Rate of decay and Half-life Questions in English

$A$ radioactive element has a half-life of $200 \ days$. The percentage of original activity remaining after $83 \ days$ is $....$ (Nearest integer).
(Given: $\text{antilog } 0.125 = 1.333$,$\text{antilog } 0.693 = 4.93$)

Assuming $1 \, \mu g$ of trace radioactive element $X$ with a half-life of $30 \ years$ is absorbed by a growing tree. The amount of $X$ remaining in the tree after $100 \ years$ is $n \times 10^{-1} \, \mu g$. Find the value of $n$. $[Given : \ln 10 = 2.303 ; \log 2 = 0.30]$

${ }^{227}Ac$ has a half-life of $22 \, years$ with respect to radioactive decay. The decay follows two parallel paths: ${ }^{227}Ac \longrightarrow { }^{227}Th$ and ${ }^{227}Ac \longrightarrow { }^{223}Fr$. If the percentage of the two daughter nuclides are $2.0$ and $98.0$,respectively,the decay constant (in $year^{-1}$) for ${ }^{227}Ac \longrightarrow { }^{227}Th$ path is closest to

The decay profiles of three radioactive species $A$,$B$ and $C$ are given below:
These profiles imply that the decay constants $k_A$,$k_B$ and $k_C$ follow the order:

After $2 \, \text{hours}$,the amount of a certain radioactive substance reduces to $1/16^{th}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is $...... \, \text{min}$.

The half-life of radioisotopic bromine $-82$ is $36 \ hours$. The fraction which remains after one day is . . . . . . $\times 10^{-2}$. (Given $\text{antilog } 0.2006 = 1.587$)

The ratio of $\frac{{}^{14}C}{{}^{12}C}$ in a piece of wood is $\frac{1}{8}$ part that of the atmosphere. If the half-life of ${}^{14}C$ is $5730 \text{ years}$,the age of the wood sample is $.....$ years.

Carbon-$14$ is used to determine the age of organic material. The procedure is based on the formation of $^{14}C$ by neutron capture in the upper atmosphere.
${ }_{7}^{14}N + { }_{0}n^1 \rightarrow { }_{6}^{14}C + { }_{1}H^1$
$^{14}C$ is absorbed by living organisms during photosynthesis. The $^{14}C$ content is constant in living organisms. Once the plant or animal dies,the uptake of carbon dioxide by it ceases and the level of $^{14}C$ in the dead being falls due to the decay which $^{14}C$ undergoes.
${ }_{6}^{14}C \rightarrow { }_{7}^{14}N + \beta^{-}$
The half-life period of $^{14}C$ is $5770$ years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda = \frac{0.693}{t_{1/2}}$.
The comparison of the $\beta^{-}$ activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method,however,ceases to be accurate over periods longer than $30,000$ years. The proportion of $^{14}C$ to $^{12}C$ in living matter is $1 : 10^{12}$.
$1.$ Which of the following options is correct?
$(A)$ In living organisms,circulation of $^{14}C$ from the atmosphere is high so the carbon content is constant in the organism.
$(B)$ Carbon dating can be used to find out the age of the Earth's crust and rocks.
$(C)$ Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay,hence the carbon content remains constant in living organisms.
$(D)$ Carbon dating cannot be used to determine the concentration of $^{14}C$ in dead beings.
$2.$ What should be the age of a fossil for meaningful determination of its age?
$(A)$ $6$ years
$(B)$ $6000$ years
$(C)$ $60,000$ years
$(D)$ It can be used to calculate any age.
$3.$ $A$ nuclear explosion has taken place,leading to an increase in the concentration of $^{14}C$ in nearby areas. $^{14}C$ concentration is $C_1$ in nearby areas and $C_2$ in areas far away. If the age of the fossil is determined to be $T_1$ and $T_2$ at the places respectively,then:
$(A)$ The age of the fossil will increase at the place where the explosion has taken place and $T_1 - T_2 = \frac{1}{\lambda} \ln \frac{C_1}{C_2}$
$(B)$ The age of the fossil will decrease at the place where the explosion has taken place and $T_1 - T_2 = \frac{1}{\lambda} \ln \frac{C_1}{C_2}$
$(C)$ The age of the fossil will be determined to be the same.
$(D)$ $\frac{T_1}{T_2} = \frac{C_1}{C_2}$

