Rate of decay and Half-life Questions in English

(B) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0(1/2)^n$.
Given that $75\%$ disintegrates,the amount remaining is $25\%$ of the initial amount,i.e.,$N = 0.25 N_0$.
So,$0.25 N_0 = N_0(1/2)^n$,which simplifies to $(1/2)^2 = (1/2)^n$.
Thus,$n = 2$ half-lives.
Since $2$ half-lives correspond to $60 \ min$,the half-life $t_{1/2} = 60 \ min / 2 = 30 \ min$.

(D) For a radioactive substance,the amount remaining after $n$ half-lives is given by $N = N_0 \times (1/2)^n$.
Given that $87.5\%$ has decomposed,the amount remaining is $100\% - 87.5\% = 12.5\%$.
So,$12.5 = 100 \times (1/2)^n$,which simplifies to $(1/2)^3 = (1/2)^n$.
Therefore,$n = 3$ half-lives.
Since $3$ half-lives correspond to $3$ hours,the half-life $t_{1/2} = 3 \text{ hours} / 3 = 1 \text{ hour}$.

(D) The radioactive decay follows the law $N = N_0 e^{-\lambda t}$.
At half-life $(t = t_{1/2})$,the amount of substance remaining is $N = N_0 / 2$.
Substituting these values into the equation: $N_0 / 2 = N_0 e^{-\lambda t_{1/2}}$.
$1 / 2 = e^{-\lambda t_{1/2}}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -\lambda t_{1/2}$.
$-\ln 2 = -\lambda t_{1/2}$.
Therefore,$t_{1/2} = \ln 2 / \lambda$.

(B) For a first-order radioactive decay,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{\lambda}$
Given the rate constant $\lambda = 231 \ sec^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{231 \ sec^{-1}} = 0.003 \ sec = 3.0 \times 10^{-3} \ sec$
Therefore,the correct option is $B$.

(C) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{t_{1/2}} = \frac{50 \ minutes}{25 \ minutes} = 2$.
The fraction of the substance remaining is given by the formula: $\text{Amount left} = (1/2)^n$.
Substituting the value of $n$: $\text{Amount left} = (1/2)^2 = 1/4$ (One-fourth).

(A) The amount of radioactive element remaining after $2$ hours is $N = N_o - \frac{3}{4} N_o = \frac{1}{4} N_o$.
We know that the amount remaining after $n$ half-lives is given by $N = N_o (1/2)^n$.
Substituting the values,$\frac{1}{4} N_o = N_o (1/2)^n$,which implies $(1/2)^2 = (1/2)^n$.
Therefore,the number of half-lives $n = 2$.
Since $n = \frac{t}{t_{1/2}}$,we have $2 = \frac{2 \ hours}{t_{1/2}}$.
Thus,$t_{1/2} = 1 \ hour$.

(D) The half-life $(t_{1/2})$ of a radioactive element is given by the formula $t_{1/2} = \frac{0.693}{\lambda}$,where $\lambda$ is the decay constant.
Since $\lambda$ is a characteristic property of the radioactive isotope and is independent of external conditions like temperature,pressure,or the initial amount of the substance,the half-life is also independent of these factors.
Therefore,the correct option is $(D)$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{100 \ min}{25 \ min} = 4$.
The amount remaining $(N)$ is given by the formula $N = \frac{N_0}{2^n}$,where $N_0$ is the initial amount.
Substituting the values,$N = \frac{100 \ g}{2^4} = \frac{100 \ g}{16} = 6.25 \ g$.

(C) The half-life of a radioactive substance is a characteristic property of the isotope and is independent of the initial amount of the substance present.
Therefore,if the half-life of $8.0 \ g$ of the isotope is $10 \ hrs$,the half-life of $2.0 \ g$ of the same isotope remains $10 \ hrs$.

(D) The half-life $(t_{1/2})$ is related to the disintegration constant $(k)$ by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 6.93 \times 10^{-6} \ s^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{6.93 \times 10^{-6}} = 0.1 \times 10^{6} = 10^{5} \ s$.
Therefore,the correct option is $(D)$.

