoms,$N_t = N_0 - N_t$,which implies $N_t = N_0 / 2$.
This means the time elapsed is equal to one half-life period $(t_{1/2})$.
Therefore,the age of the rock $t = t_{1/2} = 4.5 \times 10^9$ years.

(D) The rate of disintegration is given by $R = N \lambda$,where $\lambda$ is the decay constant.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by the formula $\lambda = \frac{0.693}{T_{1/2}}$.
Substituting $\lambda$ into the rate equation: $R = N \times \frac{0.693}{T_{1/2}}$.
Rearranging to solve for the number of atoms $N$: $N = \frac{R \times T_{1/2}}{0.693}$.
Given $R = 10^{10} \ s^{-1}$ and $T_{1/2} = 2.2 \times 10^{9} \ s$:
$N = \frac{10^{10} \times 2.2 \times 10^{9}}{0.693} = \frac{2.2 \times 10^{19}}{0.693} \approx 3.17 \times 10^{19} \ atoms$.

(C) All nuclear decays follow first-order kinetics.
$t = \frac{1}{k} \ln \frac{[A_0]}{[A]}$
Given $t_{1/2} = 6.93 \; \text{years}$,so $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.93} = 0.1 \; \text{year}^{-1}$.
To reduce it by $90 \%$,the remaining amount $[A] = 10 \% \text{ of } [A_0] = 0.1 [A_0]$.
$t = \frac{2.303}{k} \log_{10} \frac{[A_0]}{0.1 [A_0]} = \frac{2.303}{0.1} \log_{10} (10) = 23.03 \times 1 = 23.03 \; \text{years}$.

(N/A) The decay follows first-order kinetics,where the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{28.1} \ y^{-1}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.
For $t = 10 \ years$:
$10 = \frac{2.303}{0.693 / 28.1} \log \frac{1}{[R]}$
$\log [R] = -\frac{10 \times 0.693}{2.303 \times 28.1} \approx -0.1071$
$[R] = \text{antilog}(-0.1071) \approx 0.7814 \ \mu g$.
For $t = 60 \ years$:
$60 = \frac{2.303}{0.693 / 28.1} \log \frac{1}{[R]}$
$\log [R] = -\frac{60 \times 0.693}{2.303 \times 28.1} \approx -0.6425$
$[R] = \text{antilog}(-0.6425) \approx 0.2278 \ \mu g$.
Thus,$0.7814 \ \mu g$ remains after $10 \ years$ and $0.2278 \ \mu g$ remains after $60 \ years$.

(B) The half-life period of tritium $(^3H)$ is $12.33$ years.

(N/A) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 1620 \ years = 1620 \times 365 \times 24 \times 60 \ min = 8.51472 \times 10^{8} \ min$.
$\lambda = \frac{0.693}{8.51472 \times 10^{8}} \ min^{-1} = 8.1389 \times 10^{-10} \ min^{-1}$.
The number of atoms $N$ in $0.001 \ g$ of $Ra^{226}$ is $N = \frac{0.001}{226} \times 6.022 \times 10^{23} = 2.6646 \times 10^{18} \ atoms$.
The rate of decay (activity) is $A = \lambda N = (8.1389 \times 10^{-10} \ min^{-1}) \times (2.6646 \times 10^{18}) = 2.1686 \times 10^{9} \ min^{-1} \approx 2.17 \times 10^{9} \ min^{-1}$.

(B) The radioactive decay follows first-order kinetics. The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ \text{year}^{-1}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \left( \frac{N_0}{N_t} \right)$.
Here,$N_0 = 15.3$ and $N_t = 2.25$.
$t = \frac{2.303 \times 5730}{0.693} \log \left( \frac{15.3}{2.25} \right)$.
$t = 19042.7 \times \log(6.8)$.
$t = 19042.7 \times 0.8325 \approx 15853 \ \text{years}$.
Rounding to the nearest significant value,the age is approximately $1.58 \times 10^{4} \ \text{years}$.

(N/A) The decay constant $\lambda$ is given by $\lambda = \frac{0.693}{t_{1/2}}$.
First,convert $t_{1/2}$ to seconds: $t_{1/2} = 1.39 \times 10^{10} \ \text{years} \times 365 \times 24 \times 3600 \ \text{s/year} \approx 4.38 \times 10^{17} \ \text{s}$.
$\lambda = \frac{0.693}{4.38 \times 10^{17} \ \text{s}} \approx 1.58 \times 10^{-18} \ \text{s}^{-1}$.
The number of atoms $N$ in $1.0 \ \text{g}$ of $^{232}Th$ is $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1.0 \ \text{g}}{232 \ \text{g/mol}} \times 6.022 \times 10^{23} \ \text{atoms/mol} \approx 2.596 \times 10^{21} \ \text{atoms}$.
The rate of emission (activity) is $A = \lambda N = (1.58 \times 10^{-18} \ \text{s}^{-1}) \times (2.596 \times 10^{21} \ \text{atoms}) \approx 4.10 \times 10^{3} \ \text{particles/s}$.

(N/A) The decay of $C^{14}$ follows first-order kinetics. The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ \text{yr}^{-1}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \left( \frac{N_0}{N_t} \right)$.
Here,$N_0 = 15.3$ and $N_t = 11.9$.
$t = \frac{2.303 \times 5730}{0.693} \log \left( \frac{15.3}{11.9} \right)$.
$t \approx 19033.8 \times \log(1.2857) \approx 19033.8 \times 0.1091 \approx 2077 \ \text{years}$.
Rounding to the nearest value,the age is approximately $2079 \ \text{years}$.

(B) Radioactive decay is a spontaneous process where the rate of decay is directly proportional to the number of radioactive nuclei present at that time.
Mathematically,this is expressed as $-\frac{dN}{dt} = \lambda N$.
Since the rate depends on the first power of the concentration of the radioactive substance,radioactive decay reactions are always $1^{st}$ order reactions.