Modern Physics — Concepts, Formulas & Solved Numericals

What Is Modern Physics — and Why Does It Matter for Your Exam?

Modern physics is everything that came after classical mechanics broke down in the late 19th century. When scientists tried to explain blackbody radiation, the photoelectric effect, and atomic spectra using classical physics, the answers came out completely wrong. That failure gave birth to quantum mechanics, the special theory of relativity, and our entire modern understanding of matter and energy.

For Class 12 boards, this chapter carries roughly 8-10 marks. For JEE Main, you’ll see 2-3 direct questions every session. JEE Advanced tests conceptual depth — they’ll combine photoelectric effect with energy levels or ask you to reason about de Broglie wavelength in tricky contexts.

The good news: modern physics is one of the most scoring chapters in Class 12. The formulas are few, the concepts are clean, and once you internalize the physical picture, the numericals become straightforward.

Key Terms & Definitions

Photon — A quantum (packet) of electromagnetic radiation. Energy $E = h\nu$ , where $h = 6.626 \times 10^{-34}$ J·s is Planck’s constant and $\nu$ is frequency. Photons have zero rest mass but carry momentum $p = h/\lambda$ .

Work Function ( $\phi$ ) — The minimum energy needed to eject an electron from a metal surface. Different metals have different work functions: sodium ( $\phi \approx 2.3$ eV), platinum ( $\phi \approx 5.7$ eV). This determines the threshold frequency.

Threshold Frequency ( $\nu_0$ ) — The minimum frequency of incident light that can cause photoemission. Below this frequency, no electrons are emitted regardless of intensity. $\nu_0 = \phi/h$ .

Stopping Potential ( $V_0$ ) — The reverse voltage needed to stop the most energetic photoelectrons. $eV_0 = h\nu - \phi = KE_{max}$ .

de Broglie Wavelength — Every particle with momentum has an associated wave. $\lambda = h/mv = h/p$ . This is why electrons can diffract — they have wavelength.

Bohr Radius ( $a_0$ ) — The most probable radius of the electron in a hydrogen atom at ground state: $a_0 = 0.529$ Å.

Binding Energy — The energy required to completely remove a nucleon from the nucleus. Greater binding energy per nucleon = more stable nucleus.

Half-Life ( $T_{1/2}$ ) — Time for half of a radioactive sample to decay. Related to decay constant: $T_{1/2} = 0.693/\lambda$ .

Core Concepts & Methods

Photoelectric Effect

Einstein’s 1905 explanation (which won him the Nobel Prize, not relativity) established that light consists of photons. When a photon hits a metal, it either gives ALL its energy to one electron or nothing — there’s no partial transfer.

The energy equation:

KE_{max} = h\nu - \phi = h\nu - h\nu_0

eV_0 = h\nu - \phi

where $V_0$ is stopping potential, $\phi$ is work function.

Three key observations and their explanations:

Below threshold frequency — no emission, even with intense light. Intensity means more photons, not more energetic photons. If $\nu < \nu_0$ , each photon lacks the energy to free any electron.
Above threshold — emission is instantaneous. Classical wave theory predicted electrons would need time to absorb enough energy. Quantum picture: one photon hits one electron, transfer is immediate.
$KE_{max}$ depends on frequency, not intensity. More intensity means more electrons emitted (more photons = more collisions), but each electron gets the same maximum energy since individual photon energy is fixed.

Bohr’s Model of the Hydrogen Atom

Bohr postulated that electrons orbit in fixed energy levels and emit/absorb photons when jumping between levels.

E_n = -\frac{13.6}{n^2} \text{ eV}

r_n = n^2 \times 0.529 \text{ Å}

v_n = \frac{2.18 \times 10^6}{n} \text{ m/s}

The negative sign in energy is crucial — it means the electron is bound. Ground state ( $n=1$ ) is most negative, meaning most tightly bound. Ionisation energy = 13.6 eV (energy to remove electron from $n=1$ to $n=\infty$ ).