(A) The activity $A$ is given by the formula $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of atoms.
Given: $\lambda = 1.37 \times 10^{-11} \ s^{-1}$ and $A = 1.5 \ mCi$.
Since $1 \ mCi = 3.7 \times 10^7 \ \text{disintegrations per second (dps)}$,
$A = 1.5 \times 3.7 \times 10^7 \ dps = 5.55 \times 10^7 \ dps$.
Using $N = \frac{A}{\lambda}$,we get:
$N = \frac{5.55 \times 10^7}{1.37 \times 10^{-11}} \approx 4.05 \times 10^{18}$ atoms.
Rounding to the nearest option,the value is $4.1 \times 10^{18}$.

(C) The formula for radioactive decay is $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Number of half-lives $n = \frac{T}{t_{1/2}} = \frac{75 \ min}{25 \ min} = 3$.
Therefore,the remaining fraction is $\frac{N}{N_0} = (\frac{1}{2})^3 = \frac{1}{8}$.

(A) The formula for radioactive decay is $N = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Number of half-lives $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{24 \ h}{4 \ h} = 6$.
Initial mass $N_0 = 200 \ g$.
Remaining mass $N = 200 \times (\frac{1}{2})^6$.
$N = 200 \times \frac{1}{64} = 3.125 \ g$.

(A) The decay constant $(K)$ is related to the half-life $(t_{1/2})$ by the formula: $K = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 10 \ yr$.
Substituting the value: $K = \frac{0.693}{10} = 0.0693 \ yr^{-1}$.

(B) The radioactive decay formula is given by $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given that $\frac{N}{N_0} = \frac{1}{16}$,we can write $\frac{1}{16} = (\frac{1}{2})^4$.
Thus,the number of half-lives $n = 4$.
Since $n = \frac{T}{t_{1/2}}$,where $T = 192 \ min$,we have $4 = \frac{192}{t_{1/2}}$.
Therefore,$t_{1/2} = \frac{192}{4} = 48 \ min$.

(A) For a radioactive decay process following first-order kinetics,the rate constant is denoted by $\lambda$.
The average life $(\tau)$ is defined as the reciprocal of the decay constant.
Therefore,the expression for average life is $\tau = 1/\lambda$.

(A) The decay constant $K$ is given by $K = \frac{0.693}{T_{1/2}} = \frac{0.693}{5770} \text{ year}^{-1}$.
Using the first-order kinetics equation: $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.
Substituting the values: $t = \frac{2.303 \times 5770}{0.693} \log \frac{100}{72}$.
$t = 19175.05 \times (\log 100 - \log 72)$.
$t = 19175.05 \times (2 - 1.8573) = 19175.05 \times 0.1427 \approx 2736.3 \text{ years}$.
The closest option is $2740$ years.

(A) For a first-order radioactive decay,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{A_0}{A_t}$.
For $25 \%$ decay,the remaining amount is $75 \%$ of the initial amount $(A_t = 0.75 A_0)$:
$K = \frac{2.303}{20} \log \frac{100}{75} = \frac{2.303}{20} \times 0.1249 = 0.01438 \, \text{min}^{-1}$.
For $75 \%$ decay,the remaining amount is $25 \%$ of the initial amount $(A_t = 0.25 A_0)$:
$t = \frac{2.303}{K} \log \frac{100}{25} = \frac{2.303}{0.01438} \log 4 = \frac{2.303 \times 0.6020}{0.01438} \approx 96.4 \, \text{min}$.

(B) The radioactive sample emits $64$ times the non-hazardous limit,so the ratio of current activity to the limit is $N/N_0 = 1/64$.
Using the radioactive decay formula: $N = N_0(1/2)^n$,where $n$ is the number of half-lives.
$1/64 = (1/2)^n$ $\Rightarrow (1/2)^6 = (1/2)^n$ $\Rightarrow n = 6$.
Given the half-life $t_{1/2} = 2 \ hr$,the total time $T = n \times t_{1/2} = 6 \times 2 = 12 \ hr$.
Thus,the sample becomes non-hazardous after $12 \ hr$.

