Group
Group
Modern Physics Formulas for JEE
Modern Physics covers the photoelectric effect, matter waves, atomic structure, nuclear physics, and semiconductors. Solving problems in these topics requires a clear understanding of the concepts, accurate unit conversions, and careful use of physical constants.
Use this JEE Mains Formula Sheet to revise key relations and understand when each applies. Practise questions independently, then check the tables to review your approach. Mark the relations you find difficult and revisit them regularly to strengthen recall and reduce calculation errors.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$h$$ | Planck constant | $$6.626\times10^{-34}\,\text{J s}$$ |
| $$c$$ | Speed of light in vacuum | $$3.00\times10^{8}\,\text{m s}^{-1}$$ |
| $$e$$ | Magnitude of electron charge | $$1.602\times10^{-19}\,\text{C}$$ |
| $$m_e$$ | Electron rest mass | $$9.11\times10^{-31}\,\text{kg}$$ |
| $$\phi$$ | Work function of the material | J or eV (convert) |
| $$\nu_0$$ | Threshold frequency | Hz |
| $$K_{\text{max}}$$ | Maximum kinetic energy of emitted electron | J or eV |
| $$I$$ | Intensity of incident light | W m-2 |
| $$\lambda$$ | Wavelength of photon or matter wave | m |
| $$p$$ | Linear momentum | kg m s-1 |
| $$n$$ | Principal quantum number | dimensionless |
| $$Z$$ | Atomic number (protons) | dimensionless |
| $$R_H$$ | Rydberg constant for hydrogen | $$1.097\times10^{7}\,\text{m}^{-1}$$ |
| $$E_n$$ | Energy of $$n^{\text{th}}$$ orbit | J or eV |
| $$r_n$$ | Radius of $$n^{\text{th}}$$ orbit | m |
| $$\Delta m$$ | Mass defect | kg or u |
| $$B.E.$$ | Nuclear binding energy | J or MeV |
| $$A$$ | Mass number (nucleons) | dimensionless |
| $$\lambda_r$$ | Radioactive decay constant | s-1 |
| $$T_{1/2}$$ | Half-life | s |
| $$N_0,\, N(t)$$ | Nuclei count initially / at time $$t$$ | dimensionless |
| $$n_i$$ | Intrinsic carrier concentration | m-3 |
| $$E_g$$ | Energy band gap | eV |
| $$I_s$$ | Saturation current of diode | A |
| $$\eta$$ | Ideality factor (≈1 for Si) | dimensionless |
| $$V_T$$ | Thermal voltage $$kT/e$$ | ≈25.9 mV at 300 K |
Photoelectric Effect Formulas
Here you relate photon energy to electron ejection from a metal surface. JEE setters often give stopping potential or wavelength and ask for work function or $$h/e$$. Numbers are clean so a careless unit slip is the main danger, building on constants defined above.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$h\nu = \phi + K_{\text{max}}$$ | Energy conservation for one photon–one electron event | Photon strikes clean surface, single-photon ejection |
| $$K_{\text{max}} = eV_s$$ | Connects maximum kinetic energy to stopping potential $$V_s$$ | Retarding potential just prevents most energetic electrons |
| $$\nu_0 = \phi/h$$ | Threshold frequency | $$K_{\text{max}}=0$$ |
| $$\lambda_0 = c/\nu_0 = hc/\phi$$ | Threshold wavelength | --- |
| $$K_{\text{max}} = h(\nu-\nu_0)$$ | Slope-intercept form used in graphs | $$\nu\gt \nu_0$$ |
| $$I \propto \text{photoelectron count rate}$$ | Current versus intensity relation | Frequency fixed $$\gt \nu_0$$ |
Worked Example
Light of wavelength $$300\,\text{nm}$$ falls on potassium surface (work function $$2.3\,\text{eV}$$). Find stopping potential.
- Photon energy $$E = hc/\lambda = (6.626\times10^{-34})(3.00\times10^{8})/(3.00\times10^{-7}) = 6.626\times10^{-19}\,\text{J} = 4.14\,\text{eV}$$.
- Maximum kinetic energy $$K_{\text{max}} = E - \phi = 4.14 - 2.3 = 1.84\,\text{eV}$$.
- Stopping potential $$V_s = K_{\text{max}}/e = 1.84\,\text{V}$$.
Answer: 1.8 V (to two significant figures)
Shortcuts and Special Cases
- For any wavelength in nm, photon energy in eV is $$1240/\lambda_{\text{nm}}$$; safe within 1% under exam conditions.
- The slope of $$K_{\text{max}}$$ vs $$\nu$$ graph is always $$h$$, independent of material; intercept on $$\nu$$ axis is $$\nu_0$$.
- If two metals A and B have different $$\phi$$ but same incident light, $$V_{s,B}-V_{s,A} = (\phi_A-\phi_B)/e$$ quickly yields unknown $$\phi$$.
Watch Out
- Mixing J and eV in one equation; convert before subtracting.
- Using intensity to find $$K_{\text{max}}$$—it changes current, not electron energy.
- For graph questions, students reverse the axes; read what is plotted.
Wave–Particle Duality and de Broglie Wavelength Formulas
After photon quantisation comes the turn of matter waves. Here questions ask you to compare electron and proton wavelengths, or fit an electron’s de Broglie wavelength to a diffraction grating from the previous sub-topic.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\lambda = h/p$$ | de Broglie wavelength of any particle | Non-relativistic unless you use relativistic $$p$$ |
| For electron accelerated through $$V: \; \lambda = h/\sqrt{2m_e eV}$$ | Shortcut for lab electrons | $$eV \ll m_e c^2$$ (≤100 keV) |
| Numerical: $$\lambda(\text{Å}) = 12.27/\sqrt{V(\text{volt})}$$ | Electron wavelength in Ångströms | Same non-relativistic limit |
| Relativistic momentum $$p = \sqrt{(E/c)^2 - (m_0 c)^2}$$ | Use for high-energy particles | Near-relativistic speeds |
Worked Example
Find de Broglie wavelength of electrons accelerated by $$150\,\text{V}$$.
