If in $3160$ years, a radioactive substance becomes one-fourth of the original amount, find its half-life period.
Answer
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Hint: Half life of any radioactive substance is the amount of time required for the quantity by weight of the substance to reduce to half of its initial value. So after one half life, we will have $\dfrac{1}{2}$ or $50\% $ of the substance remaining. Also, after $2$ half lives, we will have $\dfrac{1}{4}$ of the substance remaining. In other words, $\dfrac{3}{4}$ of the substance will take two half lives to decay.
Complete answer:
From the question, we know that a radioactive substance becomes one-fourth of the original amount.
Let $N_0$ be the original amount of the radioactive substance and N be the radioactive substance after decay.
$
N = \dfrac{1}{4} \times {N_0} \\
\dfrac{N}{{{N_0}}} = \dfrac{1}{4} \\
$
Since half life of any radioactive substance is the amount of time required for the quantity by weight of the substance to reduce to half of its initial value. So we equate $\dfrac{1}{4}$ to ${\left( {\dfrac{1}{2}} \right)^n}$
$
\dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^n} \\
\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^n} \\
\ $
On solving we get,
$n = 2$
we found that there were $2$ half lives involved in the reaction.
we know that, total time $'T' = n \times {\left( t \right)_{\dfrac{1}{2}}}$
where ${\left( t \right)_{\dfrac{1}{2}}}$is it’s half life period.
$3160 = 2 \times {\left( t \right)_{\dfrac{1}{2}}}$
${\left( t \right)_{\dfrac{1}{2}}} = \dfrac{{3160}}{2} = 1580$ years
The half life period of the radioactive substance is $1580$ years.
Note:
A radioactive substance's half life is a characteristic constant. It measures how long it takes for a fixed quantity of a material to degrade to half its original value, resulting in the release of radiation.
Atoms that decay spontaneously make up radioactive material. They have the ability to emit alpha, beta, and gamma radiation. Since they cannot be switched off like X-ray sources, they are more difficult to monitor.
Complete answer:
From the question, we know that a radioactive substance becomes one-fourth of the original amount.
Let $N_0$ be the original amount of the radioactive substance and N be the radioactive substance after decay.
$
N = \dfrac{1}{4} \times {N_0} \\
\dfrac{N}{{{N_0}}} = \dfrac{1}{4} \\
$
Since half life of any radioactive substance is the amount of time required for the quantity by weight of the substance to reduce to half of its initial value. So we equate $\dfrac{1}{4}$ to ${\left( {\dfrac{1}{2}} \right)^n}$
$
\dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^n} \\
\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^n} \\
\ $
On solving we get,
$n = 2$
we found that there were $2$ half lives involved in the reaction.
we know that, total time $'T' = n \times {\left( t \right)_{\dfrac{1}{2}}}$
where ${\left( t \right)_{\dfrac{1}{2}}}$is it’s half life period.
$3160 = 2 \times {\left( t \right)_{\dfrac{1}{2}}}$
${\left( t \right)_{\dfrac{1}{2}}} = \dfrac{{3160}}{2} = 1580$ years
The half life period of the radioactive substance is $1580$ years.
Note:
A radioactive substance's half life is a characteristic constant. It measures how long it takes for a fixed quantity of a material to degrade to half its original value, resulting in the release of radiation.
Atoms that decay spontaneously make up radioactive material. They have the ability to emit alpha, beta, and gamma radiation. Since they cannot be switched off like X-ray sources, they are more difficult to monitor.
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