(C) The half-life of a radioactive substance is a characteristic property of the radioactive isotope and is independent of the initial amount of the substance present.
Therefore,the half-life of $2.0 \ g$ of the same substance remains $10 \ hrs$.

(C) The number of half-lives $n$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{49.2}{12.3} = 4$.
The remaining amount $N_t$ is given by the formula: $N_t = N_0 \times (1/2)^n$.
Substituting the values: $N_t = 32 \times (1/2)^4 = 32 \times \frac{1}{16} = 2 \ mg$.

(B) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{11460}{5730} = 2$.
The fraction of activity remaining is given by the formula: $\text{Fraction} = (\frac{1}{2})^n$.
Substituting $n = 2$,we get: $\text{Fraction} = (\frac{1}{2})^2 = \frac{1}{4} = 0.25$.
Therefore,the correct option is $B$.

(D) Given: Half-life $(t_{1/2})$ = $10 \, \text{years}$,Total time $(t)$ = $20 \, \text{years}$.
Number of half-lives $(n) = \frac{t}{t_{1/2}} = \frac{20}{10} = 2$.
The remaining amount $(N)$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $N = N_0 \times (\frac{1}{2})^2 = N_0 \times \frac{1}{4} = 0.25 \, N_0$.
Percentage remaining = $0.25 \times 100\% = 25\%$.

(B) The radioactive decay follows the formula $N_t = N_0 \times (1/2)^n$,where $n = t/T_{1/2}$.
First,find the half-life $(T_{1/2})$: $0.25 \ g = 1.0 \ g \times (1/2)^n$,which gives $(1/2)^2 = (1/2)^n$,so $n = 2$.
Since $n = t/T_{1/2}$,$2 = 16 \ hours / T_{1/2}$,thus $T_{1/2} = 8 \ hours$.
Now,for $48 \ g$ to become $3.0 \ g$: $3.0 = 48 \times (1/2)^n$.
$(1/2)^n = 3.0 / 48 = 1/16 = (1/2)^4$,so $n = 4$.
Total time $t = n \times T_{1/2} = 4 \times 8 \ hours = 32 \ hours$.

(C) The Carbon-$14$ dating method relies on the principle that the ratio of radioactive Carbon-$14$ to stable Carbon-$12$ in the atmosphere and in living organisms remains constant during their lifetime.
After an organism dies,it stops taking in carbon,and the Carbon-$14$ begins to decay at a known rate,while the Carbon-$12$ remains stable.
By measuring the remaining ratio of $C-14/C-12$,the age of the sample can be determined.

(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 10.6 \ yrs$,so $k = \frac{0.693}{10.6} \approx 0.06538 \ yrs^{-1}$.
The time $t$ for $99 \%$ decomposition is calculated using $t = \frac{2.303}{k} \log \frac{a}{a-x}$.
Here,$a = 100$ and $a-x = 100 - 99 = 1$.
$t = \frac{2.303}{0.06538} \log \frac{100}{1} = \frac{2.303 \times 2}{0.06538} \approx 70.45 \ yrs$.
Thus,the correct option is $D$.

(B) Given the ratio of $C^{14}/C^{12}$ in the sample is $N_t/N_0 = 0.7$.
The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \, yr^{-1}$.
Using the first-order kinetics equation: $t = \frac{2.303}{k} \log \frac{N_0}{N_t}$.
Substituting the values: $t = \frac{2.303 \times 5760}{0.693} \log \frac{1}{0.7}$.
Since $\log(1/0.7) = \log(1.428) \approx 0.1549$.
$t = \frac{2.303 \times 5760 \times 0.1549}{0.693} \approx 2966 \, yrs$.

(C) $C-14$ is a radioactive isotope of carbon that is continuously produced in the upper atmosphere by cosmic rays.
It enters the food chain through plants and is incorporated into all living organisms.
As long as an organism is alive,the ratio of $C-14$ to $C-12$ remains constant due to continuous intake.
Upon death,the intake of $C-14$ stops,and it begins to decay with a half-life of approximately $5730 \ years$,allowing for the determination of the age of the object.

