
The activity of a sample is \[64 \times {10^{ - 5}}\] Ci. Its half-life is 3 days. The activity will become \[5 \times {10^{ - 6}}\] Ci after
A. 12 days
B. 7 days
C. 18 days
D. 21 days
Answer
162.3k+ views
Hint: In the given question, we will find the days taken by the activity to become \[5 \times {10^{ - 6}}\]Ci . When the half-life is 3 days, the activity of a sample is \[64 \times {10^{ - 5}}\]Ci.
Formula Used:
\[A = {A_0}{(\dfrac{1}{2})^{t/{T_{1/2}}}}\]
where,
A= Activity
\[{A_0}\]= Activity of Sample
t= ?
T=Half-life
Complete step by step solution:
The activity of a radioactive substance is described as the number of radioactive disintegrations occurring in one second in the sample. The S.I. unit of activity is Becquerel (Bq).
Given, A= \[5 \times {10^{ - 6}}\]
\[{A_0}\] = \[64 \times {10^{ - 5}}\]
T = 3 days
We know that,
\[A = {A_0}{(\dfrac{1}{2})^{t/{T_{1/2}}}}\]
By substituting known values we get,
\[5\times {10^{ - 6}} = 64 \times {10^{ - 5}}{(\dfrac{1}{2})^{(t/3)}} \\ \]
\[\Rightarrow \dfrac{1}{{128}} = {(\dfrac{1}{2})^{t/3}}\]
\[\Rightarrow {(\dfrac{1}{2})^{7}} = {(\dfrac{1}{2})^{t/3}}\]
Upon comparing the powers we get;
$\dfrac{t}{3}=7$
$\therefore t=21\,days$
Therefore, t=21 days i.e, after 21 days the activity will become \[5 \times {10^{ - 6}}\] Ci.
Therefore the correct answer is Option D.
Additional Information:The activity of a sample of a radioactive matter is described by the number of disintegrations that takes place at its core at any given time. The activity also signifies the number of radiations emitted. One call, thus, alpha, beta, and gamma activity are the number of alpha, beta, and gamma rays that are emitted and are proportional to the number of disintegrations.
These activities are fundamental characteristics of the radioactive sample and of the kind of radiation emitted. They represent its ‘baseline radioactivity’. When the sample includes more than a single element, the total activity is the sum of the individual activity values.
The activity of a radio element is inversely proportional to its lifespan. The Longer the half-life of a certain substance, the shorter its activity. An analogy can be made with a soft-burning candle, whose lower flame burns for more time. Therefore, in uranium 238, which has a half-life of 4.5 billion years, only a single nucleus out of 65 million will decay each century.
Note: In most cases, disintegrations result in the emission of beta or an alpha ray. Then the numbers are equal. If specific decays are accompanied by the emission of gamma rays, gamma activity is launched which is proportional to the emission frequency of these gamma rays.
Formula Used:
\[A = {A_0}{(\dfrac{1}{2})^{t/{T_{1/2}}}}\]
where,
A= Activity
\[{A_0}\]= Activity of Sample
t= ?
T=Half-life
Complete step by step solution:
The activity of a radioactive substance is described as the number of radioactive disintegrations occurring in one second in the sample. The S.I. unit of activity is Becquerel (Bq).
Given, A= \[5 \times {10^{ - 6}}\]
\[{A_0}\] = \[64 \times {10^{ - 5}}\]
T = 3 days
We know that,
\[A = {A_0}{(\dfrac{1}{2})^{t/{T_{1/2}}}}\]
By substituting known values we get,
\[5\times {10^{ - 6}} = 64 \times {10^{ - 5}}{(\dfrac{1}{2})^{(t/3)}} \\ \]
\[\Rightarrow \dfrac{1}{{128}} = {(\dfrac{1}{2})^{t/3}}\]
\[\Rightarrow {(\dfrac{1}{2})^{7}} = {(\dfrac{1}{2})^{t/3}}\]
Upon comparing the powers we get;
$\dfrac{t}{3}=7$
$\therefore t=21\,days$
Therefore, t=21 days i.e, after 21 days the activity will become \[5 \times {10^{ - 6}}\] Ci.
Therefore the correct answer is Option D.
Additional Information:The activity of a sample of a radioactive matter is described by the number of disintegrations that takes place at its core at any given time. The activity also signifies the number of radiations emitted. One call, thus, alpha, beta, and gamma activity are the number of alpha, beta, and gamma rays that are emitted and are proportional to the number of disintegrations.
These activities are fundamental characteristics of the radioactive sample and of the kind of radiation emitted. They represent its ‘baseline radioactivity’. When the sample includes more than a single element, the total activity is the sum of the individual activity values.
The activity of a radio element is inversely proportional to its lifespan. The Longer the half-life of a certain substance, the shorter its activity. An analogy can be made with a soft-burning candle, whose lower flame burns for more time. Therefore, in uranium 238, which has a half-life of 4.5 billion years, only a single nucleus out of 65 million will decay each century.
Note: In most cases, disintegrations result in the emission of beta or an alpha ray. Then the numbers are equal. If specific decays are accompanied by the emission of gamma rays, gamma activity is launched which is proportional to the emission frequency of these gamma rays.
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