Q.A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The half-life of a certain radio isotope is 4 minutes. The number of radioactive nuclei at a given instant of time is $10^6$. Then the number of radioactive nuclei left 2 minutes later would be (A) $\dfrac{10^6}{2}$ (B) $10^3$ (C) $\dfrac{10^6}{\sqrt{2}}$ (D) $\sqrt{2}\times10^6$

   ›Reveal solutionSolution

   Radioactive decay formula $N=N_0(1/2)^{t/T_{1/2}}$ with a fractional number of half-lives elapsed (here, half a half-life) gives $N_0/\sqrt2$.

   Concept and Intuition

   Radioactive decay is exponential: after every half-life, exactly half the nuclei remain. When the elapsed time isn't a whole number of half-lives, we still use $N=N_0\left(\dfrac12\right)^{t/T_{1/2}}$ — the exponent need not be an integer, and here it's exactly $1/2$.

   Step-by-Step Solution

   1. $t/T_{1/2}=2/4=1/2$.
   2. $N=10^6\times\left(\dfrac12\right)^{1/2}=\dfrac{10^6}{\sqrt2}$.

   Common Mistakes …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The radioactivity of a certain radioactive element drops to 1/64 of its initial value in 30 sec. Its half-life is (A) 2 Sec (B) 4 Sec (C) 5 Sec (D) 6 Sec

   ›Reveal solutionSolution

   Since $1/64=(1/2)^6$, exactly six half-lives fit into the 30-second interval, giving a half-life of 5 seconds. Answer: (C).

   Concept and Intuition

   Radioactive decay follows $N=N_0\left(\dfrac12\right)^{t/T_{1/2}}$. When the remaining fraction is a clean power of $1/2$, the number of half-lives elapsed is just that exponent — no need for logarithms.

   Step-by-Step Solution

   1. Fraction remaining: $\dfrac{N}{N_0}=\dfrac{1}{64}$.
   2. Express as a power of $1/2$: $\dfrac{1}{64}=\dfrac{1}{2^6}=\left(\dfrac12\right)^6$.
   3. So $t/T_{1/2}=6$, i.e. 6 half-lives occur in $t=30$ s.
   4. $T_{1/2}=\dfrac{30}{6}=5$ s.

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Two radioactive materials $A_1$ and $A_2$ have half life periods 20 s and 10 s respectively. Initially a mixture of these materials contain 40 g of $A_1$ and 160 g of $A_2$. The time taken for $A_1$ and $A_2$ to become equal in the mixture is (A) 60 s (B) 80 s (C) 20 s (D) 40 s

   ›Reveal solutionSolution

   Two substances with different half-lives and different starting masses become equal in amount at a specific time found by equating their exponential decay laws.

   Concept and Intuition

   Each radioactive species decays independently following $N(t)=N_0(1/2)^{t/T_{1/2}}$. Even though $A_2$ starts with far more mass, it decays twice as fast (half-life 10 s vs 20 s), so its amount eventually falls to meet $A_1$'s. We just need the crossing time.

   Step-by-Step Solution

   1. $N_1(t)=40\left(\dfrac12\right)^{t/20}$, $N_2(t)=160\left(\dfrac12\right)^{t/10}$.
   2. Set $N_1=N_2$: $40\left(\dfrac12\right)^{t/20}=160\left(\dfrac12\right)^{t/10}$.
   3. Divide: $\left(\dfrac12\right)^{t/20-t/10}=4\Rightarrow\left(\dfrac12\right)^{-t/20}=4\Rightarrow2^{t/20}=2^2$. …

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4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.If decay constant of a radioactive element doubles, then its half-life time becomes (A) Doubles (B) Halves (C) Same (D) increased by 4 times

   ›Reveal solutionSolution

   A direct inverse-proportionality fact about radioactive decay: half-life and decay constant are reciprocally related.

   Concept and Intuition

   The decay constant $\lambda$ measures how "fast" a radioactive substance decays per unit time, while half-life $T_{1/2}$ measures how long it takes to decay to half. Because $T_{1/2}=\dfrac{\ln 2}{\lambda}$, a faster decay constant directly means a shorter half-life, in exact inverse proportion.

   Step-by-Step Solution

   1. Relation: $T_{1/2}=\dfrac{\ln 2}{\lambda}$.
   2. If $\lambda\to2\lambda$, then $T_{1/2}'=\dfrac{\ln2}{2\lambda}=\dfrac{T_{1/2}}{2}$.
   3. So the half-life becomes exactly half of its original value.

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5. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The half-life of a radioactive substance is 20 minutes. $\dfrac{1}{3}$rd part of substance has decayed in time $t_1$ and $\dfrac{2}{3}$rd part of it has decayed in time $t_2$. Then, $(t_2 - t_1)$ is nearly (A) 7 minutes (B) 14 minutes (C) 20 minutes (D) 28 minutes

   ›Reveal solutionSolution

   A neat algebraic identity makes $t_2-t_1$ exactly equal to the half-life $T$, regardless of the specific fractions (1/3 and 2/3 here) — the answer is exactly 20 minutes.

