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

(A) Let the initial mass of both radioactive elements $A$ and $B$ be $M_0$. Since their atomic weights are the same,the initial number of atoms $(N_0)_A = (N_0)_B = N_0$.
For element $A$,the amount remaining after time $t$ is $N_A = N_0 \left(\frac{1}{2}\right)^{t/10} = \frac{1}{e}$.
For element $B$,the amount remaining after time $t$ is $N_B = N_0 \left(\frac{1}{2}\right)^{t/20} = 1$.
Dividing the two equations:
$\frac{N_A}{N_B} = \frac{1/e}{1} = \frac{N_0 (1/2)^{t/10}}{N_0 (1/2)^{t/20}}$
$\frac{1}{e} = \left(\frac{1}{2}\right)^{t/10 - t/20} = \left(\frac{1}{2}\right)^{t/20}$.
Taking natural logarithm on both sides:
$\ln(e^{-1}) = \frac{t}{20} \ln(1/2)$
$-1 = \frac{t}{20} (-\ln 2)$
$1 = \frac{t}{20} (0.7)$
$t = \frac{20}{0.7} = \frac{200}{7} \ yrs$.

(B) The decay constant $\lambda$ for a process is given by $\lambda = \frac{\ln 2}{T_{1/2}}$.
For two parallel decay processes,the effective decay constant is $\lambda_{\text{eff}} = \lambda_1 + \lambda_2$.
Given $T_1 = 0.7 \ hr$ and $T_2 = 0.3 \ hr$,and $\ln 2 = 0.7$.
Then $\lambda_1 = \frac{0.7}{0.7} = 1 \ hr^{-1}$ and $\lambda_2 = \frac{0.7}{0.3} = \frac{7}{3} \ hr^{-1}$.
Effective decay constant $\lambda_{\text{eff}} = 1 + \frac{7}{3} = \frac{10}{3} \ hr^{-1}$.
The effective average life $\tau_{\text{eff}}$ is the reciprocal of the effective decay constant:
$\tau_{\text{eff}} = \frac{1}{\lambda_{\text{eff}}} = \frac{1}{10/3} = 0.3 \ hr$.
Converting to minutes: $\tau_{\text{eff}} = 0.3 \times 60 \ min = 18 \ min$.

(A) Given,half-life $t_{1/2} = 693 \ s$. Kinetic energy $K = 0.084 \ eV = 0.084 \times 1.6 \times 10^{-19} \ J$.
Using $K = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 0.084 \times 1.6 \times 10^{-19}}{1.68 \times 10^{-27}}} = \sqrt{0.16 \times 10^8} = 0.4 \times 10^4 \ m/s$.
Time taken to travel $d = 1000 \ m$ is $t = \frac{d}{v} = \frac{1000}{0.4 \times 10^4} = 0.25 \ s$.
Using the decay law $N = N_0 e^{-\lambda t}$,the fraction decayed is $\frac{N_0 - N}{N_0} = 1 - e^{-\lambda t}$.
Since $\lambda t = \frac{\ln 2}{t_{1/2}} \times t$ is very small,$1 - e^{-\lambda t} \approx \lambda t = \frac{\ln 2 \times t}{t_{1/2}}$.
Fraction decayed $= \frac{0.693 \times 0.25}{693} = 0.25 \times 10^{-3} \times 10^{-2} = 0.25 \times 10^{-5}$.

(B) Given decay constants are $\lambda_1 = 6\lambda$ and $\lambda_2 = \lambda$.
The half-life of $R_2$ is $(t_{1/2})_2 = 1.4 \times 10^{17} \,s$.
The number of nuclei remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.
For $R_1$: $N_1 = N_0 e^{-\lambda_1 t} = N_0 e^{-6\lambda t}$.
For $R_2$: $N_2 = N_0 e^{-\lambda_2 t} = N_0 e^{-\lambda t}$.
Given the ratio $\frac{N_2}{N_1} = e$.
$\frac{N_0 e^{-\lambda t}}{N_0 e^{-6\lambda t}} = e$.
$e^{-\lambda t + 6\lambda t} = e^1$.
$e^{5\lambda t} = e^1$.
Comparing the exponents: $5\lambda t = 1$, so $t = \frac{1}{5\lambda}$.
We know $\lambda = \frac{\ln 2}{(t_{1/2})_2} = \frac{0.7}{1.4 \times 10^{17}} = 0.5 \times 10^{-17} \,s^{-1}$.
Substituting $\lambda$ into the expression for $t$:
$t = \frac{1}{5 \times 0.5 \times 10^{-17}} = \frac{1}{2.5 \times 10^{-17}} = 0.4 \times 10^{17} \,s = 4 \times 10^{16} \,s$.