For hydrogen-like ions (He $^+$ , Li $^{2+}$ , etc.) with atomic number $Z$ :

E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}, \quad r_n = \frac{n^2 a_0}{Z}

Spectral Series — When electrons fall to different levels, they emit photons in different series:

Series	Transition to	Region
Lyman	$n=1$	Ultraviolet
Balmer	$n=2$	Visible
Paschen	$n=3$	Infrared
Brackett	$n=4$	Infrared

The Balmer series is the one tested most — it’s visible light and appears in PYQs regularly.

de Broglie Wave-Particle Duality

\lambda = \frac{h}{p} = \frac{h}{mv}

For a particle accelerated through potential $V$ :

\lambda = \frac{h}{\sqrt{2meV}}

For an electron: $\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$ (V in volts)

This formula connects directly to Bohr’s model. Bohr’s condition $mvr = nh/2\pi$ is exactly the condition that an integer number of de Broglie wavelengths must fit around the orbit: $n\lambda = 2\pi r$ .

Radioactivity & Nuclear Physics

Radioactive decay is a statistical process — we can never predict when one specific nucleus decays, but we can predict the average behavior of a large sample.

N(t) = N_0 e^{-\lambda t}

T_{1/2} = \frac{0.693}{\lambda}

A = \lambda N \quad \text{(Activity)}

\text{Mean life} = \tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693}

Binding Energy per Nucleon — This is the key to understanding nuclear stability. Calculate as:

\text{BE} = [\Delta m] c^2 = [Z m_p + N m_n - m_{\text{nucleus}}] \times 931.5 \text{ MeV/u}

Iron-56 has the highest binding energy per nucleon (~8.8 MeV). Elements lighter than iron gain energy by fusion; heavier elements gain energy by fission.

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Light of frequency $7 \times 10^{14}$ Hz falls on a metal with work function 2 eV. Find the stopping potential.

Solution:

Energy of incident photon:

E = h\nu = 6.626 \times 10^{-34} \times 7 \times 10^{14} = 4.638 \times 10^{-19} \text{ J}

Convert to eV: $E = \frac{4.638 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.9 \text{ eV}$

Using photoelectric equation:

eV_0 = E - \phi = 2.9 - 2.0 = 0.9 \text{ eV}

\boxed{V_0 = 0.9 \text{ V}}

Example 2 — Medium (JEE Main Level)

Q: The electron in a hydrogen atom jumps from $n = 4$ to $n = 2$ . Find the wavelength of the emitted photon.

Solution:

Energy of photon emitted:

E = E_4 - E_2 = -\frac{13.6}{16} - \left(-\frac{13.6}{4}\right) = -0.85 + 3.4 = 2.55 \text{ eV}

Convert to Joules: $E = 2.55 \times 1.6 \times 10^{-19} = 4.08 \times 10^{-19}$ J

\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.08 \times 10^{-19}} = 4.87 \times 10^{-7} \text{ m} = 487 \text{ nm}

This is in the visible range (blue-green), confirming it belongs to the Balmer series.

Example 3 — Hard (JEE Advanced Level)

Q: A radioactive sample has $N_0$ atoms at $t = 0$ . After time $T$ , the number of atoms is $N_0/4$ . In the next time interval $T$ , how many atoms decay?

Solution:

After time $T$ : $N = N_0/4$ . This means $T = 2T_{1/2}$ (two half-lives elapsed).

So $T_{1/2} = T/2$ .

After time $2T$ (four half-lives total): $N = N_0/16$

Atoms at the start of the second interval $T$ : $N_0/4$

Atoms at the end of the second interval $T$ : $N_0/16$

Atoms that decayed in interval $T$ :

\Delta N = \frac{N_0}{4} - \frac{N_0}{16} = \frac{4N_0 - N_0}{16} = \boxed{\frac{3N_0}{16}}

The trick in JEE radioactivity problems: always find the half-life first. Then express any given time as a multiple of $T_{1/2}$ . This avoids messy exponentials.

Exam-Specific Tips

CBSE Class 12 Boards

CBSE marking scheme awards 1 mark for writing the correct formula, 1 mark for substitution, and 1 mark for the final answer with units. Even if your arithmetic is wrong, you get 2/3 marks for correct method. Always write the formula first.

