JEE Main 2004 Chemical Bonding — Half Life Radioisotope Four Hours Initial

Quick Summary

Subject: Chemistry
Chapter: Chemical Bonding (Nuclear Decay)
Topic: Half-Life Calculation
Difficulty: ⭐⭐
Time: 1-2 mins
Key Formula: $$N = N_0 \left( \frac{1}{2} \right)^n$$
Answer: (A) 3.125 g
Why: The mass halves every 4 hours. Over 24 hours, 6 half-lives occur, reducing the 200 g mass to 3.125 g.

The Question

The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is:

(A) 3.125 g

(B) 2.084 g

(C) 1.042 g

(D) 4.167 g

Quick Answer

The correct answer is (A) 3.125 g.

To solve this, we first determine that 24 hours corresponds to 6 half-lives (since $24 / 4 = 6$). Applying the decay formula $$N = N_0 \left( \frac{1}{2} \right)^n$$ with $N_0 = 200 \text{ g}$ and $n = 6$, we calculate $$N = 200 \times \frac{1}{64} = 3.125 \text{ g}$$.

Why Other Options Are Incorrect

Option (B): 2.084 g
This option is incorrect because it suggests a non-integer number of half-lives or an incorrect decay calculation. The remaining mass must be a fraction of the original mass based on integer powers of 2. A value of 2.084 g does not align with the exponential decay curve for 6 half-lives.

Option (C): 1.042 g
This option is incorrect because it underestimates the remaining mass. This value would imply that 7 half-lives have passed ($200 \times \frac{1}{128} \approx 1.56$, or closer to 7.5 half-lives), which is not the case here as the time elapsed is exactly 6 half-lives.

Option (D): 4.167 g
This option is incorrect because it overestimates the remaining mass. A value of 4.167 g would correspond to approximately 5.5 half-lives ($200 \times 0.5^{5.5} \approx 4.17$), which is less than the total time available (24 hours).

Video Solution

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Understanding the Concept

Half-life is the characteristic time required for half of the radioactive nuclei in a sample to decay into another nuclear species or a different nuclear isotope. It is an exponential decay process, meaning the rate of decay is proportional to the amount of substance remaining. Unlike linear decay (where the same amount disappears in fixed intervals), radioactive decay follows a multiplicative pattern where the quantity is multiplied by $1/2$ every half-life period.

Detailed Step-by-Step Solution

Step 1: Determine the number of half-life periods

The half-life ($t_{1/2}$) of the radioisotope is 4 hours, and the total time elapsed ($t$) is 24 hours. We calculate the number of half-life periods ($n$) using the formula:
$$n = \frac{t}{t_{1/2}}$$
Substituting the given values:
$$n = \frac{24 \text{ hours}}{4 \text{ hours}} = 6$$
This means the isotope has undergone 6 complete half-life cycles.

Step 2: Apply the decay formula

The remaining mass ($N$) after $n$ half-lives is given by the exponential decay formula:
$$N = N_0 \left( \frac{1}{2} \right)^n$$
where $N_0$ is the initial mass (200 g) and $n$ is 6.

Step 3: Simplify the equation to find the final mass

Substituting the values into the formula:
$$N = 200 \left( \frac{1}{2} \right)^6$$
$$N = 200 \times \frac{1}{64}$$
$$N = \frac{200}{64}$$
$$N = 3.125 \text{ g}$$
Therefore, the mass remaining after 24 hours is 3.125 g.

Final Answer

✔ 3.125 g (Option A)

Essential Formulas for This Topic

Half-Life Formula: $$N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$$
Use this when time ($t$) and half-life ($t_{1/2}$) are given.
Number of Half-Lives: $$n = \frac{t}{t_{1/2}}$$
Use this to find how many times the substance has halved.
Decay Constant: $$k = \frac{0.693}{t_{1/2}}$$
Use this when working with logarithmic decay equations ($N = N_0 e^{-kt}$).

Common Mistakes to Avoid

Mistake 1: Reversing the Time and Half-Life

Wrong Thinking: Calculating $n = \frac{4}{24} = \frac{1}{6}$ and using this fraction in the formula.
Correct Approach: Always divide the total time by the half-life to find the number of periods ($n$). Here, $n = \frac{24}{4} = 6$.

Mistake 2: Using Linear Decay Instead of Exponential

Wrong Thinking: Subtracting a fixed amount (e.g., 50 g) every 4 hours ($200 – 50 – 50 – …$).
Correct Approach: Radioactive decay is multiplicative, not additive. You must multiply by $1/2$ (or divide by 2) at each step.

Mistake 3: Arithmetic Errors with Powers of 2

Wrong Thinking: Calculating $(1/2)^6$ incorrectly as $1/36$ or $1/32$.
Correct Approach: Memorize or calculate powers of 2 carefully: $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64$.

Key Concept Summary

Half-life is the time required for the quantity of a substance to reduce to half of its initial value.
Decay is exponential; the substance loses a constant *fraction* of its mass, not a constant *mass*.
The remaining mass is always a fraction of the initial mass ($N_0$).
Mathematically, remaining mass is $N = N_0 / 2^n$.

Golden Rule: If the time elapsed is an exact multiple of the half-life, the answer is always $N_0 / 2^n$. If not, logarithms are required to solve for $t$.

Frequently Asked Questions

Q: Can the half-life of a substance change?

A: No, the half-life of a specific radioisotope is a constant physical property that does not change under different conditions (like temperature or pressure).

Q: What happens if the time elapsed is not a multiple of the half-life?

A: You must use the general formula $$N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$$ directly, or take logarithms to solve for $n$.

Q: Is the decay process reversible?

A: No, radioactive decay is an irreversible spontaneous process.

Q: How is the decay constant ($k$) related to half-life?

A: They are inversely proportional. A shorter half-life means a higher decay constant ($k$).

Prerequisites to Solve This Question

Understanding of the definition of Half-Life.
Familiarity with exponential functions and exponents.
Basic arithmetic skills for division and powers of 2.

After Solving This, You Can:

✔ Calculate remaining quantities in radioactive decay scenarios.
✔ Distinguish between linear and exponential decay patterns.
✔ Solve for time or half-life given the other variables.

Study Tips for This Topic

For JEE Main, this is a high-yield topic. Memorize the powers of 2 up to $2^{10}$ (1024) as these frequently appear in percentage and mass reduction problems. Always ensure you are calculating the number of periods ($n$) correctly before applying the exponent.

Difficulty Rating & Exam Frequency

Difficulty: ⭐⭐ (Easy to Medium)
JEE Main Frequency: High (Appears frequently in Physical Chemistry section)
JEE Advanced Frequency: Medium (Often combined with other nuclear concepts)