   Concept and Intuition

   Radioactive decay follows $N=N_0e^{-\lambda t}$ with $\lambda=\dfrac{\ln2}{T}$. Rather than compute $t_1$ and $t_2$ and subtract numerically, it helps to notice that the ratio of remaining fractions at $t_1$ and $t_2$ is exactly what determines $t_2-t_1$, and here that ratio works out to exactly $\tfrac12$ — the definition of one half-life.

   Step-by-Step Solution

   1. At $t_1$, $1/3$ has decayed, so remaining fraction is $2/3$: $\dfrac{2}{3}=e^{-\lambda t_1}\Rightarrow t_1=\dfrac{\ln(3/2)}{\lambda}$.
   2. At $t_2$, $2/3$ has decayed, so remaining fraction is $1/3$: $\dfrac13=e^{-\lambda t_2}\Rightarrow t_2=\dfrac{\ln 3}{\lambda}$.
   3. $t_2-t_1=\dfrac{\ln3-\ln(3/2)}{\lambda}=\dfrac{\ln(3/(3/2))}{\lambda}=\dfrac{\ln2}{\lambda}$.
   4. But $\lambda=\ln2/T$, so $t_2-t_1=\dfrac{\ln2}{\ln2/T}=T=20$ minutes. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.An element X of a half-life of $1.4\times10^9$ years decays to form another stable element Y. A sample is taken from a rock that contains both X and Y in the ratio 1 : 7. If at the time of formation of the rock, Y was not present in the sample, then the age of the rock in years is (A) $4.2\times10^9$ (B) $1.4\times10^9$ (C) $0.35\times10^9$ (D) $2.8\times10^9$

   ›Reveal solutionSolution

   Radioactive dating: the X:Y ratio of 1:7 means 3 half-lives have passed, giving an age of $4.2\times10^9$ years.

   Concept and Intuition

   In radioactive decay, if none of the stable daughter product Y was present initially, every atom of Y now present must have come from decay of X. So the original amount of X equals the current $X + Y$. The fraction of X remaining, $X/(X+Y)$, follows $\left(\tfrac12\right)^n$ where $n$ is the number of half-lives elapsed.

   Step-by-Step Solution

   1. Given $X : Y = 1 : 7$, so total original amount $N_0 = X + Y = 8$ (in units of $X=1$).
   2. Remaining fraction of X: $\dfrac{N}{N_0} = \dfrac{1}{8}$.
   3. Since $\dfrac18 = \left(\dfrac12\right)^3$, exactly 3 half-lives have elapsed.
   4. Age of rock $= 3 \times t_{1/2} = 3 \times 1.4\times10^9 = 4.2\times10^9$ years. …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the half-life of a radioactive element is 12.5 hours, then the time taken to disintegrate 256 g of the substance into 1 g is (in hours) (A) 12.5 (B) 25 (C) 37.5 (D) 100

   ›Reveal solutionSolution

   Radioactive decay halves the quantity every half-life; going from 256 g to 1 g takes 8 half-lives, i.e. 100 hours.

   Concept and Intuition

   Radioactive decay follows $N = N_0 \left(\dfrac{1}{2}\right)^{t/T_{1/2}}$, where $T_{1/2}$ is the half-life. If the remaining fraction $N/N_0$ is a power of $\dfrac12$, the number of half-lives elapsed is simply that exponent, making the arithmetic clean when the numbers are chosen as powers of 2 (as here: $256 = 2^8$).

   Step-by-Step Solution

   1. Initial mass $N_0 = 256$ g, final mass $N = 1$ g.
   2. Ratio: $\dfrac{N_0}{N} = 256 = 2^8$, so the substance has undergone 8 half-lives. …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If the time taken for a radioactive substance to decay 8% to 77% is 12 minutes, then the half life of the substance in minutes is (A) 24 (B) 18 (C) 12 (D) 6

   ›Reveal solutionSolution

   Converting the given decayed-percentages to remaining fractions (92% and 23%) shows the amount drops by exactly a factor of 4 — i.e. two half-lives — over the 12 minutes, giving a half-life of 6 minutes.

   Concept and Intuition

   Radioactive decay is exponential: $N(t) = N_0\, 2^{-t/T_{1/2}}$, so the remaining fraction halves every half-life, regardless of how much has already decayed. The question states the substance decays from 8% (decayed) to 77% (decayed) in 12 minutes — meaning the remaining amount shrinks from $100\%-8\%=92\%$ down to $100\%-77\%=23\%$. The key insight is that decay ratios, not raw percentages, are what connect to whole numbers of half-lives: if the remaining amount drops by a clean factor of $2^n$, that interval spans exactly $n$ half-lives.