To determine the half-life of a radioactive element,a student plots a graph of $\ln|dN(t)/dt|$ versus $t$. Here $dN(t)/dt$ is the rate of radioactive decay at time $t$. If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16 \ \text{years}$,the value of $p$ is

$A$ certain nuclide has a half-life period of $30 \ min$. If a sample containing $600$ atoms is allowed to decay for $90 \ min$,how many atoms will remain (in $atoms$)?

What type of reaction order is followed by radioactive processes?

The unit of the decay constant for radioactive disintegration is:

$A$ radioactive isotope having $t_{1/2} = 3 \ days$ was measured after $12 \ days$. If $3 \ g$ of the isotope remains in the container,what was the initial weight of the isotope (in $g$)?

The half-life period for a radioactive substance is $15$ minutes. How many grams of this radioactive substance is decayed from $50$ g of substance after $1$ hour?

$2 \text{ g}$ of a radioactive sample having a half-life of $15$ days was synthesized on $1^{st}$ Jan $2009$. The amount of the sample left behind on $1^{st}$ March $2009$ (including both the days) is: (in $\text{ g}$)

The half-lives of two radioactive nuclides $A$ and $B$ are $1 \ min$ and $2 \ min$ respectively. Equal weights of $A$ and $B$ are taken separately and allowed to disintegrate for $4 \ min$. What will be the ratio of weights of $A$ and $B$ disintegrated?

$10 \ g$ of a radioactive element is disintegrated to $1 \ g$ in $2.303 \ \text{minutes}$. What is the half-life (in minutes) of that radioactive element?

The half-life for decay of ${}^{14}C$ by $\beta$-emission is $5730 \ yr$. The fraction of ${}^{14}C$ that decays in a sample that is $22920 \ yr$ old would be

If $n_t$ number of radioatoms are present at time $t$,the following expression will be a constant:

If in the case of a radioisotope the value of half-life $(T_{1/2})$ and decay constant $(\lambda)$ are identical in magnitude,then their value should be

The half-life period of ${}_{53}I^{125}$ is $60$ days. The radioactivity after $180$ days will be (in $\%$)

The half-life of $C^{14}$ is $5760 \ years$. For a $200 \ mg$ sample of $C^{14}$,the time taken to change to $25 \ mg$ is (in $years$)

$A$ piece of wood from an archaeological sample has $5.0 \text{ counts min}^{-1} \text{ g}^{-1}$ of $^{14}C$,while a fresh sample of wood has a count of $15.0 \text{ counts min}^{-1} \text{ g}^{-1}$. If the half-life of $^{14}C$ is $5770 \text{ yr}$,the age of the archaeological sample is:

Radioactivity of a sample $(Z=22)$ decreases by $90 \%$ after $10 \ years$. What will be the half-life of the sample (in $years$)?

The half-life of a radioactive element is $10 \ hours$. How much will be left after $4 \ hours$ in a $1 \ g$-atom sample?

The half-life of ${}^{65}Zn$ is $245 \ days$. After $x$ days,$75\%$ of original activity remained. The value of $x$ in days is . . . . . . . (Nearest integer)
(Given $: \log 3=0.4771$ and $\log 2=0.3010$ )

The half-life for radioactive decay of $^{14}C$ is $5730 \text{ years}$. An archaeological artifact containing wood had only $80\%$ of the $^{14}C$ found in a living tree. Which is the correct formula for age $(t)$ of the sample?