Question 171:
easyThe decay constant of a radio isotope is $\lambda$. If $A_1$ and $A_2$ are its activities at times $t_1$ and $t_2$ respectively, the number of nuclei which have decayed during the time $(t_1 – t_2)$:
(2010 Mains)
Activity is related to the number of nuclei by $A = \lambda N$. The number of nuclei at $t_1$ is $N_1 = A_1/\lambda$ and at $t_2$ is $N_2 = A_2/\lambda$. Number of nuclei decayed is $N_1 - N_2 = \frac{A_1 - A_2}{\lambda}$.
Question 172:
easyThe activity of a radioactive sample is measured as $N_0$ counts per minute at t = 0 and $N_0/e$ counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is:
(2010 Pre)
Using $A = A_0 e^{-\lambda t}$, we get $N_0/e = N_0 e^{-5\lambda}$, yielding $5\lambda = 1$ and $\lambda = 1/5$ per minute. The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = 5 \ln 2 = 5 \log_e 2$.
Question 173:
easyTwo radioactive materials $X_1$ and $X_2$ have decay constants $5\lambda$ and $\lambda$ respectively. Initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $1/e$ after a time:
(2008)
The number of nuclei are $N_1 = N_0 e^{-5\lambda t}$ and $N_2 = N_0 e^{-\lambda t}$. Their ratio is $N_1/N_2 = e^{-4\lambda t} = e^{-1}$. Therefore, $4\lambda t = 1$, yielding $t = \frac{1}{4\lambda}$.
Question 174:
easyTwo radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At t = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $(1/e)^2$ after a time interval:
(2007)
$N_A = N_0 e^{-5\lambda t}$ and $N_B = N_0 e^{-\lambda t}$. The ratio is $N_A/N_B = e^{-4\lambda t} = (1/e)^2 = e^{-2}$. This gives $4\lambda t = 2$, which implies $t = 1/2\lambda$.
Question 175:
easyThe half life of a radioactive sample undergoing $\alpha$-decay is $1.4 \times 10^{17} s$. If the number of nuclei in the sample is $2.0 \times 10^{21}$, the activity of the sample is nearly.
(2020-Covid)
Activity $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$. Substituting the values gives $A \approx \frac{0.693}{1.4 \times 10^{17}} \times 2.0 \times 10^{21} \approx 0.99 \times 10^4 Bq$, which is nearly $10^4 Bq$.
Question 176:
easyFor a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
(2018)
Remaining nuclei $N = 600 - 450 = 150$. The fraction remaining is $150/600 = 1/4 = (1/2)^2$. Since 2 half-lives have passed, the total time is $t = 2 \times 10 = 20$ minutes.
Question 177:
easyRadioactive material ‘A’ has decay constant ‘$8\lambda$’ and material ‘B’ has decay constant ‘$\lambda$’. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material ‘A’ to that of ‘B’ will be $1/e$?
(2017-Delhi)
$N_A = N_0 e^{-8\lambda t}$ and $N_B = N_0 e^{-\lambda t}$. Ratio $N_A/N_B = e^{-7\lambda t} = e^{-1}$. Therefore, $7\lambda t = 1$, which gives $t = 1/7\lambda$.
Question 178:
easyThe half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is:
(2016 – II)
Remaining nuclei at $t_1$ is $100 - 40 = 60\%$. Remaining at $t_2$ is $100 - 85 = 15\%$. The ratio is $15/60 = 1/4 = (1/2)^2$, meaning 2 half-lives have passed. Time $= 2 \times 30 = 60$ minutes.
Question 179:
easyA radio isotope $X$ with a half life of $1.4 \times 10^9$ years decays to $Y$ which is stable. A sample of the rock from a cave was found to contain $X$ and $Y$ in the ratio $1 : 7$. The age of the rock is:
(2014)
Ratio $X/Y = 1/7$ implies $X/(X+Y) = 1/8 = (1/2)^3$. Thus, 3 half-lives have elapsed. The age of the rock is $3 \times 1.4 \times 10^9 = 4.2 \times 10^9$ years.
Question 180:
easyThe half life of a radioactive isotope ‘$X$’ is 20 years. It decays to another element ‘$Y$’ which is stable. The two elements ‘$X$’ and ‘$Y$’ were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be:
(2013)
The fraction of remaining radioactive isotope is $X/(X+Y) = 1/(1+7) = 1/8 = (1/2)^3$. Three half-lives have passed, so age $t = 3 \times 20 = 60$ years.