KCET · Physics · Nuclear Physics
A radio-active elements has half-life of 15 years. What is the fraction that will decay in 30 years?
- A \(0.25\)
- B \(0.5\)
- C \(0.75\)
- D \(0.85\)
Answer & Solution
Correct Answer
(C) \(0.75\)
Step-by-step Solution
Detailed explanation
Given, half-life, \(T_{1 / 2}=15\) years
Time, \(t=30\) years
\(\therefore\) Number of half-life, \(n=\frac{t}{T}=\frac{30}{15}=2\)
The number of nuclei left undecayed in 30 years or 2 half-lives is
\(\begin{aligned}
\qquad N=N_{0}\left(\frac{1}{2}\right)^{n} \Rightarrow \frac{N}{N_{0}} &=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \\
\therefore \text { Fraction of decayed element } &=\left(1-\frac{N}{N_{0}}\right)=1-\frac{1}{4} \\
&=\frac{3}{4} \text { or } 0.75
\end{aligned}\)
Time, \(t=30\) years
\(\therefore\) Number of half-life, \(n=\frac{t}{T}=\frac{30}{15}=2\)
The number of nuclei left undecayed in 30 years or 2 half-lives is
\(\begin{aligned}
\qquad N=N_{0}\left(\frac{1}{2}\right)^{n} \Rightarrow \frac{N}{N_{0}} &=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \\
\therefore \text { Fraction of decayed element } &=\left(1-\frac{N}{N_{0}}\right)=1-\frac{1}{4} \\
&=\frac{3}{4} \text { or } 0.75
\end{aligned}\)
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