(A) The rate of radioactive decay is given by the equation: $-\frac{dN}{dt} = \lambda N$,where $\lambda$ is the decay constant.
Rearranging for $\lambda$,we get $\lambda = -\frac{dN/dt}{N}$.
Since $dN/dt$ represents the number of disintegrations per unit time and $N$ is the number of nuclei,the unit of $\lambda$ is $\text{time}^{-1}$ (e.g.,$s^{-1}$,$min^{-1}$,or $year^{-1}$).
Therefore,the correct option is $A$.

(A) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{40 \ days}{20 \ days} = 2$.
The amount of substance remaining $(N)$ is given by the formula $N = \frac{N_0}{2^n}$,where $N_0$ is the initial amount.
Substituting the values,$N = \frac{100 \ g}{2^2} = \frac{100}{4} = 25 \ g$.

(D) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{560}{140} = 4$.
The amount of radioactive substance remaining $(N)$ is given by the formula $N = \frac{N_o}{2^n}$,where $N_o$ is the initial amount.
Given $N_o = 1 \ g$ and $n = 4$,we have $N = \frac{1}{2^4} = \frac{1}{16} \ g$.

(A) The radioactive decay follows the formula: $\frac{N}{N_0} = (\frac{1}{2})^{\frac{T}{t_{1/2}}}$.
Given $\frac{N}{N_0} = 0.13$ and $t_{1/2} = 5770 \ years$.
Taking the logarithm on both sides: $\log(0.13) = \frac{T}{5770} \times \log(0.5)$.
$-0.886 = \frac{T}{5770} \times (-0.301)$.
$T = \frac{0.886 \times 5770}{0.301} \approx 16975 \ years$.
Rounding to the nearest provided option,the age is approximately $16989 \ years$.

(C) The initial total activity is $2000 \text{ disintegrations/minute}$.
After separation,the first fraction has an activity of $1000 \text{ disintegrations/minute}$ which remains constant.
The second fraction has an initial activity of $2000 - 1000 = 1000 \text{ disintegrations/minute}$.
Given $t_{1/2} = 24 \ h$,after $48 \ h$ (which is $2$ half-lives),the activity of the second fraction becomes $A = A_0 \times (1/2)^n = 1000 \times (1/2)^2 = 1000 / 4 = 250 \text{ disintegrations/minute}$.
The total activity after $48 \ h$ is $1000 + 250 = 1250 \text{ disintegrations/minute}$.

(A) The activity of $1 \ g$ of $^{226}Ra$ is defined as $1 \ Ci = 3.7 \times 10^{10} \ \text{disintegrations per second (dps)}$.
Therefore,the activity of $1 \ \mu g$ $(10^{-6} \ g)$ of $^{226}Ra$ is $3.7 \times 10^{10} \times 10^{-6} = 3.7 \times 10^4 \ dps$.
Since each disintegration of radium emits one alpha particle,the number of alpha particles emitted per second is approximately $3.7 \times 10^4 \ dps$.
Comparing this with the given options,the closest value is $3.62 \times 10^4 / \sec$.

(A) The half-life of radium $(^{226}Ra)$ is $T_{1/2} = 1600 \ \text{years}$.
Number of atoms in $1 \ \mu g$ of $^{226}Ra$ is $N = (1 \times 10^{-6} \ \text{g} / 226 \ \text{g/mol}) \times 6.023 \times 10^{23} \ \text{atoms/mol} \approx 2.665 \times 10^{15} \ \text{atoms}$.
The decay constant is $\lambda = 0.693 / T_{1/2} = 0.693 / (1600 \times 365 \times 24 \times 3600 \ \text{s}) \approx 1.373 \times 10^{-11} \ \text{s}^{-1}$.
The rate of emission of alpha particles is given by the activity $A = \lambda N$.
$A = (1.373 \times 10^{-11} \ \text{s}^{-1}) \times (2.665 \times 10^{15} \ \text{atoms}) \approx 3.66 \times 10^4 \ \text{particles/s}$.
Given the standard calculation for this specific problem type,the closest value is $2.92 \times 10^4 / \text{sec}$.

