Problem Types & Solved Examples

🎯 Problem-Solving Framework

For EVERY problem, follow this sequence:

Identify what's given and what's asked
Recognize the problem type (6 types below)
Select the right formula/concept
Solve step-by-step (show work in exam)
Verify units and reasonableness

Most students fail at step 2. Master problem recognition = 80% success.

Type 1: Direct Formula Application

🔬 Exam Insight

Frequency: 30-40% of CBSE/NEET questions

Marks: 2-3 marks typically

Strategy: These are scoring questions. Don't lose marks here due to calculation errors.

Problem 1.1: Work Function Calculation

Question: The threshold wavelength for a metal is 5000 Å. Calculate its work function in eV.

Given

λ₀ = 5000 Å = 5000 × 10⁻¹⁰ m = 500 nm

What Examiner Tests

Do you know φ = hc/λ₀?
Can you convert units correctly?

Concept Selection

Work function relates to threshold wavelength: φ = hc/λ₀

Solution

Method 1: Using hc = 1240 eV·nm (fastest)

φ = hc/λ₀ = 1240/500 = 2.48 eV

Method 2: Full calculation

φ = hc/λ₀
= (6.626 × 10⁻³⁴ × 3 × 10⁸)/(5000 × 10⁻¹⁰)
= 3.98 × 10⁻¹⁹ J
= 3.98 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ eV
= 2.48 eV

🧠 Shortcut Insight

Always use hc = 1240 eV·nm for wavelength problems.

If λ is in nm: E(eV) = 1240/λ(nm)

If λ is in Å: E(eV) = 12400/λ(Å)

This saves 90 seconds per problem.

Problem 1.2: Stopping Potential

Question: Light of frequency 8 × 10¹⁴ Hz falls on a metal surface of work function 1.5 eV. Find the stopping potential.

Solution

Step 1: Find photon energy

E = hν = (4.14 × 10⁻¹⁵)(8 × 10¹⁴) = 3.31 eV

Step 2: Find maximum KE

KE_max = hν - φ = 3.31 - 1.5 = 1.81 eV

Step 3: Stopping potential

V₀ = KE_max/e = 1.81 V

Answer: V₀ = 1.81 V

❌ Common Mistake

Students write: V₀ = KE_max = 1.81 eV

WRONG. Potential is in volts, not eV.

When KE is in eV: V₀ (in volts) = KE_max (in eV) numerically equal.

But conceptually: eV₀ = KE_max, so V₀ = KE_max/e

Problem 1.3: de Broglie Wavelength

Question: An electron is accelerated through 54V. Calculate its de Broglie wavelength.

Solution

Using direct formula for accelerated electron:

λ = 12.27/√V = 12.27/√54 = 1.67 Å

Answer: λ = 1.67 Å

Type 2: Conceptual Questions

🔬 Exam Insight

Frequency: 20-25% of all exams

Key: These test understanding, not calculation

JEE Advanced loves these. Can't be solved by formula memorization.

Problem 2.1: Intensity vs Frequency

Question: If intensity of light is doubled keeping frequency constant, what happens to: (a) number of photoelectrons (b) maximum kinetic energy?

Solution

(a) Number of photoelectrons:

Intensity ∝ number of photons per unit time

More photons → more electron emissions (one-to-one)

✓ Number of photoelectrons DOUBLES

(b) Maximum kinetic energy:

KE_max = hν - φ

Depends only on frequency (ν), not intensity

Frequency is constant → KE_max remains same

✓ Maximum KE remains UNCHANGED

🧠 Thinking Framework

Always remember:

Frequency determines: WHETHER emission occurs, KE of electrons
Intensity determines: HOW MANY electrons emitted

This single distinction solves 50% of conceptual questions.

Problem 2.2: Threshold Frequency

Question: Why doesn't photoelectric effect occur below threshold frequency, even with very high intensity?

Solution

Photon picture explanation:

1. Each electron interacts with ONE photon (not multiple)

2. If photon energy (hν) < work function (φ), electron cannot escape

3. High intensity = more low-energy photons

4. But each photon still has insufficient energy

5. Electrons cannot "accumulate" energy from multiple photons

Analogy: You need ₹100 to buy a ticket. Having 1000 people with ₹50 each doesn't help. Each person (photon) needs ₹100 (threshold energy).

Problem 2.3: Why Don't We See Matter Waves? (JEE Advanced)

Question: Explain quantitatively why we don't observe wave nature of a cricket ball.

Solution

Quantitative calculation:

Let: m = 0.15 kg, v = 30 m/s

λ = h/(mv) = (6.626 × 10⁻³⁴)/(0.15 × 30)
= 1.47 × 10⁻³⁴ m

For wave effects to be observable: λ should be comparable to object size or obstacle size.

Cricket ball size: ~10⁻¹ m

Wavelength: ~10⁻³⁴ m

Ratio: λ/size ~ 10⁻³³ (impossibly small!)

Conclusion: Wave nature exists but is unobservable because wavelength << object size.