(B) The activity of a radioactive sample at time $t$ is given by $A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$,where $T_{1/2}$ is the half-life.
Alternatively,using the general decay law $A = A_0 \left(\frac{1}{n}\right)^{t/t_0}$,where the activity becomes $\frac{1}{n}$ of its initial value in time $t_0$.
Given that the activity becomes $\frac{1}{3}$ of its initial value in $t_0 = 20 \text{ hours}$,we have $\frac{A}{A_0} = \left(\frac{1}{3}\right)^{t/t_0}$.
For $t = 80 \text{ hours}$,the fraction remaining is $\frac{A}{A_0} = \left(\frac{1}{3}\right)^{80/20}$.
$\frac{A}{A_0} = \left(\frac{1}{3}\right)^4 = \frac{1}{81}$.

(D) The initial amount of the sample is $N_0 = 100 \text{ g}$.
The final amount of the sample is $N = 3.125 \text{ g}$.
The half-life $T_{1/2} = 103 \text{ years}$.
Using the radioactive decay formula: $N = N_0 \left( \frac{1}{2} \right)^n$,where $n$ is the number of half-lives.
$3.125 = 100 \left( \frac{1}{2} \right)^n$
$\left( \frac{1}{2} \right)^n = \frac{3.125}{100} = \frac{1}{32}$
Since $\frac{1}{32} = \left( \frac{1}{2} \right)^5$,we have $n = 5$.
The total time taken is $t = n \times T_{1/2} = 5 \times 103 \text{ years} = 515 \text{ years}$.

(B) Given the decay rate formula: $A(t) = A_0 2^{-(t/t_0)}$,where $t_0$ is the half-life.
We are given $A_1$ at $t_1 = 120 \ h$ and $A_2$ at $t_2 = 200 \ h$.
The ratio is given as $\frac{A_1}{A_2} = 16$.
Substituting the expressions for $A_1$ and $A_2$:
$\frac{A_0 2^{-(120/t_0)}}{A_0 2^{-(200/t_0)}} = 16$
$2^{-(120/t_0) + (200/t_0)} = 2^4$
$2^{(80/t_0)} = 2^4$
Equating the exponents:
$\frac{80}{t_0} = 4$
$t_0 = \frac{80}{4} = 20 \ h$.
Thus,the half-life is $20 \ h$.

(A) Given decay constant,$\lambda = 7.9 \times 10^{-10} / s$.
Activity of the sample,$A = 55.3 \times 10^{11} \text{ disintegration/second}$.
We know that the relationship between activity $A$ and the number of nuclei $N$ is given by the formula:
$A = \lambda N$
To find the number of nuclei $N$,we rearrange the formula:
$N = \frac{A}{\lambda}$
Substituting the given values:
$N = \frac{55.3 \times 10^{11}}{7.9 \times 10^{-10}}$
$N = 7.0 \times 10^{21}$
Therefore,the number of nuclei at that instant is $7.0 \times 10^{21}$.

(B) The decay constant for the first process is $\lambda_1 = \frac{\ln 2}{T_1}$ and for the second process is $\lambda_2 = \frac{\ln 2}{T_2}$.
Since the nucleus decays by two processes,the effective decay constant is $\lambda = \lambda_1 + \lambda_2$.
Substituting the expressions for decay constants,we get $\frac{\ln 2}{T} = \frac{\ln 2}{T_1} + \frac{\ln 2}{T_2}$,which simplifies to $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2}$.
Thus,the effective half-life $T$ is given by $T = \frac{T_1 T_2}{T_1 + T_2}$.
Given $T_1 = 5 \times 10^3 \text{ yr}$ and $T_2 = 10^5 \text{ yr} = 100 \times 10^3 \text{ yr}$.
$T = \frac{(5 \times 10^3) \times (100 \times 10^3)}{5 \times 10^3 + 100 \times 10^3} = \frac{500 \times 10^6}{105 \times 10^3} = \frac{500000}{105} \approx 4761.9 \text{ yr}$.
Rounding to the nearest integer,we get $4762 \text{ yr}$.

(B) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$.
Given $t = 40 \ h$ and $T_{1/2} = 10 \ h$,we have $n = \frac{40}{10} = 4$.
The remaining fraction of the initial radioactivity is given by $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting $n = 4$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.