High-weightage topics for boards:

Photoelectric effect (definition + one numerical): 3-4 marks every year
Bohr model energy level calculation: 2-3 marks
Radioactive decay law numerical: 3 marks
Nuclear binding energy: 2 marks

For 2-mark conceptual questions, CBSE often asks: “Why does $KE_{max}$ not depend on intensity?” Write two clear sentences explaining photon-electron one-to-one interaction.

JEE Main

Modern physics carries 2-3 questions per session, worth 8 marks. Based on patterns from 2023-2024:

de Broglie wavelength with kinetic energy or accelerating voltage: appeared in JEE Main 2024 Session 1
Stopping potential vs frequency graph interpretation
Hydrogen atom transitions and spectral series identification
Radioactive decay with mean life vs half-life distinction

JEE Main 2024 had a question asking for the ratio of de Broglie wavelengths of a proton and an alpha particle accelerated through the same potential. The key: $\lambda \propto 1/\sqrt{mq}$ . For $\alpha$ : mass $= 4m_p$ , charge $= 2e$ . Ratio works out to $\sqrt{8} = 2\sqrt{2}$ .

JEE Advanced

Advanced tests your ability to connect concepts. Expect questions that:

Combine photoelectric effect with Bohr model (photon emitted in transition = photon that causes emission from another metal)
Ask for the number of possible spectral lines when electrons fall from level $n$ to ground state: answer is $n(n-1)/2$
Involve binding energy and Q-value calculations for nuclear reactions

Common Mistakes to Avoid

Mistake 1: Confusing intensity with frequency in photoelectric effect. More intensity = more electrons emitted. Higher frequency = electrons with more kinetic energy. These are completely separate effects. Students often mix this up in 1-mark theory questions.

Mistake 2: Using $\nu_0 = \phi/h$ with $\phi$ in eV without converting. $h$ in SI is $6.626 \times 10^{-34}$ J·s. If $\phi$ is given in eV, either convert $\phi$ to Joules ( $\times 1.6 \times 10^{-19}$ ), or use $h = 4.136 \times 10^{-15}$ eV·s. Mixing units here gives completely wrong answers.

Mistake 3: Forgetting the negative sign in $E_n = -13.6/n^2$ eV. When an electron absorbs a photon and jumps up, energy increases (becomes less negative). When it falls down and emits, energy decreases. The sign tells you the direction. Students who ignore it get transition energies with wrong signs.

Mistake 4: Confusing half-life and mean life. Mean life $\tau = T_{1/2}/0.693 \approx 1.44 \, T_{1/2}$ . Mean life is always greater than half-life. CBSE once asked “find the fraction remaining after time equal to mean life” — answer is $1/e \approx 37\%$ , not $50\%$ .

Mistake 5: Wrong mass defect calculation — using atomic mass instead of nuclear mass. In binding energy problems, if you’re using atomic masses (which include electron masses), you need to account for this consistently. Either subtract electron masses explicitly, or use the fact that for neutral atoms, the electron masses cancel out in the mass defect formula.

Practice Questions

Q1. A photon has wavelength 400 nm. Find its energy in eV. (Take $hc = 1240$ eV·nm)

$E = hc/\lambda = 1240/400 = 3.1$ eV

Note: $hc = 1240$ eV·nm is a standard value worth memorizing. It makes wavelength-to-energy conversions instant.

Q2. The work function of a metal is 3.45 eV. What is the maximum wavelength of light that can cause photoemission?

Threshold wavelength: $\lambda_0 = hc/\phi = 1240/3.45 = 359$ nm

This is in the UV region — visible light (400-700 nm) cannot eject electrons from this metal.

Q3. An electron is accelerated through 100 V. Find its de Broglie wavelength.

Using the shortcut formula for electrons: $\lambda = 1.227/\sqrt{V}$ nm

$\lambda = 1.227/\sqrt{100} = 1.227/10 = 0.1227$ nm $= 1.227$ Å

This is comparable to atomic spacing in crystals — which is why electron diffraction experiments work.

Q4. Hydrogen atom is in the $n = 3$ state. How many different spectral lines can be emitted as it returns to ground state?

Possible transitions: $3 \to 2$ , $3 \to 1$ , $2 \to 1$ — that’s 3 lines.

General formula: from level $n$ , possible lines $= n(n-1)/2 = 3 \times 2/2 = 3$ .