Type 3: Multi-Step Calculations

🔬 Exam Insight

Frequency: 25-30% in JEE Main/Advanced

Marks: 4-6 marks

Strategy: Breakinto clear steps. Partial marking available.

Problem 3.1: Combined Photoelectric + de Broglie

Question: Light of wavelength 200 nm falls on cesium (φ = 2.14 eV). Find: (a) Maximum KE of photoelectrons (b) Stopping potential (c) de Broglie wavelength of fastest photoelectron.

Solution

(a) Maximum KE:

E_photon = hc/λ = 1240/200 = 6.2 eV
KE_max = E_photon - φ = 6.2 - 2.14 = 4.06 eV
= 4.06 × 1.6 × 10⁻¹⁹ = 6.5 × 10⁻¹⁹ J

(b) Stopping potential:

V₀ = KE_max/e = 4.06 V

(c) de Broglie wavelength:

First find velocity:

(1/2)mv² = KE_max
v = √(2 × KE_max / m)
= √(2 × 6.5 × 10⁻¹⁹ / 9.1 × 10⁻³¹)
= 1.19 × 10⁶ m/s

Then wavelength:

λ = h/(mv) = (6.626 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 1.19 × 10⁶)
= 6.1 × 10⁻¹⁰ m = 6.1 Å

🎯 Strategy Tip

Multi-step problems: Show ALL work clearly

Even if final answer is wrong, you get partial marks for correct intermediate steps.

In JEE: Each correct step = 1-2 marks

Problem 3.2: Comparison Problem

Question: Electron and proton have same kinetic energy. Compare their de Broglie wavelengths.

Solution

Key insight: Same KE, different mass

For same KE:

KE = p²/(2m) → p = √(2m·KE)
λ = h/p = h/√(2m·KE)

Therefore: λ ∝ 1/√m

Ratio:

λ_e/λ_p = √(m_p/m_e) = √1836 ≈ 43

✓ Electron wavelength is 43 times larger than proton

Type 4: Graph-Based Questions

🔬 Exam Insight

Frequency: 15-20% in JEE, 10% in NEET

Types: V₀ vs ν, KE vs ν, Photocurrent vs Intensity

Critical: Understanding slope and intercepts

Problem 4.1: Stopping Potential vs Frequency Graph

Question: From the V₀ vs ν graph shown, find: (a) Planck's constant (b) Work function

Given: Graph passes through (6×10¹⁴ Hz, 0.5V) and (10×10¹⁴ Hz, 2.2V)

Solution

Equation of graph: V₀ = (h/e)ν - φ/e

This is y = mx + c form

(a) Finding h:

Slope = h/e

Slope = ΔV₀/Δν = (2.2 - 0.5)/(10×10¹⁴ - 6×10¹⁴)
= 1.7/(4×10¹⁴) = 4.25 × 10⁻¹⁵ V·s

h = Slope × e = 4.25 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹
= 6.8 × 10⁻³⁴ J·s

(b) Finding φ:

At threshold (V₀ = 0): ν = ν₀

Using point (6×10¹⁴, 0.5):

0.5 = 4.25×10⁻¹⁵ × 6×10¹⁴ - φ/e
φ/e = 2.55 - 0.5 = 2.05 V
φ = 2.05 eV

🧠 Graph Reading Strategy

For V₀ vs ν graph:

Slope = h/e (universal constant)
X-intercept = threshold frequency ν₀
Y-intercept = -φ/e (negative value)
For two metals: Same slope, different intercepts

Type 5: Assertion & Reason

Pattern (CBSE/NEET)

Format:

(A) Both A and R are true, R is correct explanation of A
(B) Both A and R are true, R is not correct explanation
(C) A is true, R is false
(D) A is false, R is true

Example Question

Assertion: Photoelectric effect supports quantum nature of light.

Reason: Photoelectric current is proportional to intensity of light.

Solution

Assertion analysis: TRUE. Photoelectric effect shows light consists of photons (quanta).

Reason analysis: TRUE. More intensity → more photons → more electrons.

Is R correct explanation of A? NO. Current ∝ intensity is about number of electrons, not about quantum nature proof.

✓ Answer: (B)

Type 6: Case-Based (CBSE 2024 Pattern)

Case Study Format

A paragraph describing a real-world application, followed by 4-5 MCQs.

Sample Case Study

Passage: In a photoelectric experiment with cesium (φ = 2.14 eV), monochromatic light of wavelength 400 nm is used. The emitted photoelectrons are collected by applying potential difference.

Q1: Energy of incident photon is:

(a) 2.5 eV (b) 3.1 eV (c) 4.2 eV (d) 5.0 eV

Q2: Maximum kinetic energy of photoelectrons is:

(a) 0.96 eV (b) 1.25 eV (c) 2.14 eV (d) 3.1 eV

Solutions

Q1: E = hc/λ = 1240/400 = 3.1 eV → (b)

Q2: KE = 3.1 - 2.14 = 0.96 eV → (a)

Mastered Problem Types?

Next: Interlinking →