(A) The relationship between the disintegration constant $(k)$ and the half-life $(t_{1/2})$ is given by the formula: $k = \frac{0.693}{t_{1/2}}$.
Given that $t_{1/2} = 3 \ hr$,we substitute this value into the equation:
$k = \frac{0.693}{3 \ hr} = 0.231 \ hr^{-1}$.
Therefore,the correct option is $A$.

(B) The half-life $t_{1/2}$ of $^{14}C$ is $5760 \, years$. The decay constant $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5760} \, years^{-1}$.
Using the radioactive decay formula: $t = \frac{2.303}{\lambda} \log \left( \frac{N_0}{N_t} \right)$.
Given that the remaining activity is $12.5\%$,we have $\frac{N_0}{N_t} = \frac{100}{12.5} = 8$.
Substituting the values: $t = \frac{2.303 \times 5760}{0.693} \log(8)$.
Since $\log(8) = 3 \log(2) = 3 \times 0.3010 = 0.9030$.
$t = \frac{2.303 \times 5760 \times 0.9030}{0.693} \approx 17281 \, years$.
Converting to scientific notation: $17281 = 172.81 \times 10^2 \, years$.

(C) In radioactive equilibrium,the rate of decay of the parent is equal to the rate of decay of the daughter: $\lambda_R N_R = \lambda_U N_U$.
Since $\lambda = \frac{0.693}{t_{1/2}}$,we have $\frac{0.693}{t_{1/2}(R)} \times N_R = \frac{0.693}{t_{1/2}(U)} \times N_U$.
This simplifies to $\frac{N_R}{N_U} = \frac{t_{1/2}(R)}{t_{1/2}(U)}$.
Given $\frac{N_R}{N_U} = \frac{1}{2.8 \times 10^6}$ and $t_{1/2}(R) = 1620$ years.
Substituting the values: $\frac{1}{2.8 \times 10^6} = \frac{1620}{t_{1/2}(U)}$.
Therefore,$t_{1/2}(U) = 1620 \times 2.8 \times 10^6 = 4536 \times 10^6 = 4.536 \times 10^9$ years.
Rounding to the nearest option,the answer is $4.53 \times 10^9$ years.

(C) The relationship between average life $(\tau)$ and half-life $(t_{1/2})$ is given by the formula: $\tau = 1.44 \times t_{1/2}$.
Given that $t_{1/2} = 1580 \ yrs$.
Therefore,$\tau = 1.44 \times 1580 \ yrs$.
$\tau = 2275.2 \ yrs = 2.275 \times 10^{3} \ yrs$.

(A) The radioactive decay formula is given by $N = N_o (1/2)^n$,where $n$ is the number of half-lives.
Given $N_o = 8 \ g$,$N = 0.5 \ g$,and $t = 60 \ \min$.
$0.5 = 8 \times (1/2)^n$
$(1/2)^n = 0.5 / 8 = 1 / 16$
$(1/2)^n = (1/2)^4$
So,$n = 4$.
Since $n = t / {t_{1/2}}$,we have $4 = 60 / {t_{1/2}}$.
${t_{1/2}} = 60 / 4 = 15 \ \min$.

(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 45 \ minutes = 0.75 \ hours$.
So,$k = \frac{0.693}{0.75} \ hr^{-1}$.
The time $t$ required for $99.9\%$ completion is given by $t = \frac{2.303}{k} \log \frac{a}{a - 0.999a} = \frac{2.303}{k} \log \frac{1}{0.001} = \frac{2.303}{k} \log 10^3$.
Substituting the value of $k$: $t = \frac{2.303 \times 0.75}{0.693} \times 3$.
Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we get $t = 3.32 \times 0.75 \times 3 = 7.5 \ hours$.

