Physics Optics and Modern - How to Solve: Dual Nature of Matter (Photoelectric Effect, de Broglie Wavelength) – IIT JEE Guide

How to Solve: Dual Nature of Matter (Photoelectric Effect, de Broglie Wavelength) – IIT JEE Guide

Introduction

Mastering the dual nature of matter unlocks 5-10 marks in IIT JEE (Main + Advanced) every year—enough to push you into the top 1%. It’s also the foundation for quantum mechanics, electron microscopes, and solar panels. If you can solve photoelectric effect and de Broglie wavelength problems fast and error-free, you’ll outscore 90% of students.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Energy of a photon: ( E = h\nu ) (Planck’s equation)
2. Work function (φ): Minimum energy needed to eject an electron from a metal.
3. Kinetic energy (KE): ( KE = \frac{1}{2}mv^2 )

If these are unclear, stop now and review them first.

KEY TERMS & FORMULAS

1. Photoelectric Effect (Einstein’s Equation)

Formula: [ h\nu = \phi + KE_{\text{max}} ] Variables: - ( h ) = Planck’s constant (( 6.626 \times 10^{-34} \, \text{Js} )) – MEMORISE THIS - ( \nu ) = Frequency of incident light (Hz) - ( \phi ) = Work function of the metal (J or eV) - ( KE_{\text{max}} ) = Maximum kinetic energy of ejected electrons (J or eV)

Other key formulas: - Threshold frequency (( \nu_0 )): ( \phi = h\nu_0 ) – MEMORISE THIS - Stopping potential (( V_0 )): ( eV_0 = KE_{\text{max}} ) – MEMORISE THIS

2. de Broglie Wavelength

Formula: [ \lambda = \frac{h}{p} = \frac{h}{mv} ] Variables: - ( \lambda ) = de Broglie wavelength (m) - ( p ) = Momentum (( mv )) (kg·m/s) - ( m ) = Mass of the particle (kg) - ( v ) = Velocity of the particle (m/s)

Special cases: - For electrons accelerated through potential ( V ): [ \lambda = \frac{h}{\sqrt{2meV}} ] – MEMORISE THIS (Derived from ( KE = eV = \frac{1}{2}mv^2 ), then ( p = \sqrt{2meV} ))

STEP-BY-STEP METHOD

For Photoelectric Effect Problems

Step 1: Identify what’s given and what’s asked. - Given: Usually ( \nu ), ( \phi ), or ( V_0 ). - Asked: ( KE_{\text{max}} ), ( \nu_0 ), ( \lambda ), or ( V_0 ).

Step 2: Write down Einstein’s equation: [ h\nu = \phi + KE_{\text{max}} ]

Step 3: If ( \phi ) is in eV, convert to Joules (1 eV = ( 1.6 \times 10^{-19} ) J).

Step 4: Solve for the unknown. - If ( KE_{\text{max}} ) is asked, rearrange: [ KE_{\text{max}} = h\nu - \phi ] - If ( V_0 ) is asked, use: [ eV_0 = KE_{\text{max}} ]

Step 5: Check units! Ensure all energies are in Joules (or all in eV).

For de Broglie Wavelength Problems

Step 1: Identify the particle (electron, proton, etc.) and its mass and velocity. - Electron mass (( m_e )) = ( 9.11 \times 10^{-31} ) kg – MEMORISE THIS - Proton mass (( m_p )) = ( 1.67 \times 10^{-27} ) kg – MEMORISE THIS

Step 2: If velocity (( v )) is not given, find it using: - ( KE = \frac{1}{2}mv^2 ) (if KE is given) - ( eV = \frac{1}{2}mv^2 ) (if accelerated through potential ( V ))

Step 3: Calculate momentum (( p = mv )).

Step 4: Plug into de Broglie formula: [ \lambda = \frac{h}{p} ]

Step 5: If the particle is an electron accelerated through ( V ) volts, use the shortcut: [ \lambda = \frac{12.27}{\sqrt{V}} \, \text{Å} ] – MEMORISE THIS

WORKED EXAMPLES

Example 1 – Basic (Photoelectric Effect)

Problem: Light of wavelength ( 300 \, \text{nm} ) falls on a metal with work function ( 2.5 \, \text{eV} ). Find the maximum kinetic energy of ejected electrons.

