Problem Types & Solved Examples

Master all 6 problem patterns with examiner insights

Type 1: Direct Formula

Easy

Problem 1.1: Half-Life Calculation

Given: A radioactive substance has 10⁸ atoms initially. After 20 days, only 2.5 × 10⁷ atoms remain. Find the half-life.

What examiner tests: Can you identify the decay formula and solve for T½ when N₀, N, and t are given?

Concept Selection: Use N = N₀/(2^n) where n = t/T½. Since N₀/N = 4, we have 2^n = 4, so n = 2.

Step-by-Step Solution:

Step 1: Identify given values

N₀ = 10⁸ atoms
N = 2.5 × 10⁷ atoms
t = 20 days
Find: T½ = ?

Step 2: Use the half-life form

N = N₀/(2^n) where n = t/T½

Step 3: Calculate the ratio

N₀/N = 10⁸/(2.5 × 10⁷) = 4 = 2²

Step 4: Find number of half-lives

2^n = 4 → n = 2 half-lives

Step 5: Calculate T½

n = t/T½ → T½ = t/n = 20/2 = 10 days

Shortcut: When you see N₀/N = 2, 4, 8, 16... (powers of 2), immediately use the half-life form. It's faster than exponential form!

Type 2: Conceptual

Medium

Problem 2.1: Nuclear Stability

Question: Why is Fe-56 the most stable nucleus? Explain using the binding energy curve. Can energy be obtained by fusing two Fe-56 nuclei?

What examiner tests: Do you understand BE/A curve and can explain stability qualitatively? This is favorite CBSE 3-mark theory question.

Complete Answer:

Part 1: Why Fe-56 is most stable

Fe-56 has the highest binding energy per nucleon (BE/A ≈ 8.8 MeV)
Higher BE/A means nucleons are more tightly bound
More energy required to break Fe-56 nucleus apart
It sits at the peak of the BE/A vs A curve

Part 2: Energy from BE/A curve logic

Light nuclei (A < 56): Moving right on curve → BE/A increases → Fusion releases energy
Heavy nuclei (A > 56): Moving left on curve → BE/A increases → Fission releases energy
At Fe-56 (peak): Can't move higher on curve → Neither fusion nor fission releases energy

Part 3: Can we fuse two Fe-56?

No, energy is ABSORBED, not released.

Reasoning: 2Fe-56 → X (where A ≈ 112)
• Fe-56 is at the peak (BE/A ≈ 8.8 MeV)
• Nucleus with A ≈ 112 is right of peak (BE/A ≈ 8.5 MeV, lower)
• Lower BE/A means less stable
• Energy must be supplied (endothermic)

This is why stars stop at iron and can't produce heavier elements through fusion.

Key Principle: Energy is released only when you move TOWARD the peak of BE/A curve, whether from left (fusion) or right (fission).

Type 3: Multi-Step

Hard

Problem 3.1: Successive Decay Problem

Given: ²³⁸U decays to ²³⁴Th with half-life 4.5 × 10⁹ years. ²³⁴Th decays to ²³⁴Pa with half-life 24 days. A sample initially has only ²³⁸U (10²⁰ atoms). Find the number of ²³⁴Th atoms after 48 days.

What examiner tests: Can you handle decay chains? Do you recognize steady-state approximation? This is JEE Advanced territory.

Most students forget that Th is being produced AND decaying simultaneously. You can't use simple decay formula directly!

Detailed Solution:

Step 1: Analyze the problem

U-238 → Th-234 (T½ = 4.5 × 10⁹ years)
Th-234 → Pa-234 (T½ = 24 days)
Time = 48 days = 2 × T½ of Th-234

Step 2: Simplification

Since U-238 half-life is HUGE compared to 48 days, we can assume U-238 remains constant at 10²⁰ atoms. Effectively, U-238 produces Th-234 at a constant rate.

Key Insight: When parent half-life >> time period, parent acts as constant source. Th-234 reaches equilibrium between production and decay.

Step 3: Rate equations

dN_Th/dt = λ_U × N_U - λ_Th × N_Th

(Production from U) - (Decay of Th)

Step 4: Equilibrium approximation

Since T½(Th) << T½(U), after few half-lives of Th, production rate ≈ decay rate:
λ_U × N_U ≈ λ_Th × N_Th

Step 5: Calculate

N_Th = (λ_U/λ_Th) × N_U = (T½_Th/T½_U) × N_U

N_Th = (24 days)/(4.5 × 10⁹ × 365 days) × 10²⁰

N_Th ≈ 1.46 × 10¹⁰ atoms

When to use equilibrium: If time >> T½ of intermediate product AND T½ (parent) >> T½ (daughter), assume equilibrium (production rate = decay rate).

