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Question: An atomic battery wrist watch uses \({{61}^{147}}Pm\,\) as a source of \(\,\beta -\,\) particle \(\,...

An atomic battery wrist watch uses 61147Pm {{61}^{147}}Pm\, as a source of  β− \,\beta -\, particle  (t1/2=2.62years) \,({{t}_{1/2}}=2.62years)\, for energy required for its operation. The time it would take for the rates of  β− \,\beta -\,emission in the battery to reduced to  5\,5%\, of its original value is :
a. 0.2490.249 years
b. 0.1390.139 years
c. 0.1940.194 years
d. None of these

Explanation

Solution

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= 2.303×log⁡×aa−xtsec⁡−1k=\,\dfrac{2.303\times \log \times \dfrac{a}{a-x}}{t}{{\sec }^{-1}}
Where,  a= \,a=\,initial concentration
 a−x= \,a-x=\,final concentration at time  t \,t\,
 t= \,t=\,time taken
k=0.693t12 sec⁡−1k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}\,{{\sec }^{-1}}
Where, t12={{t}_{\dfrac{1}{2}}}=Half life of the reaction
k=k=rate constant

Complete step by step answer:
Let us first understand what is beta decay;
Beta decay (β−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= 2.303×log⁡×aa−xtsec⁡−1k=\,\dfrac{2.303\times \log \times \dfrac{a}{a-x}}{t}{{\sec }^{-1}}
Where,  a= \,a=\,initial concentration
a−x=a-x=final concentration at time tt
t=t=time taken
We can find the rate constant by the equation of half-life of the first order reaction;
k=0.693t12 sec⁡−1k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}\,{{\sec }^{-1}}
Where, t12={{t}_{\dfrac{1}{2}}}=Half life of the reaction
k=k=rate constant
Now, let us look into the given data;
Here, we have t12=2.62years{{t}_{\dfrac{1}{2}}}=2.62years
Therefore,  k=0.6932.62=0.264sec⁡−1 \,k=\dfrac{0.693}{2.62}=0.264{{\sec }^{-1}}\,
Also, we take a=100,x=5∴a−x=95 \,a=100,x=5\therefore a-x=95\,because, it is given that the change in concentration is  5\,5%\,so the total can be considered as  100 \,100\,.
Now, substituting this in the first equation we get;
 0.264sec⁡−1=2.303×log⁡×10095tsec⁡−1 \,0.264{{\sec }^{-1}}=\dfrac{2.303\times \log \times \dfrac{100}{95}}{t}{{\sec }^{-1}}\,
From this after calculation, we get t=0.194yearst=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.