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Question: Half-life period of \({U^{237}}\) is \(2.5 \times {10^5}\) years. In how much time will the amount o...

Half-life period of U237{U^{237}} is 2.5×1052.5 \times {10^5} years. In how much time will the amount of U237{U^{237}} remaining be only 25%25\% of the original amount?
(A) 2.5×1052.5 \times {10^5} years
(B) 1.25×1051.25 \times {10^5} years
(C) 5×1055 \times {10^5} years
(D) 106{10^6} years

Explanation

Solution

The time required for a quantity to reduce to half of its initial value is known as half-life. It is used to describe how fast an unstable atom undergoes radioactive decay or how long stable atoms survive. Half-life is calculated from the measurements on change in the mass of the nuclide and the time taken for it to occur.
Formula used: N=(12)nN∘N = {\left( {\dfrac{1}{2}} \right)^n}{N_ \circ }
Where N∘={N_ \circ } = Initial amount of substance
N=N = Initial amount of substance
n=Time requiredhalf - lifen = \dfrac{{{\text{Time required}}}}{{{\text{half - life}}}}

Complete answer:
Nuclear decay generally occurs when the nucleus of an atom is unstable and during this process, energy is spontaneously emitted in the form of radiation. After this, the nucleus changes into the nucleus of one or more elements. The daughter nuclei are more stable and have lower mass and energy than the parent nucleus.
Given
Half-life period of U237{U^{237}} is 2.5×1052.5 \times {10^5} years
Let the initial amount of the isotope be N∘=100{N_ \circ } = 100
The final amount of the isotope will be N=25N = 25
Using the formula,
N=(12)nN∘N = {\left( {\dfrac{1}{2}} \right)^n}{N_ \circ }
⇒25=(12)n×100\Rightarrow 25 = {\left( {\dfrac{1}{2}} \right)^n} \times 100
⇒25100=(12)n\Rightarrow \dfrac{{25}}{{100}} = {\left( {\dfrac{1}{2}} \right)^n}
⇒14=(12)n\Rightarrow \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^n}
⇒(12)2=(12)n\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^n}
⇒\Rightarrow n=2n = 2
Now, the time taken T=n×half - lifeT = n \times {\text{half - life}}
=2×2.5×105= 2 \times 2.5 \times {10^5}
⇒5×105\Rightarrow 5 \times {10^5}years
In 5×1055 \times {10^5} years, the amount of U237{U^{237}} remaining will be only 25%25\% of the original amount.
Therefore, option C is the correct answer.

Note:
Half-life is the characteristic property of unstable atomic nuclei and the particular way in which they decay. Alpha decay and beta decay are slower processes than the gamma decay. Half-lives for gamma decay are too short to measure. The half-lives of various radioactive isotopes may range from a few microseconds to some billions of years.