Question 361:
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 362:
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 363:
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 364:
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 365:
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 366:
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.
Question 367:
easyThe half life of a radioactive nucleus is 50 days. The time interval $(t_2 – t_1)$ between the time $t_2$ when 2 / 3 of it has decayed and the time $t_1$ when 1 / 3 of it had decayed is:
(2012 Mains)
At $t_1$, fraction remaining is $1 - 1/3 = 2/3$. At $t_2$, fraction remaining is $1 - 2/3 = 1/3$. The ratio of remaining nuclei is $(1/3)/(2/3) = 1/2$. This represents exactly one half-life, which is 50 days.
Question 368:
easyA radioactive sample with a half life of 1 month has the label: ‘Activity = 2 micro curies on $1-8-1991$’. What would be its activity two months earlier?
(1988)
Going backward in time by two months corresponds to adding 2 half-lives to the activity. The activity was $2 \times 2^2 = 2 \times 4 = 8$ micro curies.
Question 369:
easyWhat happens to the mass number and atomic number of an element when it emits $\gamma$-radiation?
(2020-Covid)
Gamma radiation involves the emission of high-energy photons when a nucleus transitions from a higher energy state to a lower one. This process does not alter the number of protons or neutrons, so the mass number and atomic number remain unchanged.
Question 370:
easy$\alpha$-particle consists of :
(2019)
An $\alpha$-particle is identical to a helium-4 nucleus, which contains exactly 2 protons and 2 neutrons, with no electrons.