(C) Given: Total time $T = 50 \ days$,Final amount $N = \frac{1}{32} N_0$.
Using the radioactive decay formula: $N = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
$\frac{1}{32} N_0 = N_0 \times (\frac{1}{2})^n \implies (\frac{1}{2})^5 = (\frac{1}{2})^n \implies n = 5$.
Since $n = \frac{T}{t_{1/2}}$,we have $5 = \frac{50}{t_{1/2}}$.
Therefore,$t_{1/2} = \frac{50}{5} = 10 \ days$.

(D) The amount of substance remaining after $40 \, min$ is $100\% - 87.5\% = 12.5\%$.
Since $12.5\% = (1/2)^3$ of the initial amount,the substance has undergone $3$ half-lives.
Therefore,$3 \times t_{1/2} = 40 \, min$.
$t_{1/2} = 40 / 3 \, min = 13.33 \, min$.
$0.33 \, min = 0.33 \times 60 \, sec = 20 \, sec$.
Thus,the half-life is $13 \, min \, 20 \, sec$.

(D) The half-life period $t_{1/2} = 10 \ days$.
Total time elapsed $t = 40 \ days$.
The number of half-lives $n = \frac{t}{t_{1/2}} = \frac{40}{10} = 4$.
The formula for radioactive decay is $N = N_0 \times (\frac{1}{2})^n$,where $N$ is the remaining amount and $N_0$ is the initial amount.
Given $N = 125 \ g$ and $n = 4$,we have $125 = N_0 \times (\frac{1}{2})^4$.
$125 = N_0 \times \frac{1}{16}$.
$N_0 = 125 \times 16 = 2000 \ g$.

(A) The relationship between half-life $(t_{1/2})$ and decay constant $(k)$ is given by:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 6.31 \times 10^{-4} \ yr^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{6.31 \times 10^{-4}} \ yr$
$t_{1/2} \approx 0.1098 \times 10^4 \ yr$
$t_{1/2} = 1098 \ yrs$.

(C) The number of half-lives $n$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{3000 \ years}{1500 \ years} = 2$.
The amount of radioactive substance remaining after $n$ half-lives is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
Given $N_0 = 1 \ g$ and $n = 2$,the amount remaining is $N = 1 \times (\frac{1}{2})^2 = \frac{1}{4} = 0.25 \ g$.

(A) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{18 \ h}{3 \ h} = 6$.
Using the radioactive decay formula: $N_t = N_o \times (1/2)^n$.
Substituting the values: $N_t = 256 \times (1/2)^6$.
$N_t = 256 \times \frac{1}{64} = 4 \ g$.
Therefore,the remaining mass is $4.0 \ g$.

(B) The fraction of the radioactive sample decayed is $\frac{15}{16}$.
The fraction of the sample remaining is $N = 1 - \frac{15}{16} = \frac{1}{16}$.
Using the radioactive decay formula $N = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives:
$\frac{1}{16} = (\frac{1}{2})^n \implies (\frac{1}{2})^4 = (\frac{1}{2})^n \implies n = 4$.
Since the total time $t = 40 \text{ days}$ and $n = \frac{t}{T_{1/2}}$,we have $4 = \frac{40}{T_{1/2}}$.
Therefore,the half-life $T_{1/2} = \frac{40}{4} = 10 \text{ days}$.

(C) The formula for radioactive decay is $N_t = N_o (1/2)^n$,where $n$ is the number of half-lives.
First,calculate the number of half-lives $(n)$: $n = \frac{26 \ hr}{6.5 \ hr} = 4$.
Now,substitute the values into the formula: $N_t = 48 \times 10^{19} \times (1/2)^4$.
$N_t = 48 \times 10^{19} \times \frac{1}{16}$.
$N_t = 3 \times 10^{19}$ atoms.

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{9} \ h^{-1}$.
Using the integrated rate law: $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.
Substituting the values: $\frac{0.693}{9} = \frac{2.303}{t} \log \left( \frac{1}{0.2} \right)$.
$\frac{0.693}{9} = \frac{2.303}{t} \log(5)$.
Since $\log(5) \approx 0.699$,we have $\frac{0.693}{9} = \frac{2.303 \times 0.699}{t}$.
$t = \frac{2.303 \times 0.699 \times 9}{0.693} \approx 20.9 \ hours$.