Rate of decay and Half-life Questions in English

(B) The rate of disintegration $(r)$ of a radioactive substance is given by the equation $r = \lambda \cdot N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms present.
Since the rate of disintegration is directly proportional to the number of atoms $(r \propto N)$,if the amount of substance (number of atoms $N$) is increased three times,the number of atoms disintegrated per unit time will also increase three times.

(B) The number of half-lives $n$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{24}{8} = 3$.
The remaining mass $N$ is given by the formula $N = \frac{N_0}{2^n}$,where $N_0$ is the initial mass.
Substituting the values,$N = \frac{40}{2^3} = \frac{40}{8} = 5 \ g$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{90}{30} = 3$.
The number of remaining atoms $(N)$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$,where $N_0$ is the initial number of atoms.
Substituting the values: $N = 600 \times (\frac{1}{2})^3 = 600 \times \frac{1}{8} = 75 \ atoms$.

(D) The rate of radioactive disintegration is given by the formula: $Rate = \lambda \times N$,where $\lambda$ is the decay constant and $N$ is the number of radioactive atoms present.
Since the rate is directly proportional to the number of atoms $(Rate \propto N)$,if the quantity of the radioactive element is doubled,the number of atoms $N$ becomes $2N$.
Therefore,the new rate becomes $\lambda \times (2N) = 2 \times (\lambda \times N)$,which is double the original rate.
Thus,the correct option is $(D)$.

(C) The disintegration constant $(k)$ is related to the half-life $(t_{1/2})$ by the formula: $k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1000 \ s$.
Substituting the value: $k = \frac{0.693}{1000 \ s} = 0.000693 \ s^{-1} = 6.93 \times 10^{-4} \ s^{-1}$.

(B) The disintegration constant $(K)$ is related to the half-life $(t_{1/2})$ by the formula: $K = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 1600 \ years$.
$K = \frac{0.693}{1600} = 4.33 \times 10^{-4} \ year^{-1}$.

(C) The relationship between the disintegration constant $(K)$ and the half-life period $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{K}$
Given $K = 0.58 \, hr^{-1}$,
$t_{1/2} = \frac{0.693}{0.58} \approx 1.2 \, hr$.
Therefore,the correct option is $C$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{16}{8} = 2$.
The remaining mass $(N)$ is given by the formula $N = \frac{N_0}{2^n}$,where $N_0$ is the initial mass.
Substituting the values: $N = \frac{16.0}{2^2} = \frac{16.0}{4} = 4.0 \ g$.

(B) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{80}{20} = 4$.
The fraction of the substance remaining after $n$ half-lives is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting $n = 4$,we get $\frac{N}{N_0} = (\frac{1}{2})^4 = \frac{1}{16}$.

(A) The formula for radioactive decay is $N = (1/2)^n \times N_o$,where $n$ is the number of half-lives.
Given $N_o = 1.0 \, g = 1000 \, mg$ and $N = 125 \, mg$.
$(1/2)^n = 125 / 1000 = 1/8$.
Since $1/8 = (1/2)^3$,we have $n = 3$.
The number of half-lives $n = \text{Total time} / t_{1/2}$.
$3 = 24 \, hours / t_{1/2}$.
Therefore,$t_{1/2} = 24 / 3 = 8 \, hours$.

(B) The radioactive decay follows first-order kinetics: $N_t = N_0 e^{-\lambda t}$.
For the first interval: $N_t = N_0 / 10$ at $t = 15 \ min$.
So,$1/10 = e^{-\lambda(15)} \implies \ln(10) = 15\lambda \implies \lambda = \ln(10) / 15$.
For the total time $T$ to reach $1/100$ of the original amount: $N_T = N_0 / 100$.
$1/100 = e^{-\lambda T} \implies \ln(100) = \lambda T$.
Since $\ln(100) = 2 \ln(10)$,we have $2 \ln(10) = (\ln(10) / 15) \times T$.
Solving for $T$: $T = 2 \times 15 = 30 \ min$.
The additional time required is $T - 15 = 30 - 15 = 15 \ min$.