(C) The rate of decay is given by $\frac{dN}{dt} = \lambda N$,where $\lambda = \frac{0.693}{T_{1/2}}$.
First,convert the half-life $T_{1/2} = 1620$ years into seconds:
$T_{1/2} = 1620 \times 365 \times 24 \times 3600 \approx 5.11 \times 10^{10} \ s$.
Next,calculate the number of atoms $N$ in $1 \ g$ of $Ra^{226}$:
$N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1}{226} \times 6.023 \times 10^{23} \approx 2.665 \times 10^{21}$ atoms.
Now,calculate the activity $\frac{dN}{dt} = \frac{0.693}{T_{1/2}} \times N$:
$\frac{dN}{dt} = \frac{0.693}{5.11 \times 10^{10}} \times 2.665 \times 10^{21} \approx 3.61 \times 10^{10}$ decays per second.

(D) The number of nuclei remaining at time $t$ is given by the radioactive decay law: $N(t) = N_0 e^{-\lambda t}$.
For material $X_1$ with decay constant $10 \lambda$: $N_1 = N_0 e^{-10 \lambda t}$.
For material $X_2$ with decay constant $\lambda$: $N_2 = N_0 e^{-\lambda t}$.
Given that the ratio $N_1 / N_2 = 1 / e$ at time $t$:
$\frac{N_1}{N_2} = \frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = e^{-10 \lambda t + \lambda t} = e^{-9 \lambda t}$.
We are given $\frac{N_1}{N_2} = e^{-1}$.
Equating the exponents: $-9 \lambda t = -1$.
Therefore,$t = \frac{1}{9 \lambda}$.

(A) The number of radioactive nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
Average life $\tau$ is defined as $\tau = 1/\lambda$.
At time $t = \tau$,the number of remaining nuclei is $N(\tau) = N_0 e^{-\lambda(1/\lambda)} = N_0 e^{-1} \approx N_0 / 2.718 \approx 0.368 N_0$.
The number of nuclei that have decayed is $N_{decayed} = N_0 - N(\tau) = N_0 - 0.368 N_0 = 0.632 N_0$.
Since $0.632 N_0 > 0.5 N_0$,more than half of the active nuclei decay in one average lifetime.

(A) The law of radioactive decay is given by $N = N_0 e^{-\lambda t}$.
Activity $A$ is defined as the rate of decay of nuclei,given by $A = -\frac{dN}{dt}$.
Differentiating $N$ with respect to $t$,we get:
$\frac{dN}{dt} = -\lambda N_0 e^{-\lambda t} = -\lambda N$.
Therefore,the magnitude of activity is $A = |-\frac{dN}{dt}| = \lambda N$.
This equation $A = \lambda N$ represents a linear relationship between $N$ and $A$,where $\lambda$ is the decay constant.
Since $A$ is directly proportional to $N$ $(A \propto N)$,the graph of $N$ versus $A$ will be a straight line passing through the origin. Thus,option $A$ is correct.

(B) The decay process is $X \xrightarrow{\alpha} Y \xrightarrow{\beta} Z$.
Let $N_X$ and $N_Y$ be the number of nuclei of $X$ and $Y$ present at any time $t$.
The rate of change of the number of daughter nuclei $Y$ is given by: $\frac{dN_Y}{dt} = \lambda_X N_X - \lambda_Y N_Y$.
Since the rate of emission of $\beta$-particles $(\lambda_Y N_Y)$ is constant over several months,we have $\frac{dN_Y}{dt} \approx 0$.
This implies $\lambda_X N_X = \lambda_Y N_Y$.
The rate of emission of $\alpha$-particles is $\lambda_X N_X$,and the rate of emission of $\beta$-particles is $\lambda_Y N_Y$.
Given that the rate of $\beta$-emission is $10^{7} / \text{h}$,the rate of $\alpha$-emission must also be $10^{7} / \text{h}$.
Therefore,the number of $\alpha$-particles emitted in one hour is $10^{7}$.

