The half-life of radioactive sample is 5.01 days. Calculate the decay constant.

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Q. The half-life of radioactive sample is 5.01 days. Calculate the decay constant.

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Concept understanding — Half Life

Half-Life: The Heartbeat of Radioactive Decay

Imagine you have a giant jar of popcorn kernels, and every minute, exactly half of the kernels left in the jar pop. You start with 1000 kernels. After one minute, 500 are left. After two minutes, 250. After three minutes, 125. After four minutes, about 62. And so on.

That constant "halving time" — the fixed interval it takes for half of whatever remains to disappear — is the core idea of half-life.

Radioactive decay works the same way, except the "popping" is a nucleus spontaneously transforming into a different nucleus by emitting radiation. The key insight: each nucleus has the same fixed probability of decaying per second, regardless of how old it is or how many other nuclei are around. This is a purely random, memoryless process.

Note

Because decay is random and memoryless, the half-life is a constant for a given isotope. It does not depend on how much of the substance you started with. A gram of carbon-14 has the same half-life as a tonne of carbon-14.

The Precise Statement

The half-life, denoted $T_{1/2}$ , is the time required for exactly half of the radioactive nuclei in a sample to decay.

If you start with $N_0$ nuclei, after one half-life you have $\frac{N_0}{2}$ left. After two half-lives, you have $\frac{N_0}{4}$ left. After three, $\frac{N_0}{8}$ , and so on.

Mathematically, the number of nuclei remaining after time $t$ follows an exponential decay law:

N(t) = N_0 e^{-\lambda t}

where $\lambda$ is the decay constant — the probability per unit time that a given nucleus will decay. The half-life is the value of $t$ that makes $N(t) = N_0/2$ :

\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}

Cancelling $N_0$ and taking natural logs:

\ln\left(\frac{1}{2}\right) = -\lambda T_{1/2}

-\ln 2 = -\lambda T_{1/2}

T_{1/2} = \frac{\ln 2}{\lambda}

T_{1/2} = \frac{0.693}{\lambda}

The number $0.693$ is just $\ln 2$ to three decimal places. This formula is the exact bridge between the decay constant (a microscopic probability) and the half-life (a macroscopic, measurable time).

Why "Independent of Initial Amount"?

This is the most counterintuitive part for beginners. Suppose you have two samples of the same isotope: one with 1 million atoms and one with 10 atoms. The half-life is identical for both. …

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The decay constant follows directly from the half-life: $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5.01\ \text{days}} \approx 0.1383\ \text{day}^{-1} \approx 1.6\times10^{-6}\ \text{s}^{-1}$ .

1. Relation between half-life and decay constant. The half-life $T_{1/2}$ of a radioactive sample is the time in which half of the nuclei present decay. It is related to the decay constant $\lambda$ (the probability per unit time that a given nucleus decays) by

$T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda} \quad\Rightarrow\quad \lambda = \dfrac{0.693}{T_{1/2}}$

2. Substituting the given half-life. Here $T_{1/2} = 5.01$ days, so

$\lambda = \dfrac{0.693}{5.01\ \text{days}} = 0.1383\ \text{day}^{-1}$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$_0$
(a) $\dfrac{N_0}{4}$
(b) $\dfrac{N_0}{2}$
(c) $\dfrac{N_0}{8}$
(d) $\dfrac{N_0}{\sqrt{2}}$
›Reveal solutionSolution

Substituting $t=T_{1/2}/2$ into the exponential decay law gives $N=N_0/\sqrt2$ .

Working

The radioactive decay law is

$N = N_0 e^{-\lambda t} = N_0\left(\dfrac12\right)^{t/T_{1/2}}$

(using $\lambda=\ln2/T_{1/2}$ ).

At $t = \dfrac{T_{1/2}}{2}$ : …

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CBSE 2024Set A1 markMCQ
$Q. Half-life of radioactive substance is (A) 0.6931 \times λ (B) log 10^2 / λ (C) 0.6931/λ (D) Average age/0.6931$
›Reveal solutionSolution

Half-life of a radioactive substance is T½ = 0.6931/λ → option (C).

Radioactive decay follows N = N₀e^(−λt). The half-life is the time for the number of nuclei to fall to half, so putting N = N₀/2:

…

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CBSE 2023Set F1 markMCQ
$Q.The half-life of a radioactive isotope (210Bi) is 5 days. The fraction of the nuclei undecayed at the end of 20 days will be (A) 1/2 (B) 1/4 (C) 1/5 (D) 1/16$
›Reveal solutionSolution

20 days = 4 half-lives ⇒ fraction left = (½)⁴ = 1/16.