Type 4: Graph-Based

Medium

Problem 4.1: ln(N) vs t Graph

Question: A graph of ln(N) vs time (t) for a radioactive sample is a straight line with slope -0.0693 day⁻¹. Find (a) decay constant λ, (b) half-life T½, and (c) time when N = N₀/10.

What examiner tests: Can you extract information from linearized decay plots? NEET loves this pattern.

Complete Solution:

Concept: Linearization of exponential decay

N = N₀ e^(-λt)

Taking natural log both sides:

ln(N) = ln(N₀) - λt

This is equation of straight line: y = c + mx
where y = ln(N), x = t, slope m = -λ, intercept c = ln(N₀)

(a) Find λ:

slope = -λ = -0.0693 day⁻¹

λ = 0.0693 day⁻¹

(b) Find T½:

T½ = 0.693/λ = 0.693/0.0693 = 10 days

(c) Find t when N = N₀/10:

N/N₀ = 1/10 = e^(-λt)

ln(1/10) = -λt → -ln(10) = -0.0693t

t = ln(10)/0.0693 = 2.303/0.0693 ≈ 33.24 days

Graph Recognition:
• ln(N) vs t → straight line (slope = -λ)
• N vs t → exponential decay curve
• log₁₀(N) vs t → straight line (slope = -λ/2.303)
Master all three forms!

Type 5: Assertion & Reason

Easy

Problem 5.1: NEET Pattern

Assertion (A): Nuclear forces are short-range forces.
Reason (R): Nuclear forces operate only over distances of the order of nuclear size (~10⁻¹⁵ m).

Options:

(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is NOT the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Analysis:

Check Assertion (A):

"Nuclear forces are short-range forces" → TRUE
Nuclear forces are effective only up to ~10⁻¹⁵ m. Beyond this distance, they become negligible.

Check Reason (R):

"Nuclear forces operate only over distances ~10⁻¹⁵ m" → TRUE
This is the defining characteristic of nuclear forces.

Is R the correct explanation of A?

YES. Statement R directly explains WHY nuclear forces are called "short-range." The fact that they operate only over nuclear size distances (~10⁻¹⁵ m) is precisely what makes them short-range.

Answer: (a) Both A and R are true, and R is the correct explanation of A

NEET Strategy for A&R Questions:
1. First verify each statement independently (True/False)
2. If both true, check logical connection
3. Ask: "Does R explain WHY A is true?" or "Is R just another fact?"
4. Don't rush—read carefully (2-3 marks secured with proper analysis)

Type 6: Case-Based

Medium

Problem 6.1: Carbon Dating Application

Case Study: Carbon-14 dating is used to determine the age of ancient artifacts. Living organisms maintain a constant C-14/C-12 ratio by exchanging carbon with the atmosphere. When an organism dies, C-14 intake stops, and existing C-14 decays with a half-life of 5730 years.

Question: An ancient wooden artifact shows C-14 activity of 8 decays/min per gram of carbon, while living wood shows 16 decays/min per gram. Calculate the age of the artifact.

What examiner tests: Can you apply radioactive decay to real-world scenarios? CBSE loves case-based questions (4-5 marks).

Step-by-Step Solution:

Step 1: Understanding the data

A₀ (living wood) = 16 decays/min per gram
A (artifact) = 8 decays/min per gram
T½ = 5730 years
Find: Age t = ?

Step 2: Identify the pattern

Activity ratio: A/A₀ = 8/16 = 1/2
This means activity has halved → exactly 1 half-life has passed!

Shortcut Recognition: When A/A₀ = 1/2, 1/4, 1/8, etc., you're at exactly n half-lives. No complex calculation needed!

Step 3: Calculate age

A = A₀ e^(-λt) = A₀ e^(-0.693t/T½)

Since A/A₀ = 1/2:

1/2 = e^(-0.693t/5730)

ln(1/2) = -0.693t/5730

-0.693 = -0.693t/5730

t = 5730 years

OR using the shortcut:

A/A₀ = 1/2 = 1/(2¹) → n = 1 half-life

t = n × T½ = 1 × 5730 = 5730 years

Final Answer: The artifact is approximately 5730 years old.

Common Errors in C-14 Dating:
• Using N instead of A in formula (activity and number decay identically)
• Forgetting that C-14/C-12 ratio ∝ Activity
• Not recognizing simple half-life multiples (wasting time on complex calculations)

Problem-Solving Framework:

Given: List all data with units
What examiner tests: Identify the concept being tested
Concept selection: Choose the right formula/approach
Solution: Solve step-by-step with clear logic
Shortcut insight: Learn faster methods for exam speed

This framework works for ALL problem types. Practice until it becomes automatic.

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