(D) The amount of substance remaining after $75\%$ disintegration is $100\% - 75\% = 25\%$.
Since $25\% = (1/2)^2$,the substance has undergone $2$ half-lives.
Given that $2$ half-lives $= 30 \, \text{min}$,the half-life $t_{1/2}$ is calculated as:
$t_{1/2} = \frac{30 \, \text{min}}{2} = 15 \, \text{min}$.
Therefore,the correct option is $D$.

(A) The relationship between average life $(\tau)$ and half-life $(t_{1/2})$ is given by $\tau = \frac{t_{1/2}}{0.693}$.
Given $t_{1/2} = 69.3 \text{ minutes}$.
Therefore,$\tau = \frac{69.3}{0.693} = 100 \text{ minutes}$.

(A) The radioactive decay follows the formula $N = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
For the first case: $1.25 = 10 \times (1/2)^n \implies (1/2)^n = 1.25/10 = 1/8 = (1/2)^3$.
Thus,$n = 3$ half-lives correspond to $15 \ days$.
Therefore,the half-life period $T_{1/2} = 15/3 = 5 \ days$.
For the second case: $N_0 = 1 \ kg = 1000 \ g$ and $N = 500 \ g$.
$500 = 1000 \times (1/2)^n \implies (1/2)^n = 500/1000 = 1/2 = (1/2)^1$.
Thus,$n = 1$ half-life is required.
Time taken = $n \times T_{1/2} = 1 \times 5 = 5 \ days$.

(D) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{12}{3} = 4$.
The relationship between the initial amount $(N_0)$ and the remaining amount $(N)$ is given by $N = N_0 \times (\frac{1}{2})^n$.
Rearranging for $N_0$,we get $N_0 = N \times 2^n$.
Substituting the given values,$N_0 = 3 \times 2^4 = 3 \times 16 = 48 \ g$.

(C) The decay constant $\lambda$ is given by the formula: $\lambda = \frac{2.303}{t} \log \left( \frac{N_0}{N} \right)$.
Given $t = 2.303 \ s$ and $\frac{N}{N_0} = \frac{1}{10}$,so $\frac{N_0}{N} = 10$.
Substituting the values: $\lambda = \frac{2.303}{2.303} \log(10) = 1 \times 1 = 1 \ s^{-1}$.
The half-life period $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{\lambda}$.
Therefore,$t_{1/2} = \frac{0.693}{1} = 0.693 \ s$.

(D) Given,half-life $(t_{1/2})$ = $60 \ minutes = 1 \ hour$.
Total time $(T)$ = $3 \ hours$.
Number of half-lives $(n)$ = $\frac{T}{t_{1/2}} = \frac{3 \ hours}{1 \ hour} = 3$.
The fraction of substance remaining is given by $\frac{N}{N_o} = (\frac{1}{2})^n$.
Substituting $n = 3$,we get $\frac{N}{N_o} = (\frac{1}{2})^3 = \frac{1}{8}$.
Percentage remaining = $\frac{1}{8} \times 100 = 12.5\%$.
Therefore,the correct option is $(D)$.

(B) The intensity of radiation is proportional to the amount of radioactive substance present.
Let the initial intensity be $I_0$ and the safe level be $I_s$.
Given $I_0 = 64 \times I_s$.
Using the radioactive decay formula $I = I_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
We need $I = I_s$,so $I_s = 64 \times I_s \times (\frac{1}{2})^n$.
$\frac{1}{64} = (\frac{1}{2})^n$.
Since $64 = 2^6$,we have $(\frac{1}{2})^6 = (\frac{1}{2})^n$,which gives $n = 6$.
Total time $t = n \times T_{1/2} = 6 \times 2 \ hr = 12 \ hr$.

(B) The half-life $(t_{1/2})$ of a radioactive sample is given by the formula $t_{1/2} = \frac{\ln 2}{\lambda}$.
The mean life $(\tau)$ of a radioactive sample is the reciprocal of the decay constant,given by $\tau = \frac{1}{\lambda}$.
Therefore,the half-life and mean life are $\frac{\ln 2}{\lambda}$ and $\frac{1}{\lambda}$ respectively.