Number of half-lives elapsed:

$n = \frac{t}{T_{1/2}} = \frac{20}{5} = 4.$

Fraction of nuclei remaining undecayed: …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Mean life of a radioactive sample is 100 second (s). Then its half life is$
(1) 6.93 s
(2) 0.693 s
(3) 69.3 s
(4) 100 s
›Reveal solutionSolution

Mean life and half life of a radioactive sample are related by a factor of ln 2.

…

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CBSE 2022Set I1 markMCQ
$Q. The relation between half-life time T_1/2 and decay constant is (A) T_1/2 = 0.693/λ (B) T_1/2 = λ/0.693 (C) T_1/2 = 0.693λ (D) T_1/2 = 0.693λ^2$
›Reveal solutionSolution

Half-life T_1/2 = 0.693/λ = ln 2 / λ.

Radioactive decay follows N = N₀e^(−λt). At the half-life, N = N₀/2:

$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \Rightarrow e^{\lambda T_{1/2}} = 2 \Rightarrow \lambda T_{1/2} = \ln 2$

…

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CBSE 2020Set XS1 mark
$Q. Write the relationship between the average life and half-life of a radioactive substance.$
›Reveal solutionSolution

$\tau = T_{1/2}/0.693 = 1.44\,T_{1/2}$ .

Definitions in terms of decay constant $\lambda$ :

$\text{mean life }\tau=\frac{1}{\lambda},\qquad \text{half-life }T_{1/2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda}.$

Relation. Dividing, …

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CBSE 2019Set ANNUAL1 markMCQ
$\dfrac{1}{16}$
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
›Reveal solutionSolution

Converting each undecayed fraction into a number of half-lives elapsed in the same 8-hour interval gives half-life periods of 2 h for A and 1 h for B, a ratio of 2:1.

For a radioactive sample, the undecayed fraction remaining after $n$ half-lives is $\left(\dfrac{1}{2}\right)^n$ .

For substance A, the undecayed fraction after 8 hours is $\dfrac{1}{16}$ . Since $\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$ , exactly $n_A = 4$ half-lives have elapsed in 8 hours. So the half-life of A is $T_{1/2,A} = \dfrac{8\text{ h}}{4} = 2$ h.

…

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CBSE 2019Set ANNUAL1 mark
$Q. Write the relation between Half-Life and Mean-Life of radio active element.$
›Reveal solutionSolution

Mean life $\tau$ and half-life $T_{1/2}$ are related by $T_{1/2} = 0.693\,\tau$ , i.e. $\tau = 1.44\,T_{1/2}$ .

Concept. Both quantities are tied to the decay constant $\lambda$ : $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$ and $\tau = \dfrac{1}{\lambda}$ .

Relation. Dividing, …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The time in which radioactive substance becomes half of its initial amount is called -$
(a) average life
(b) half - life
(c) time - period
(d) decay constant
›Reveal solutionSolution

The time to fall to half the initial amount is the half-life.

By definition, the half-life T₁/₂ is the time in which the number of undecayed nuclei of a radioactive substance reduces to one-half of its initial value. It relates to the decay constant λ by

…

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CBSE 2018Set ANNUAL1 markMCQ
${}_{84}Po^{218}$
(a) 12 minutes
(b) 6 minutes
(c) infinity
(d) 9 minutes
›Reveal solutionSolution

The undecayed mass is $1/8$ of the original, corresponding to 3 half-lives, i.e. 9 minutes.

Step 1: Mass remaining undecayed $= 1\text{ g} - 0.875\text{ g} = 0.125\text{ g}$ .

Step 2: Fraction remaining $= \dfrac{0.125}{1} = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$ . …

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CBSE 2016Set ANNUAL1 markMCQ
$N_0$
(a) $\dfrac{N_0}{8}$
(b) $\dfrac{N_0}{16}$
(c) $\dfrac{N_0}{2}$
(d) $\dfrac{N_0}{4}$
›Reveal solutionSolution

After 3 half-lives ( $15\,\text{yr}/5\,\text{yr}=3$ ), the substance left is $N_0(1/2)^3 = N_0/8$ .

Radioactive decay law.

$N = N_0 \left(\dfrac12\right)^{t/T}$

where $T$ is the half-life and $t$ is elapsed time.

Step 1 -- number of half-lives.

$n = \dfrac{t}{T} = \dfrac{15\,\text{years}}{5\,\text{years}} = 3$

Step 2 -- apply the formula. …

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Q.The half-life of radioactive sample is 5.01 days. Calculate the decay constant.

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