Q5. A radioactive element has half-life 30 minutes. What fraction remains after 2 hours?

2 hours = 120 minutes = 4 half-lives

Fraction remaining $= (1/2)^4 = 1/16$

So $1/16$ of the original sample remains, meaning $15/16$ has decayed.

Q6. Find the ionisation energy of $He^+$ (singly ionised helium, $Z = 2$ ) in its ground state.

For hydrogen-like ion: $E_n = -13.6 Z^2/n^2$ eV

Ground state ( $n=1$ ) of He $^+$ : $E_1 = -13.6 \times 4/1 = -54.4$ eV

Ionisation energy $= 54.4$ eV (energy needed to remove the electron)

Q7. The stopping potential for a metal is 1.5 V when light of frequency $6 \times 10^{14}$ Hz falls on it. Find the work function in eV.

Energy of photon: $E = h\nu = 4.136 \times 10^{-15} \times 6 \times 10^{14} = 2.48$ eV

(Using $h = 4.136 \times 10^{-15}$ eV·s for direct eV answer)

Work function: $\phi = E - eV_0 = 2.48 - 1.5 = 0.98$ eV

This corresponds roughly to cesium, which is used in photocells.

Q8. A nucleus has mass defect of 0.03 u. Find its binding energy. ( $1 \text{ u} = 931.5 \text{ MeV}$ )

Binding energy $= \Delta m \times 931.5 = 0.03 \times 931.5 = 27.95$ MeV $\approx 28$ MeV

If this were a nucleus with 4 nucleons, binding energy per nucleon $= 7$ MeV — reasonable for a light nucleus.

FAQs

Why can’t classical physics explain the photoelectric effect?

Classical wave theory predicts that given enough time, any frequency of light should eject electrons — the wave just needs to deliver enough energy. But experiments showed a hard cutoff at threshold frequency with no time delay. Einstein’s photon model explained this perfectly: energy comes in discrete packets, and each packet either has enough energy or it doesn’t.

What is the physical meaning of de Broglie wavelength?

It’s the wavelength associated with the quantum mechanical wave that describes the particle. It doesn’t mean the particle is literally a wave — it means the particle exhibits wave-like behavior (interference, diffraction) with that characteristic length. Heavy objects have incredibly tiny de Broglie wavelengths, which is why we don’t see quantum effects in everyday life.

Why is the ground state energy of hydrogen negative?

The energy scale is set with a free electron (infinitely far from the nucleus) at zero. Bound states have lower energy than a free electron, so they’re negative. The more negative the energy, the more tightly bound the electron — that’s why the ground state ( $n=1$ , most negative) is most stable.

Does intensity affect stopping potential?

No. Stopping potential depends only on the maximum kinetic energy of emitted electrons, which depends only on photon frequency and work function. Increasing intensity sends more photons, which ejects more electrons — but each electron still gets the same maximum energy. This is a classic CBSE 1-mark conceptual question.

What is the difference between nuclear fission and fusion in terms of binding energy?

Both release energy by moving toward a more stable state (higher binding energy per nucleon). In fission, a heavy nucleus (like U-235) splits into medium-mass fragments with higher BE/nucleon — energy released. In fusion, light nuclei (like H isotopes) combine to form helium — again higher BE/nucleon, energy released. Iron is at the peak — neither its fission nor its fusion releases energy.

How is mean life different from half-life in radioactive decay?

Half-life is the time for exactly half the atoms to decay. Mean life ( $\tau$ ) is the average time a single atom lives before decaying — it works out to $1/\lambda$ , which equals $T_{1/2}/0.693 \approx 1.44 \, T_{1/2}$ . After one mean life, the fraction remaining is $1/e \approx 37\%$ , not $50\%$ .

Why does the Balmer series appear in the visible range?

Transitions to $n = 2$ involve moderate energy differences — not enough to produce X-rays, not too little to fall in the infrared. The $3 \to 2$ transition gives 1.89 eV (656 nm, red), the $4 \to 2$ gives 2.55 eV (487 nm, blue-green), and the $5 \to 2$ gives 2.86 eV (434 nm, violet). All these land in the visible range of 400-700 nm, which is why Balmer was identified first — it was literally visible to early spectroscopists.