Modern Physics is one of those JEE Physics areas where conceptual clarity and formula application can convert directly into marks.

Topics such as the photoelectric effect, radioactive decay, and de Broglie wavelength may look different, but JEE questions often test the same core skill: identify the physical principle, choose the correct equation, and substitute carefully.

For JEE aspirants, the objective should not be to memorise every formula separately. Instead, understand what each equation means, when it applies, and how to recognise common numerical patterns.

This guide focuses on three high-value areas:

1. Einstein’s photoelectric equation and its applications
2. Radioactive decay calculations
3. Ten common solved numerical patterns

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1. Photoelectric Effect: Start With Einstein’s Equation

The photoelectric effect occurs when electromagnetic radiation of sufficient frequency falls on a metal surface and causes electrons to be emitted.

Einstein explained the phenomenon by treating light as packets of energy called photons.

The energy of one photon is:E=hf

where:

• E = photon energy
• h = Planck’s constant
• f = frequency

If the photon provides enough energy to overcome the metal’s work function, the remaining energy becomes the maximum kinetic energy of the emitted electron.

Therefore:Kmax​=hf−ϕ​

This is the Einstein photoelectric equation.

Here, ϕ represents the work function of the metal.

Since:ϕ=hf0​

where f0​ is the threshold frequency,

we can also write:Kmax​=h(f−f0​)​

This form is particularly useful in JEE numerical problems.

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Application 1: Finding Maximum Kinetic Energy

Suppose light of frequency f falls on a metal having work function ϕ.

Immediately write:Kmax​=hf−ϕ

If the answer is required in electron volts, remember:1eV=1.6×10−19J

Example

A metal has a work function of 2eV. Light of photon energy 5eV is incident on it. Find the maximum kinetic energy.

Using:Kmax​=E−ϕKmax​=5−2Kmax​=3eV​

The key is recognising that the excess photon energy becomes kinetic energy.

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Application 2: Threshold Frequency

At threshold frequency, the emitted electron has zero maximum kinetic energy.

Therefore:hf0​=ϕ

So:f0​=hϕ​​

If the work function is given in electron volts, convert it into joules before using SI units.

Quick JEE Shortcut

If the photon energy is already expressed in eV:Kmax​(eV)=Ephoton​(eV)−ϕ(eV)

No joule conversion is needed.

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Application 3: Stopping Potential

The maximum kinetic energy of the photoelectrons can also be related to the stopping potential Vs​:Kmax​=eVs​​

Therefore:hf−ϕ=eVs​​

This equation connects photon frequency, work function and stopping potential.

For two different frequencies:e(V2​−V1​)=h(f2​−f1​)

Therefore:eh​=ΔfΔVs​​​

This relationship is commonly tested conceptually and numerically.

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Application 4: Intensity vs Frequency

This is a classic conceptual trap.

Increasing frequency

Increasing frequency increases:Ephoton​=hf

Therefore, it increases the maximum kinetic energy of emitted electrons.

Increasing intensity

Increasing intensity means more photons are incident per unit time.

If the frequency is already above threshold, increasing intensity generally increases the number of emitted electrons, and therefore the photocurrent.

It does not directly increase the maximum kinetic energy.

Remember:

Frequency → energy per photon → Kmax​

Intensity → number of photons → photocurrent

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Application 5: Work Function and Threshold Wavelength

Since:ϕ=hf0​

and:f0​=λ0​c​

we obtain:ϕ=λ0​hc​​

where λ0​ is the threshold wavelength.

Therefore:λ0​=ϕhc​​

This is useful when a question gives wavelength instead of frequency.

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2. Radioactive Decay Calculations

Radioactive decay is another area where JEE questions frequently become formula-based.

The fundamental radioactive decay equation is:N=N0​e−λt​

where:

• N0​ = initial number of radioactive nuclei
• N = number remaining after time t
• λ = decay constant
• t = time

The activity is:A=λN​

Therefore:A=A0​e−λt

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Half-Life

The half-life T1/2​ is the time required for half the radioactive nuclei to decay.

The relationship between decay constant and half-life is:T1/2​=λln2​​

or:λ=T1/2​0.693​​

This is one of the most important formulas to remember.

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Radioactive Decay Without Exponentials

If the question involves whole numbers of half-lives, use:N=N0​(21​)n​

where:n=T1/2​t​

This is usually much faster than using the exponential equation.

Example

A radioactive sample has a half-life of 5 hours. What fraction remains after 20 hours?

Number of half-lives:n=520​=4

Therefore:N0​N​=(21​)4N0​N​=161​​

So 6.25% remains.

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Activity and Half-Life

Because:A=λN

and λ remains constant for a particular radioactive substance:A∝N

Therefore, after one half-life:A=2A0​​

After two:A=4A0​​

After three:A=8A0​​

This makes activity questions very quick when the number of half-lives is obvious.

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Mean Life

The mean life τ is related to decay constant by:τ=λ1​​

Since:T1/2​=λ0.693​

we get:T1/2​=0.693τ​

or:τ=0.693T1/2​​​

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3. Ten Solved Numerical Patterns for JEE

The best way to prepare Modern Physics is to recognise recurring numerical structures.

Pattern 1: Photon Energy From Frequency

Question

Find the energy of a photon having frequency 5×1014Hz.

Use:E=hf

Taking:h=6.63×10−34Js

Therefore:E=(6.63×10−34)(5×1014)E=3.315×10−19J​

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Pattern 2: Photon Energy From Wavelength

Question

Find the energy of a photon having wavelength 400nm.

