Answer
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Hint: To solve this problem we will be using the radioactive decay equation which describes statistical behavior of large numbers of nuclide (in a group) rather than individual ones. Further by using the concept of one mean life and putting the value of time t at mean life in the decay equation we will calculate the required answer.
Formula used:
$\Rightarrow {{N}_{t}}={{N}_{\circ }}{{e}^{-\lambda t}}$
At mean life,
$\Rightarrow t=\dfrac{1}{\lambda }$
Complete answer::
Radioactivity is a type of exponential decay in which total number of nuclides at some time is calculated with the help of rate decay equation, and it is dependent on total number of nuclides present initially, mean time of nuclides and decay constant.
Decay constant $\lambda $ is different for different radioactive materials so the rate of decay is different for each radioactive material. The rate decay equation is given by,
$\Rightarrow {{N}_{t}}={{N}_{\circ }}{{e}^{-\lambda t}}$
Here, ${{N}_{\circ }}$(The value of N or total number of nuclides at t=0) and negative sign indicates that the N is decreases as time increases and P(survival) will be given by nuclides present at time t divided by total number of nuclides at t=0 which is given by,
$\Rightarrow {{P}_{S}}=\dfrac{N}{{{N}_{\circ }}}$
$\Rightarrow {{P}_{S}}=\dfrac{{{N}_{\circ }}{{e}^{-\lambda t}}}{{{N}_{\circ }}}={{e}^{-\lambda t}}$
We know that at one mean time,
$\Rightarrow t=\dfrac{1}{\lambda }$
So by putting this value in above equation we have,
$\Rightarrow {{P}_{S}}={{e}^{-\lambda \dfrac{1}{\lambda }}}={{e}^{-1}}$
$\Rightarrow {{P}_{S}}=\dfrac{1}{e}$
$\therefore $The probability of survival of radioactive nucleus for one mean life is $\dfrac{1}{e}$ .
Hence, option (A) is correct.
Note:
The SI unit of radioactive activity is Becquerel (Bq) in the name of scientist Becquerel and one Bq is defined as the disintegration per second. Another unit of radioactivity is curie (Ci) and, 1 curie (Ci) =$3.7\times {{10}^{10}}Bq$, radioactivity with low value is measured in terms of disintegrations per minute (dpm).
Formula used:
$\Rightarrow {{N}_{t}}={{N}_{\circ }}{{e}^{-\lambda t}}$
At mean life,
$\Rightarrow t=\dfrac{1}{\lambda }$
Complete answer::
Radioactivity is a type of exponential decay in which total number of nuclides at some time is calculated with the help of rate decay equation, and it is dependent on total number of nuclides present initially, mean time of nuclides and decay constant.
Decay constant $\lambda $ is different for different radioactive materials so the rate of decay is different for each radioactive material. The rate decay equation is given by,
$\Rightarrow {{N}_{t}}={{N}_{\circ }}{{e}^{-\lambda t}}$
Here, ${{N}_{\circ }}$(The value of N or total number of nuclides at t=0) and negative sign indicates that the N is decreases as time increases and P(survival) will be given by nuclides present at time t divided by total number of nuclides at t=0 which is given by,
$\Rightarrow {{P}_{S}}=\dfrac{N}{{{N}_{\circ }}}$
$\Rightarrow {{P}_{S}}=\dfrac{{{N}_{\circ }}{{e}^{-\lambda t}}}{{{N}_{\circ }}}={{e}^{-\lambda t}}$
We know that at one mean time,
$\Rightarrow t=\dfrac{1}{\lambda }$
So by putting this value in above equation we have,
$\Rightarrow {{P}_{S}}={{e}^{-\lambda \dfrac{1}{\lambda }}}={{e}^{-1}}$
$\Rightarrow {{P}_{S}}=\dfrac{1}{e}$
$\therefore $The probability of survival of radioactive nucleus for one mean life is $\dfrac{1}{e}$ .
Hence, option (A) is correct.
Note:
The SI unit of radioactive activity is Becquerel (Bq) in the name of scientist Becquerel and one Bq is defined as the disintegration per second. Another unit of radioactivity is curie (Ci) and, 1 curie (Ci) =$3.7\times {{10}^{10}}Bq$, radioactivity with low value is measured in terms of disintegrations per minute (dpm).
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