Question

Three fourth of the active decays in a radioactive sample in \[3/4\] sec. The half-life of the sample is
A. \[\dfrac{1}{2}\] sec
B. \[1\] sec
C. \[\dfrac{3}{8}\] sec
D. \[\dfrac{3}{4}\] sec

Answer 1

Hint: In the given question, we need to find half-life of the sample. For this, we will use the formula of the amount of substance that will decay. In that we will put \[\left( {\dfrac{N}{{{N_0}}}} \right) = \dfrac{1}{4}\] to find the value of the half-life of the sample.

Formula used:
The amount of substance that will decay is given by
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n}\]
Here, \[{N_0}\] is the amount of substance that will initially decay and \[N\] is the quantity that still remains and its decay has not taken place after a time \[t\].
Also, \[t = n({T_{1/2}})\].

Complete step by step solution:
We know that the amount of substance that will decay is,
\[\left( {\dfrac{N}{{{N_0}}}} \right) = {\left( {\dfrac{1}{2}} \right)^n} \\ \]
Here, three fourths of the active decays in a radioactive sample in \[3/4\] sec. So, we get
\[\dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^n}\]
By simplifying, we get
\[{\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^n} \\ \]
By comparing, we get \[n = 2\]. Also, we know that,
\[t = n({T_{1/2}}) \]
Thus, we get
\[\dfrac{3}{4} = 2({T_{1/2}}) \\ \]
By simplifying, we get
\[({T_{1/2}}) = \dfrac{3}{8}\] sec
Thus the half-life of the sample is \[\dfrac{3}{8}\] sec.

Therefore, the correct option is C.

Additional Information: Radioactivity is the spontaneous emission of energy and subatomic particles by certain types of materials. In essence, it is a property of individual atomic nuclei. An unstable nucleus will spontaneously decay, or change into a more stable structure, but this will only happen in a handful of precise ways by ejecting particular particles or kinds of electromagnetic radiation.

Note: Many students make mistakes in writing the formula that still remains and its decay has not taken place after a time \[t\]. Also, the value of half-life of the sample totally depends on the value of \[n\]. If they make mistakes here, then the final result will be wrong. So, it is necessary to do the calculations carefully.