(A) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{60 \ h}{20 \ h} = 3$.
The remaining amount $(N)$ is given by the formula $N = N_o \times (1/2)^n$.
Substituting the values,$N = N_o \times (1/2)^3 = N_o \times (1/8)$.
Therefore,the fraction remaining is $1/8$ of the initial amount.

(A) The initial amount of the sample is $12 \ g$.
Since $6 \ g$ decays in $1 \ hr$,the half-life $(t_{1/2})$ of the substance is $1 \ hr$.
After the first hour,the remaining amount of the sample is $12 \ g - 6 \ g = 6 \ g$.
In the next hour (the second half-life),half of the remaining $6 \ g$ will decay.
Amount decayed in the next hour = $\frac{1}{2} \times 6 \ g = 3 \ g$.

(C) The decay equation is given by $N = N_0 e^{-\lambda t}$ or $\lambda = \frac{2.303}{t} \log \frac{N_0}{N}$.
Given $t = 4.2 \ days$ and $N = 0.798 \ N_0$,we have $\frac{N_0}{N} = \frac{1}{0.798} \approx 1.253$.
$\lambda = \frac{2.303}{4.2} \log(1.253) \approx 0.5483 \times 0.0979 \approx 0.0537 \ days^{-1}$.
The half-life $t_{1/2}$ is given by $t_{1/2} = \frac{0.693}{\lambda}$.
$t_{1/2} = \frac{0.693}{0.0537} \approx 12.9 \ days$.
Comparing with the given options,the closest value is $12.83$ $days$.

(A) The half-life $(t_{1/2})$ of a radioactive substance is a characteristic property of the radioactive isotope and is independent of the initial amount or mass of the sample.
Therefore,if the half-life of $2.0 \ g$ of the substance is $7 \ days$,the half-life of $1 \ g$ of the same substance will also be $7 \ days$.

(A) The number of half-lives $(n)$ is calculated as: $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{80 \text{ years}}{20 \text{ years}} = 4$.
The activity after $n$ half-lives is given by the formula: $A = \frac{A_0}{2^n}$.
Given $A_0 = 8000 \text{ dis/min}$ and $n = 4$,we have:
$A = \frac{8000}{2^4} = \frac{8000}{16} = 500 \text{ dis/min}$.

(B) The half-life $(t_{1/2})$ of a radioactive isotope is a nuclear property that is independent of external physical conditions such as temperature,pressure,or chemical state. Therefore,heating the sample will not change its half-life. The correct option is $B$.

(C) The half-life $t_{1/2}$ is given as $2.95 \text{ days}$.
First,convert the half-life into seconds: $t_{1/2} = 2.95 \times 24 \times 60 \times 60 \ s = 254880 \ s$.
The decay constant $\lambda$ is calculated using the formula $\lambda = \frac{0.693}{t_{1/2}}$.
Substituting the values: $\lambda = \frac{0.693}{254880} \approx 2.7 \times 10^{-6} \ s^{-1}$.

(B) The formula for half-life $(t_{1/2})$ of a first-order reaction or radioactive decay is given by:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 2.31 \times 10^{-4} \ yr^{-1}$.
Substituting the value:
$t_{1/2} = \frac{0.693}{2.31 \times 10^{-4}} = 0.3 \times 10^4 \ yrs$
$t_{1/2} = 3.0 \times 10^3 \ yrs$.

(A) The formula for radioactive decay is $N = N_0(1/2)^n$,where $n$ is the number of half-lives.
Given,half-life $t_{1/2} = 10 \ days$ and total time $t = 40 \ days$.
Number of half-lives $n = t / t_{1/2} = 40 / 10 = 4$.
Remaining amount $N = 125 \ mg = 0.125 \ g$.
Substituting the values: $0.125 = N_0(1/2)^4$.
$0.125 = N_0(1/16)$.
$N_0 = 0.125 \times 16 = 2 \ g$.

(A) The half-life $(t_{1/2})$ of a radioactive substance is a characteristic property of the isotope and is independent of the initial amount of the substance present.
Since the half-life for $10 \ g$ is $10 \ days$,the half-life for $20 \ g$ will also be $10 \ days$.

