A radioactive-decay-style first order process has a half-life of $30\ ext{years}$. What fraction of the original quantity remains after $90\ ext{years}$?

Question

A radioactive-decay-style first order process has a half-life of $30\ 	ext{years}$. What fraction of the original quantity remains after $90\ 	ext{years}$?

Accepted Answer

Concept understanding — Half Life of a Reaction

Half Life of a Reaction

Imagine you have a pile of 1000 coins, and every minute, exactly half of the coins that are still there vanish. After one minute, 500 coins remain. After another minute, 250 remain. Then 125, then about 62, and so on. The time it takes for the pile to shrink from 1000 to 500 is the same as the time it takes to shrink from 500 to 250, or from 250 to 125. That constant time interval is the half life.

In chemistry, a reaction's half life ( $t_{1/2}$ ) is the time required for the concentration of a reactant to fall to exactly one-half of its initial value. It is a simple, intuitive way to describe how fast a reaction proceeds — the shorter the half life, the faster the reaction.

The Precise Definition

For any reaction, if you start with an initial concentration $[A]_0$ , the half life $t_{1/2}$ is the time at which:

[A] = \frac{1}{2}[A]_0

That is all. But here is the crucial point: the half life is not a universal constant — it depends on the order of the reaction. For different reaction orders, the half life behaves very differently.

Half Life for a First Order Reaction

For a first order reaction, the rate law is:

\text{Rate} = k[A]

The integrated rate equation is:

[A] = [A]_0 e^{-kt}

Set $[A] = \frac{1}{2}[A]_0$ and solve for $t$ :

\frac{1}{2}[A]_0 = [A]_0 e^{-k t_{1/2}}

Cancel $[A]_0$ :

\frac{1}{2} = e^{-k t_{1/2}}

Take natural logarithms:

\ln\left(\frac{1}{2}\right) = -k t_{1/2}

-\ln 2 = -k t_{1/2}

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

t_{1/2} = \frac{0.693}{k}

This is the key result: for a first order reaction, the half life is independent of the initial concentration. Whether you start with 1 M or 0.001 M, the time to halve the concentration is exactly the same. This is a unique property of first order reactions — no other order behaves this way.

Why Does This Matter?

The constancy of $t_{1/2}$ for first order reactions is what makes radioactive decay predictable. Carbon-14 dating works because the half life of $^{14}\text{C}$ is always 5730 years, regardless of how much carbon is present. The same principle applies to many chemical reactions, especially decompositions and isomerisations.

Tip

For a first order reaction, if you know the half life, you can find the rate constant instantly: $k = 0.693 / t_{1/2}$ . This is often the fastest way to get $k$ from experimental data.

Contrast with Other Orders

For a zero order reaction, the half life depends on the initial concentration:

$t_{1/2} = \frac{[A]_0}{2k}$ …

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The number of half-lives elapsed is $\dfrac{90\ \text{years}}{30\ \text{years}} = 3$ . Since a first order process's quantity is multiplied by $1/2$ after every half-life, the fraction remaining after 3 half-lives is $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8} = …

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

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Calculate the half-life period of a first order reaction whose rate constant is 200 s^-1.$
›Reveal solutionSolution

For a first-order reaction, the half-life is a constant, related to the rate constant by t(1/2) = 0.693/k, independent of the starting concentration.

For a first-order reaction, the half-life formula is derived from the integrated rate law ln([R]0/[R]) = kt, setting [R] = [R]0/2:

t(1/2) = ln(2)/k = 0.693/k

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The half life period for first order reaction is ________ of its initial concentration.$
(a) Dependent
(b) Independent
(c) Property
(d) None of these
›Reveal solutionSolution

For a first order reaction, $t_{1/2} = 0.693/k$ , which contains no concentration term, so the half-life is independent of the initial concentration.

For a first order reaction, integrating the rate law gives $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ . Setting $[A] = [A]_0/2$ at $t = t_{1/2}$ gives $t_{1/2} = \dfrac{0.693}{k}$ . Since this expression depends only on the rate constant $k$ (which is fixed at a given temperature) and not on $[A]_0$ , the half-life of a first order reaction is the same regardless of how much r …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The half life period of a radioactive element is 140 days. After 560 days 1 g of element will be reduced to :$
(a) $\left(\dfrac{1}{8}\right)g$
(b) $\left(\dfrac{1}{2}\right)g$
(c) $\left(\dfrac{1}{16}\right)g$
(d) $\left(\dfrac{1}{4}\right)g$
›Reveal solutionSolution

Radioactive decay is a first-order process, so the amount remaining after any whole number of half-lives is simply the initial amount halved that many times; 560 days is exactly 4 half-lives of this 140-day element.

Step 1 — number of half-lives elapsed: $n = \dfrac{\text{total time}}{t_{1/2}} = \dfrac{560\ \text{days}}{140\ \text{days}} = 4$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The half-life of a first-order reaction depends on :$
(a) Initial concentration
(b) Temperature only
(c) Rate constant
(d) Both
(a) and (c)
›Reveal solutionSolution
First-order half-life t(1/2) = 0.693/k depends only on k, not on initial concentration. Answer: (c) Rate constant.

For a first-order reaction k = (2.303/t) log([A]0/[A]). Setting [A] = [A]0/2 gives t(1/2) = 0.693/k.

