$A$ radioactive isotope decays at such a rate that after $96 \ min$,only $\frac{1}{8}$ of the original amount remains. The half-life of this nuclide in minutes is:

  • A
    $12$
  • B
    $24$
  • C
    $32$
  • D
    $48$

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Similar Questions

In the case of a radioisotope,the value of $T_{1/2}$ and $\lambda$ are identical in magnitude. The value is

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.
$(D)$ $\frac{T_1}{T_2} = \frac{C_1}{C_2}$

The unit for the radioactive decay constant is:

$t_{1/2}$ of $^{232}Th$ is $1.39 \times 10^{10} \ \text{years}$. Calculate the number of $\alpha$-particles emitted by $1.0 \ \text{g}$ of $^{232}Th$ per second.

$A$ 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 $g$)?

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