Use:E=λhc​

Convert:400nm=400×10−9m

Therefore:E=400×10−9(6.63×10−34)(3×108)​E≈4.97×10−19J​

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Pattern 3: Maximum Kinetic Energy

Question

Light of photon energy 6eV falls on a metal having work function 2.5eV. Find Kmax​.Kmax​=E−ϕ=6−2.5Kmax​=3.5eV​

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Pattern 4: Stopping Potential

Question

The maximum kinetic energy of photoelectrons is 4eV. Find the stopping potential.

Using:Kmax​=eVs​

When energy is expressed in eV:Vs​=4V​

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Pattern 5: Threshold Frequency

Question

A metal has work function 3.0eV. Find its threshold frequency.

Convert work function:ϕ=3(1.6×10−19)ϕ=4.8×10−19J

Using:f0​=hϕ​f0​=6.63×10−344.8×10−19​f0​≈7.24×1014Hz​

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Pattern 6: Number of Nuclei After Several Half-Lives

Question

A sample initially contains 8000 nuclei. Its half-life is 10 days. How many nuclei remain after 30 days?

Number of half-lives:n=1030​=3

Therefore:N=8000(21​)3N=1000​

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Pattern 7: Finding Half-Life From Remaining Fraction

Question

A radioactive sample reduces to 1/8 of its original amount in 15 hours. Find the half-life.

We know:81​=(21​)3

Therefore, 15 hours corresponds to 3 half-lives.T1/2​=315​T1/2​=5 hours​

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Pattern 8: Finding Decay Constant

Question

A radioactive substance has a half-life of 10 hours. Find its decay constant.

Use:λ=T1/2​0.693​

Therefore:λ=100.693​λ=0.0693h−1​

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Pattern 9: Activity After Two Half-Lives

Question

The initial activity of a sample is 800Bq. Find its activity after two half-lives.

After one half-life:A=2800​=400Bq

After two:A=2400​A=200Bq​

Or directly:A=800(21​)2=200Bq

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Pattern 10: de Broglie Wavelength

The de Broglie wavelength of a particle is:λ=ph​​

For a particle of mass m moving with velocity v:λ=mvh​​

Question

Find the de Broglie wavelength of an electron moving with velocity v.

Using:λ=mvh​

For an electron:m=9.1×10−31kg

Therefore:λ=(9.1×10−31)v6.63×10−34​​

The important part is recognising that momentum determines the wavelength.

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de Broglie Wavelength: The JEE Shortcut

If a charged particle is accelerated through potential difference V, its kinetic energy is:K=qV

For non-relativistic motion:K=2mp2​

Therefore:p=2mqV​

Using:λ=ph​

we get:λ=2mqV​h​​

For an electron:λ=2meV​h​​

A commonly used approximate form is:λ(A˚)=V​12.27​​

for an electron accelerated through V volts, within the non-relativistic approximation.

This formula can save considerable time in numerical questions.

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How to Recognise the Right Formula

Instead of memorising every equation independently, connect the formulas to the information given in the question.

Given	Think
Frequency f	E=hf
Wavelength λ	E=hc/λ
Work function ϕ	Kmax​=hf−ϕ
Stopping potential	Kmax​=eVs​
Threshold frequency	ϕ=hf0​
Threshold wavelength	ϕ=hc/λ0​
Half-life	N=N0​(1/2)n
Decay constant	λ=0.693/T1/2​
Activity	A=λN
Particle momentum	λ=h/p
Particle accelerated through V	λ=h/2mqV​

This kind of formula recognition is particularly useful under JEE time pressure.

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Common JEE Mistakes in Modern Physics

Mistake 1: Confusing Intensity With Frequency

Remember:

Frequency controls photon energy.

Intensity controls the number of photons.

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Mistake 2: Forgetting the Work Function

Don’t write:Kmax​=hf

The correct equation is:Kmax​=hf−ϕ​

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Mistake 3: Mixing Joules and Electron Volts

If your quantities are in eV, you can often perform the calculation directly in eV.

If using h in SI units, convert the energy into joules.

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Mistake 4: Using the Wrong Radioactive Formula

If the time is an exact multiple of the half-life, use:N=N0​(21​)n

It is faster than using:N=N0​e−λt

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Mistake 5: Forgetting That Activity Also Decays

Since:A=λN

activity decreases in the same proportion as the number of undecayed nuclei.

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Mistake 6: Confusing Mean Life and Half-Life

Remember:τ=λ1​​

while:T1/2​=λ0.693​​

Therefore:T1/2​=0.693τ​

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A Smart Revision Strategy for Modern Physics

Do not revise Modern Physics by reading the chapter repeatedly.

Use a three-step cycle.

Step 1: Build a Formula Sheet

Write down:

• Einstein equation
• Threshold frequency
• Stopping potential
• Photon energy
• Radioactive decay
• Half-life
• Mean life
• Activity
• de Broglie wavelength

Step 2: Group Questions by Pattern

Solve 5–10 questions each on:

• Photoelectric effect
• Stopping potential
• Radioactive decay
• Half-life
• Activity
• de Broglie wavelength

Step 3: Maintain a Mistake List

For every incorrect question, record:

Concept → Mistake → Correct approach

This is much more useful than simply marking the answer and moving on.

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Final Takeaway

Modern Physics becomes much easier when you stop treating it as a collection of unrelated formulas.

Think in terms of three physical ideas:

Photon Energy

E=hf=λhc​​

Photoelectric Effect

Kmax​=hf−ϕ=eVs​​

Radioactive Decay

N=N0​e−λt​

and:T1/2​=λ0.693​​

Matter Waves

λ=ph​​

Once these relationships are clear, many JEE questions become pattern-recognition exercises rather than lengthy calculations.

The best way to prepare is simple:

Understand the concept → recognise the pattern → apply the formula → check the units → move on.

That combination of conceptual clarity and speed can make Modern Physics one of the more manageable scoring areas in JEE Physics.