   Step-by-Step Solution

   1. Remaining fraction initially: $100\%-8\% = 92\%$.
   2. Remaining fraction finally: $100\%-77\% = 23\%$.
   3. Ratio of initial to final remaining amount: $\dfrac{92}{23} = 4$.
   4. Since $4 = 2^2$, the remaining amount has halved exactly twice, so 12 minutes corresponds to 2 half-lives. …

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9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.If the half-life of a radioactive substance is 32 hours, then the fraction of the substance decayed in 4 days is (A) $\dfrac{1}{16}$ (B) $\dfrac{1}{8}$ (C) $\dfrac{15}{16}$ (D) $\dfrac{7}{8}$

   ›Reveal solutionSolution

   Four days is exactly 3 half-lives (32 h each), so $7/8$ of the substance has decayed.

   Concept and Intuition

   In each half-life, exactly half of the remaining substance decays. After $n$ half-lives, the remaining fraction is $(1/2)^n$, so the decayed fraction is $1-(1/2)^n$. The key step is simply converting the elapsed time into a whole number of half-lives.

   Step-by-Step Solution

   1. Half-life $T_{1/2}=32$ hours.
   2. Elapsed time $=4$ days $=4\times24=96$ hours.
   3. Number of half-lives: $n = \dfrac{96}{32}=3$.
   4. Fraction remaining after 3 half-lives: $\left(\dfrac12\right)^3=\dfrac18$.
   5. Fraction decayed: $1-\dfrac18=\dfrac78$.

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The half life of a radioactive substance is 10 minutes. If $n_1$ and $n_2$ are the number of atoms decayed in 20 and 30 minutes respectively, then $n_1 : n_2 =$ (A) $7:8$ (B) $1:2$ (C) $6:7$ (D) $3:4$

    ›Reveal solutionSolution

    Tests computing the number of atoms decayed (not remaining) after successive half-lives; the ratio is $6:7$.

    Concept and Intuition

    After $k$ half-lives, the remaining fraction is $(1/2)^k$, so the decayed fraction is $1-(1/2)^k$. Both times here are measured from $t=0$, so $n_1$ and $n_2$ are cumulative decayed amounts, not amounts decayed in separate intervals.

    Step-by-Step Solution

    1. Take $N_0$ atoms initially, half-life $T=10$ min.
    2. At $t=20$ min $=2T$: remaining $=N_0/4$, so decayed $n_1 = N_0 - N_0/4 = \tfrac34 N_0$.
    3. At $t=30$ min $=3T$: remaining $=N_0/8$, so decayed $n_2 = N_0 - N_0/8 = \tfrac78 N_0$. …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If the half-life of a radioactive material is 10 years, then the percentage of the material decayed in 30 years is (A) 87.5 (B) 78.5 (C) 58.7 (D) 85.7

    ›Reveal solutionSolution

    Radioactive decay halves the quantity every half-life; after $n$ half-lives the fraction remaining is $(1/2)^n$. Here $n=3$, so 87.5% has decayed.

    Concept and Intuition

    Radioactive decay is a first-order process: after every half-life $T_{1/2}$, exactly half of the currently remaining material decays, not half of the original amount each time. So the surviving fraction shrinks geometrically: $1, 1/2, 1/4, 1/8, \ldots$ after $0, 1, 2, 3$ half-lives.

    Step-by-Step Solution

    1. Number of half-lives elapsed: $n = \dfrac{30\ \text{years}}{10\ \text{years}} = 3$.
    2. Fraction of material remaining: $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} = 0.125 = 12.5\%$. …

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12. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.The time gap between 44% decay and 93% decay of a radioactive substance is 81 minutes. The half life of the radioactive substance in minutes is (A) 18 (B) 54 (C) 27 (D) 9

    ›Reveal solutionSolution

    Going from 56% remaining to 7% remaining is a drop by a factor of 8 = $2^3$, i.e. exactly 3 half-lives span the given 81 minutes, so the half-life is 27 minutes.

    Concept and Intuition

    Radioactive decay is exponential: $N(t) = N_0 (1/2)^{t/T_{1/2}}$. If we know the fraction remaining at two different times, the ratio of those fractions tells us directly how many half-lives separate the two times, since each half-life divides the remaining amount by exactly 2.

    Step-by-Step Solution

    1. After 44% decay, fraction remaining $= 1-0.44 = 0.56 = 56\%$.
    2. After 93% decay, fraction remaining $= 1-0.93 = 0.07 = 7\%$.
    3. Ratio of remaining fractions: $\dfrac{0.56}{0.07} = 8 = 2^3$.
    4. So the time interval (81 minutes) corresponds to exactly 3 half-lives: $81 = 3\,T_{1/2}$. …

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