(B) The total time elapsed from February $1$ to July $1$ is $28 + 31 + 30 + 31 + 30 = 150 \, days$.
Using the radioactive decay formula: $N = N_0 (1/2)^n$,where $n$ is the number of half-lives.
$0.25 = 8 \times (1/2)^n$
$(1/2)^n = 0.25 / 8 = 1 / 32 = (1/2)^5$.
Therefore,$n = 5$.
Since $n = \text{total time} / t_{1/2}$,we have $5 = 150 / t_{1/2}$.
$t_{1/2} = 150 / 5 = 30 \, days$.

(D) Given initial amount $N_o = 10 \ g$,final amount $N = 1.25 \ g$,and half-life $t_{1/2} = 1620 \ years$.
Using the radioactive decay formula: $N = N_o (1/2)^n$,where $n$ is the number of half-lives $(n = t / t_{1/2})$.
Substituting the values: $1.25 = 10 \times (1/2)^n$.
$1.25 / 10 = (1/2)^n \implies 0.125 = (1/2)^n \implies 1/8 = (1/2)^n$.
Since $1/8 = (1/2)^3$,we get $n = 3$.
Now,calculating the total time: $t = n \times t_{1/2} = 3 \times 1620 = 4860 \ years$.

(D) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{480}{120} = 4$.
Using the radioactive decay formula $N = \frac{N_0}{2^n}$,where $N_0 = 4 \ g$ and $n = 4$:
$N = \frac{4}{2^4} = \frac{4}{16} = 0.25 \ g$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{28 \ years}{7 \ years} = 4$.
Using the radioactive decay formula $N = \frac{N_0}{2^n}$,where $N_0 = 1 \ g$ and $n = 4$:
$N = \frac{1}{2^4} = \frac{1}{16} = 0.0625 \ g$.
Therefore,the amount remaining is $0.0625 \ g$.

(C) 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,$n = 3$.
Since $n = \frac{t}{t_{1/2}}$,we have $3 = \frac{96}{t_{1/2}}$.
Thus,$t_{1/2} = \frac{96}{3} = 32 \ min$.

(A) The formula for radioactive decay is $N_t = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $N_0 = 100 \ mg$,$N_t = 25 \ mg$,and $t_{1/2} = 5760 \ years$.
$25 = 100 \times (\frac{1}{2})^n$
$(\frac{1}{2})^n = \frac{25}{100} = \frac{1}{4} = (\frac{1}{2})^2$
Therefore,$n = 2$.
Total time $t = n \times t_{1/2} = 2 \times 5760 = 11520 \ years$.

(C) The half-life $(t_{1/2})$ is defined as the time required for a radioactive substance to decay to half $(50 \%)$ of its initial amount.
Given that $t_{1/2} = 100 \ yrs$,the time taken for the element to disintegrate to $50 \%$ of its mass is exactly one half-life.
Therefore,the time required is $100 \ yrs$.

(B) The average life $(\tau)$ of a radioactive element is defined as the reciprocal of its decay constant or disintegration constant $(\lambda)$.
Mathematically,$(\tau = \frac{1}{\lambda})$.
Therefore,the correct option is $(B)$.

(B) The fraction of the substance remaining is given by the formula $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{16}$,we have $\frac{1}{16} = (\frac{1}{2})^n$.
Since $\frac{1}{16} = (\frac{1}{2})^4$,it follows that $n = 4$.
The total time required is $T = n \times t_{1/2} = 4 \times 30 \ min = 120 \ min$.

(D) The radioactive decay follows first-order kinetics.
According to the law of radioactive decay,the amount of substance remaining after time $t$ is given by $N_t = N_0 \times (1/2)^{t/T_{1/2}}$.
For the substance to decay completely,$N_t$ must be $0$.
Mathematically,this occurs only as $t \to \infty$.
Therefore,the time required for complete decay is infinite.

(C) The half-life period $(T_{1/2})$ is defined as the time required for the concentration of a reactant to be reduced to half of its initial value.
After one half-life period,the amount of substance remaining is given by the formula $N = N_0 \times (1/2)^n$,where $n$ is the number of half-lives.
For $n = 1$,$N = N_0 \times (1/2)^1 = N_0/2$.
Therefore,after one half-life,$1/2$ of the initial substance is left.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{15 \ days}{5 \ days} = 3$.
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: $N = \frac{20 \ g}{2^3} = \frac{20 \ g}{8} = 2.5 \ g$.