(D) Given,half-life $t_{1/2} = 3$ days.
Number of active nuclei remaining after time $t$ is given by $N(t) = N_0 (1/2)^{t/t_{1/2}}$.
At the start of the third day ($t = 2$ days),the number of nuclei remaining is $N_1 = N_0 (1/2)^{2/3} = N_0 / (2^{2/3}) = N_0 / (4^{1/3})$.
Given $\sqrt[3]{0.25} \approx 0.63$,we have $N_1 = N_0 \times 0.63$.
At the end of the third day ($t = 3$ days),the number of nuclei remaining is $N_2 = N_0 (1/2)^{3/3} = 0.5 N_0$.
The number of nuclei that decay during the third day is the difference between the nuclei present at the start and end of the day:
$\Delta N = N_1 - N_2 = 0.63 N_0 - 0.5 N_0 = 0.13 N_0$.
Thus,the fraction of the initial nuclei that decay on the third day is $0.13$.

(A) The half-life of Radon-$222$ is $T_{1/2} = 3.8 \text{ days}$.
Total time elapsed is $t = 19 \text{ days}$.
The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}} = \frac{19}{3.8} = 5$.
The remaining quantity $N$ is given by the formula $N = N_0 \left(\frac{1}{2}\right)^n$,where $N_0$ is the initial amount.
Given $N_0 = 0.064 \ kg$ and $n = 5$,we have:
$N = 0.064 \times \left(\frac{1}{2}\right)^5$
$N = 0.064 \times \frac{1}{32}$
$N = 0.002 \ kg$.
Thus,the quantity of Radon-$222$ left after $19$ days is $0.002 \ kg$.

(A) The radioactive decay law is given by $N(t) = N_{0} e^{-\lambda t}$,where $\lambda = \frac{\ln 2}{T}$.
Let $t_{1}$ be the time when the number of atoms is $N_{0}/2$ and $t_{2}$ be the time when the number of atoms is $N_{0}/10$.
For $N(t_{1}) = N_{0}/2$,we have $\frac{N_{0}}{2} = N_{0} e^{-\lambda t_{1}} \Rightarrow e^{\lambda t_{1}} = 2 \Rightarrow \lambda t_{1} = \ln 2$.
For $N(t_{2}) = N_{0}/10$,we have $\frac{N_{0}}{10} = N_{0} e^{-\lambda t_{2}} \Rightarrow e^{\lambda t_{2}} = 10 \Rightarrow \lambda t_{2} = \ln 10$.
The time interval required is $\Delta t = t_{2} - t_{1} = \frac{\ln 10 - \ln 2}{\lambda} = \frac{\ln(10/2)}{\lambda} = \frac{\ln 5}{\lambda}$.
Since $\lambda = \frac{\ln 2}{T}$,we substitute $\lambda$ to get $\Delta t = \frac{\ln 5}{(\ln 2 / T)} = T \frac{\ln 5}{\ln 2} = T \log_{2} 5$.

(B) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For substance $A$,$N_A = N_0 e^{-5 \lambda t}$.
For substance $B$,$N_B = N_0 e^{-\lambda t}$.
Given the ratio $\frac{N_A}{N_B} = (1/e)^2 = e^{-2}$.
Substituting the expressions:
$\frac{N_0 e^{-5 \lambda t}}{N_0 e^{-\lambda t}} = e^{-2}$
$e^{-5 \lambda t + \lambda t} = e^{-2}$
$e^{-4 \lambda t} = e^{-2}$
Equating the exponents:
$-4 \lambda t = -2$
$t = \frac{2}{4 \lambda} = \frac{1}{2 \lambda}$.

(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$, where $N(t)$ is the amount remaining at time $t$.
For $20 \%$ decay, the amount remaining is $N_1 = 80 \% \text{ of } N_0 = 0.8 N_0$. Thus, $0.8 N_0 = N_0 e^{-\lambda t_1} \Rightarrow e^{-\lambda t_1} = 0.8$.
For $80 \%$ decay, the amount remaining is $N_2 = 20 \% \text{ of } N_0 = 0.2 N_0$. Thus, $0.2 N_0 = N_0 e^{-\lambda t_2} \Rightarrow e^{-\lambda t_2} = 0.2$.
Dividing the two equations: $\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}} = \frac{0.8}{0.2} = 4$.
This simplifies to $e^{\lambda(t_2 - t_1)} = 4$.
Taking the natural logarithm on both sides: $\lambda(t_2 - t_1) = \ln(4) = 2 \ln(2)$.
Since $\lambda = \frac{\ln 2}{T_{1/2}}$, we have $\frac{\ln 2}{T_{1/2}}(t_2 - t_1) = 2 \ln 2$.
Therefore, $t_2 - t_1 = 2 \times T_{1/2} = 2 \times 20 \text{ min} = 40 \text{ min}$.