This expression contains only k, so the half-life is independent of the initial concentration. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. A reaction is found to be 50% complete in 20 minutes and 75% complete in 30 minutes. The order of the reaction is -$
(a) 0
(b) 0.5
(c) 1
(d) 2
›Reveal solutionSolution
Checking the data against the rate laws: 50% in 20 min and 75% in 30 min are consistent with a constant zero-order rate constant, so the order is 0.

We test which order fits both data points.

Zero order: $[A]_0 - [A] = k t$ , i.e. amount reacted $\propto t$ .

50% reacted in 20 min: $k(20) = 0.50[A]_0 \Rightarrow k = 0.025[A]_0\ \text{min}^{-1}$ .

75% reacted in 30 min: $k(30) = 0.75[A]_0 \Rightarrow k = 0.025[A]_0\ \text{min}^{-1}$ . …
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CBSE 2025Set ANNUAL1 mark
$k = 5.5\times10^{-14}\ \text{s}^{-1}$
›Reveal solutionSolution

The half-life of a first-order reaction is obtained directly from $t_{1/2}=0.693/k$ .

Calculation

For a first-order reaction, the half-life is independent of initial concentration and given by:

$t_{1/2} = \frac{0.693}{k}$

Substituting $k = 5.5\times10^{-14}\ \text{s}^{-1}$ :

$t_{1/2} = \frac{0.693}{5.5\times10^{-14}} = 0.126\times10^{14}\ \text{s} = 1.26\times10^{13}\ \text{s}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The half life of a first order reaction is 69.3 s. Its rate constant is -$
(i) 10⁻² s⁻¹
(ii) 10⁻⁴ s⁻¹
(iii) 10 s⁻¹
(iv) 10² s⁻¹
›Reveal solutionSolution

For a first order reaction, $k = \dfrac{0.693}{t_{1/2}}$ ; substituting $t_{1/2}=69.3\text{ s}$ gives $k = 10^{-2}\,\text{s}^{-1}$ .

For a first order reaction, the half-life is related to the rate constant by:

$t_{1/2} = \dfrac{0.693}{k}$

Rearranging for $k$ : …

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CBSE 2024Set B1 markMCQ
$Q. A first order reaction completes 75% in 16 minutes. How much time will it take to complete 50%?$
(a) 8 minutes
(b) 32 minutes
(c) 24 minutes
(d) 4 minutes
›Reveal solutionSolution

For a first-order reaction, 75% completion corresponds to exactly two half-lives, so the time for 50% completion is half of 16 minutes = 8 minutes.

For a first-order reaction, the integrated rate law is:

$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$

When 75% reacts, 25% of the reactant is left, so $[A]_0/[A] = 100/25 = 4$ :

$k = \frac{2.303}{16}\log 4 = \frac{2.303}{16}(2\log 2)$

When 50% reacts, $[A]_0/[A] = 2$ , and the time taken is $t_{50\%}$ :

$k = \frac{2.303}{t_{50\%}}\log 2$

Since $k$ is the same for both:

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Value of t1/2 for first order reaction is given as$
(a) t1/2 = [R0] / 2k
(b) t1/2 = 0.693 / k
(c) t1/2 = 6.023 / k
(d) t1/2 = -k / 0.693
›Reveal solutionSolution

The half-life of a first order reaction is derived from its integrated rate law and comes out to 0.693/k, independent of initial concentration.

Integrated first order rate law: k = (1/t) ln([R0]/[R])

At t = t1/2, [R] = [R0]/2:

k = (1/t1/2) ln([R0]/([R0]/2)) = (1/t1/2) ln 2

…

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CBSE 2023Set BZ1 markMCQ
$2.0\times10^{-3}\ s^{-1}$
(a) $3.465\times10^{3}\ s$
(b) $3.465\times10^{2}\ s$
(c) $3.465\times10^{-1}\ s$
(d) $3.465\times10^{-2}\ s$
›Reveal solutionSolution

For a first-order reaction $t_{1/2}=\dfrac{0.693}{k}=3.465\times10^{2}\ s$ .

Concept: The half-life of a first-order reaction is independent of the initial concentration and equals $t_{1/2}=\dfrac{0.693}{k}$ .

Calculation: …

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CBSE 2023Set F1 markMCQ
$Q. The half life of a first order reaction$
(a) does not depend upon rate constant
(b) does not depend upon initial concentration
(c) depends upon initial concentration
(d) all of these
›Reveal solutionSolution

For a first-order reaction, t1/2 = 0.693/k, so it is independent of initial concentration — option (B).

For a first-order reaction the half-life is:

t1/2 = 0.693 / k

This expression contains only the rate constant k, so the half-life depends on k (hence indirectly on temperature) but is completely INDEPENDENT of the initial concentration of the reactant. Whatever the starting amount, the time to fall to half is the same. Option (a) is wrong because t1/2 does depe …

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CBSE 2023Set ANNUAL1 mark
$1000°C$
›Reveal solutionSolution

For a first-order reaction, $t_{1/2}=0.693/k$ ; with $k=9.9\ s^{-1}$ , this gives about 0.07 seconds.

For a first-order reaction, the half-life is independent of the initial concentration and given by:

$t_{1/2} = \frac{0.693}{k}$

Substituting $k = 9.9\ s^{-1}$ : …

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Q.A radioactive-decay-style first order process has a half-life of 30 years30\ \text{years}30 years. What fraction of the original quantity remains after 90 years90\ \text{years}90 years?

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Q.A radioactive-decay-style first order process has a half-life of $30\ \text{years}$ . What fraction of the original quantity remains after $90\ \text{years}$ ?

Unlock everything free for 14 days