(A) The half-life period $(t_{1/2})$ of a radioactive isotope is a characteristic property of the isotope itself.
It is independent of the initial amount or concentration of the substance.
Therefore,the half-life of $0.5 \ g$ of the same substance remains the same as that of $2.0 \ g$,which is $20 \ hr$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{24 \ h}{8 \ h} = 3$.
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,$N = \frac{1 \ mg}{2^3} = \frac{1}{8} \ mg$.

(C) The half-life $(t_{1/2})$ of a radioactive substance is defined as the time required for the amount of the substance to be reduced to half of its initial value.
Given that the half-life of $_{92}U^{238}$ is $4.5 \times 10^9$ years,it follows by definition that after $4.5 \times 10^9$ years,the amount of $_{92}U^{238}$ will be reduced to half of its present amount.
Therefore,the correct option is $(C)$.

(C) The rate of disintegration $(r)$ is given by the formula $r = \lambda N$,where $\lambda = \frac{0.693}{t_{1/2}}$ and $N$ is the number of atoms.
First,calculate the decay constant $\lambda$ in seconds$^{-1}$:
$\lambda = \frac{0.693}{1600 \times 365 \times 24 \times 60 \times 60} \approx 1.37 \times 10^{-11} \ s^{-1}$.
Next,calculate the number of atoms $N$ in $1 \ g$ of radium:
$N = \frac{\text{mass}}{\text{atomic weight}} \times N_A = \frac{1}{226} \times 6.023 \times 10^{23} \approx 2.665 \times 10^{21} \ \text{atoms}$.
Finally,calculate the rate of disintegration $r$:
$r = (1.37 \times 10^{-11}) \times (2.665 \times 10^{21}) \approx 3.65 \times 10^{10} \ dps$.
This value is approximately $3.7 \times 10^{10} \ dps$.

(C) The radioactive decay follows the formula $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given $\frac{N}{N_0} = \frac{1}{16}$,we have $(\frac{1}{2})^n = \frac{1}{16} = (\frac{1}{2})^4$.
Thus,$n = 4$.
The total time $t = n \times t_{1/2} = 4 \times 6 \ months = 24 \ months$.
Since $12 \ months = 1 \ year$,$24 \ months = 2 \ years$.

(B) Given that the magnitude of half-life $(T_{1/2})$ is equal to the decay constant $(\lambda)$,we have $T_{1/2} = \lambda$.
We know the relationship between half-life and decay constant is $T_{1/2} = \frac{0.693}{\lambda}$.
Substituting $T_{1/2} = \lambda$ into the equation,we get $\lambda = \frac{0.693}{\lambda}$.
This implies $\lambda^2 = 0.693$.
Therefore,$\lambda = \sqrt{0.693} = (0.693)^{1/2}$.
Since $T_{1/2} = \lambda$,the value is $(0.693)^{1/2}$.

(C) The number of half-lives $(n)$ is calculated as $n = \frac{\text{Total time}}{\text{Half-life}} = \frac{3 \text{ days}}{1 \text{ day}} = 3$.
The amount remaining $(N)$ is given by the formula $N = N_o \times (1/2)^n$.
Substituting $n = 3$,we get $N = N_o \times (1/2)^3 = N_o \times 1/8$.
Therefore,$1/8$ of the original amount remains.

(B) The radioactive decay follows the formula: $N = N_0 \times (\frac{1}{2})^n$,where $n$ is the number of half-lives.
Given that the radioactivity is half of the fresh wood,we have $\frac{N}{N_0} = \frac{1}{2}$.
Substituting this into the equation: $\frac{1}{2} = (\frac{1}{2})^n$,which implies $n = 1$.
The age of the sample $(t)$ is calculated as $t = n \times t_{1/2}$.
Given $t_{1/2} = 6000 \ years$,we get $t = 1 \times 6000 = 6000 \ years$.

(A) The correct option is $A$.
For a first-order reaction,the half-life period is given by the formula $t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{k}$.
Thus,$t_{1/2} = \text{constant} \times k^{-1}$.