Q127 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Chemistry
A radioactive substance has a decay constant $10^{-1}$ per year. What percentage of substance undergoing decay in 13.86 years . ?
- A$25$
- B$50$
- Correct answerC$75$
- D$85$
Solution
Given decay constant $\lambda = 10^{-1} \text{ year}^{-1} = 0.1 \text{ year}^{-1}$.
Time $t = 13.86 \text{ years}$.
The half-life of the radioactive substance is given by $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.1} = 6.93 \text{ years}$.
The number of half-lives elapsed in time $t$ is $n = \dfrac{t}{T_{1/2}} = \dfrac{13.86}{6.93} = 2$.
The amount of substance remaining after $n$ half-lives is $N = \dfrac{N_0}{2^n} = \dfrac{N_0}{2^2} = \dfrac{N_0}{4} = 0.25 N_0$.
The percentage of substance remaining is $25\%$.
The percentage of substance that has undergone decay is $100\% - 25\% = 75\%$.
Answer: $75$
Time $t = 13.86 \text{ years}$.
The half-life of the radioactive substance is given by $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.1} = 6.93 \text{ years}$.
The number of half-lives elapsed in time $t$ is $n = \dfrac{t}{T_{1/2}} = \dfrac{13.86}{6.93} = 2$.
The amount of substance remaining after $n$ half-lives is $N = \dfrac{N_0}{2^n} = \dfrac{N_0}{2^2} = \dfrac{N_0}{4} = 0.25 N_0$.
The percentage of substance remaining is $25\%$.
The percentage of substance that has undergone decay is $100\% - 25\% = 75\%$.
Answer: $75$
Hence, the correct answer is option: 'C'.