Question

An atomic battery wrist watch uses ${{61}^{147}}Pm\,$ as a source of $\,\beta -\,$ particle $\,({{t}_{1/2}}=2.62years)\,$ for energy required for its operation. The time it would take for the rates of $\,\beta -\,$emission in the battery to reduced to $\,5%\,$ of its original value is :
a. $0.249$ years
b. $0.139$ years
c. $0.194$ years
d. None of these

Answer 1

Hint:Here, negative beta decay is happening. It usually follows first order kinetics. Also, here we have been provided with half-life of the reaction.In nuclear physics, the term half -life is widely used to explain how easily unstable atoms undergo radioactive decay, or how long stable atoms last.
Formula used:
Rate constant of first order reactions $k=\,\dfrac{2.303\times \log \times \dfrac{a}{a-x}}{t}{{\sec }^{-1}}$
Where, $\,a=\,$initial concentration
$\,a-x=\,$final concentration at time $\,t\,$
$\,t=\,$time taken
\[k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}\,{{\sec }^{-1}}\]
Where, ${{t}_{\dfrac{1}{2}}}=$Half life of the reaction
$k=$rate constant

Complete step by step answer:
Let us first understand what is beta decay;
Beta decay $(\beta -decay)$ is a method of radioactive decay in nuclear physics in which a beta particle (fast energetic electron or positron) is released from an atomic nucleus and the initial nuclide is converted into an isobar.
Also let us understand what us half- life as well;
The half-life of a reaction is the time taken to minimise the original value of the reactant concentration to half.
Now, let us move into the calculations here;
Rate constant of first order reactions $k=\,\dfrac{2.303\times \log \times \dfrac{a}{a-x}}{t}{{\sec }^{-1}}$
Where, $\,a=\,$initial concentration
$a-x=$final concentration at time $t$
$t=$time taken
We can find the rate constant by the equation of half-life of the first order reaction;
$k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}\,{{\sec }^{-1}}$
Where, ${{t}_{\dfrac{1}{2}}}=$Half life of the reaction
$k=$rate constant
Now, let us look into the given data;
Here, we have ${{t}_{\dfrac{1}{2}}}=2.62years$
Therefore, $\,k=\dfrac{0.693}{2.62}=0.264{{\sec }^{-1}}\,$
Also, we take$\,a=100,x=5\therefore a-x=95\,$because, it is given that the change in concentration is $\,5%\,$so the total can be considered as $\,100\,$.
Now, substituting this in the first equation we get;
$\,0.264{{\sec }^{-1}}=\dfrac{2.303\times \log \times \dfrac{100}{95}}{t}{{\sec }^{-1}}\,$
From this after calculation, we get $t=0.194years$
Therefore, for this question option c is the correct answer.

Note:
A reduction in the number of radioactive nuclei per unit time is the rate of decay, or activity, of a sample of a radioactive material.Beta decay is a relatively sluggish process as compared to other modes of radioactivity, such as gamma or alpha decay.For beta decay, half-lives are never less than a few milliseconds.