Solution: Step 1: Convert wavelength to frequency. [ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{300 \times 10^{-9}} = 1 \times 10^{15} \, \text{Hz} ]

Step 2: Convert work function to Joules. [ \phi = 2.5 \, \text{eV} = 2.5 \times 1.6 \times 10^{-19} = 4 \times 10^{-19} \, \text{J} ]

Step 3: Use Einstein’s equation. [ KE_{\text{max}} = h\nu - \phi ] [ KE_{\text{max}} = (6.626 \times 10^{-34})(1 \times 10^{15}) - 4 \times 10^{-19} ] [ KE_{\text{max}} = 6.626 \times 10^{-19} - 4 \times 10^{-19} ] [ KE_{\text{max}} = 2.626 \times 10^{-19} \, \text{J} ]

Step 4: Convert to eV (optional). [ KE_{\text{max}} = \frac{2.626 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.64 \, \text{eV} ]

What we did and why: We used Einstein’s photoelectric equation to relate photon energy to electron energy. The key was unit consistency—converting eV to Joules before plugging into the formula.

Example 2 – Medium (de Broglie Wavelength)

Problem: An electron is accelerated through a potential difference of ( 100 \, \text{V} ). Find its de Broglie wavelength.

Solution: Step 1: Use the shortcut formula for electrons. [ \lambda = \frac{12.27}{\sqrt{V}} \, \text{Å} ] [ \lambda = \frac{12.27}{\sqrt{100}} = 1.227 \, \text{Å} ]

Alternative (if you forget the shortcut): Step 1: Find velocity using ( eV = \frac{1}{2}mv^2 ). [ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 100}{9.11 \times 10^{-31}}} ] [ v = 5.93 \times 10^6 \, \text{m/s} ]

Step 2: Calculate momentum. [ p = mv = (9.11 \times 10^{-31})(5.93 \times 10^6) = 5.4 \times 10^{-24} \, \text{kg·m/s} ]

Step 3: Use de Broglie formula. [ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.4 \times 10^{-24}} = 1.227 \times 10^{-10} \, \text{m} = 1.227 \, \text{Å} ]

What we did and why: We used two methods to confirm the answer. The shortcut formula saves time in exams, but the step-by-step method ensures understanding.

Example 3 – Exam-Style (Disguised Problem)

Problem: A photon of energy ( 4.5 \, \text{eV} ) ejects an electron from a metal surface. The stopping potential is ( 2 \, \text{V} ). Find the work function of the metal.

Solution: Step 1: Understand what’s given. - Photon energy (( h\nu )) = ( 4.5 \, \text{eV} ) - Stopping potential (( V_0 )) = ( 2 \, \text{V} )

Step 2: Relate stopping potential to ( KE_{\text{max}} ). [ eV_0 = KE_{\text{max}} ] [ KE_{\text{max}} = 2 \, \text{eV} ]

Step 3: Use Einstein’s equation. [ h\nu = \phi + KE_{\text{max}} ] [ 4.5 = \phi + 2 ] [ \phi = 2.5 \, \text{eV} ]

What we did and why: The problem disguised ( KE_{\text{max}} ) as stopping potential. The key was recognizing that ( eV_0 = KE_{\text{max}} ).

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for dual nature of matter in IIT JEE:

1. Photoelectric effect:
2. Einstein’s equation: ( h\nu = \phi + KE_{\text{max}} ).
3. Stopping potential: ( eV_0 = KE_{\text{max}} ).
4. Threshold frequency: ( \phi = h\nu_0 ).

5. Always convert eV to Joules (1 eV = ( 1.6 \times 10^{-19} ) J).

6. de Broglie wavelength:

7. Formula: ( \lambda = \frac{h}{mv} ).
8. For electrons accelerated through ( V ) volts: ( \lambda = \frac{12.27}{\sqrt{V}} \, \text{Å} ).

9. Memorise electron mass: ( 9.11 \times 10^{-31} \, \text{kg} ).

10. Common traps:

11. Don’t mix eV and Joules.
12. Don’t forget to convert ( \lambda ) to ( \nu ).
13. If stopping potential is given, ( KE_{\text{max}} = eV_0 ).

That’s it. Now go crush those 10 marks!