- Use shortcut: $$\lambda(\text{Å}) = 12.27/\sqrt{150} = 12.27/12.25 = 1.00\,\text{Å}$$.
- Convert to nm: $$0.100\,\text{nm}$$.
Answer: 0.10 nm
Shortcuts and Special Cases
- Momentum of thermal neutron at 300 K: $$\lambda \approx 1.8\,\text{Å}$$—standard value, worth memorising.
- If mass increases 2000× (proton vs electron) for same kinetic energy, wavelength scales as $$1/\sqrt{m}$$, not linearly.
Watch Out
- Using $$\lambda = h/mv$$ directly when speed is relativistic; instead compute $$p$$ first.
- Dropping the square root in the electron shortcut; check algebra.
- Units confusion between Å and nm; 1 Å = 0.1 nm.
Bohr Model of Hydrogen Atom Formulas
With duality in place, Bohr quantised angular momentum and derived radii, energies, and velocities. Examiners love integer ratios: energy required to ionise from $$n=2$$, speed at $$n=3$$ etc. Builds directly on $$h$$ and $$m_e$$ from earlier sections.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$r_n = n^2 a_0/Z$$ | Radius of $$n^{\text{th}}$$ orbit; $$a_0=0.529\,\text{Å}$$ | Hydrogen-like ions |
| $$v_n = Z\alpha c / n$$ | Speed of electron in $$n^{\text{th}}$$ orbit (with $$\alpha=1/137$$) | Non-relativistic orbit $$n\gt Z\alpha^{-1}$$ |
| $$E_n = -13.6 Z^2 / n^2$$ eV | Total energy of the orbit | Binding energy convention |
| $$\mu = m_e M /(m_e+M)$$ | Reduced mass correction for nucleus mass $$M$$ | High-precision Qs |
| Ionisation energy $$E_{\infty}-E_n = 13.6 Z^2 / n^2$$ eV | Energy to free electron from $$n$$ | --- |
Worked Example
Calculate speed of electron in third orbit of He+ (single-electron ion).
- $$Z=2,\; n=3$$.
- $$v_n = Z\alpha c / n = (2)(1/137)(3.00\times10^8)/3 = (2/137)(1.00\times10^8) = 1.46\times10^6\,\text{m s}^{-1}$$.
Answer: $$1.5\times10^{6}\,\text{m s}^{-1}$$
Shortcuts and Special Cases
- Radius ratio $$r_m/r_n = (m/n)^2$$—useful to cancel $$a_0$$ quickly.
- Energy difference between consecutive levels for hydrogen $$\Delta E = 13.6(1/n^2 - 1/(n+1)^2)$$.
Watch Out
- For He+ forgetting to multiply by $$Z^2$$ in energy.
- Using Bohr for multi-electron atoms—only hydrogenic ions allowed.
- Missing reduced mass when question explicitly asks for “precise value”.
Atomic Spectra and Series Formulas
Bohr’s transitions create spectral lines. JEE links wavelength of Lyman or Balmer lines to initial and final $$n$$. The topic flows smoothly from the previous orbit energies.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\dfrac1\lambda = R_H Z^2 \left( \dfrac1{n_1^2} - \dfrac1{n_2^2} \right)$$ | Wavenumber of emitted/absorbed photon | $$n_2\gt n_1$$ |
| Lyman: $$n_1=1$$, Balmer: $$n_1=2$$, Paschen: $$n_1=3$$ | Series identifiers | --- |
| First line of series: $$n_2=n_1+1$$ | Shortest wavelength in emission? No, longest—remember reciprocal. | --- |
| Series limit $$\lambda_{\text{min}}$$: $$n_2\to\infty$$ | $$1/\lambda_{\text{min}} = R_H Z^2 / n_1^2$$ | Ionisation limit |
Worked Example
Find wavelength of first Balmer line of hydrogen.
- $$n_1=2,\; n_2=3$$.
- $$1/\lambda = R_H(1/2^2 - 1/3^2)=1.097\times10^{7}(1/4 - 1/9)=1.097\times10^{7}(5/36)=1.522\times10^{6}\,\text{m}^{-1}$$.
- $$\lambda = 657\,\text{nm}$$.
Answer: 656–657 nm (H-alpha)
Shortcuts and Special Cases
- Series limit for Lyman of H: $$\lambda_{\min}=91.2\,\text{nm}$$—worth memorising for quick ratio questions.
- Wavelength ratio between first Balmer and first Lyman lines is $$\approx 7.18$$; seen in past JEE Chapter-wise PYQ compilations.
Watch Out
- Confusing shortest vs longest wavelength in a series.
- Plugging $$Z^2$$ for neutral hydrogen (Z=1), trivial but many mis-type.
- Using nm when answer key expects Å; double-check units asked.
X-Rays Formulas
An electron beam hitting a metal target produces continuous bremsstrahlung and characteristic lines. JEE typically asks for cutoff wavelength or shifts in $$K_\alpha$$ lines. This extends photon concepts to higher energies.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\lambda_{\min} = \dfrac{hc}{eV} = \dfrac{1240\,\text{nm eV}}{V}$$ | Cutoff wavelength for bremsstrahlung | Electron energy fully converts into one photon |
| Characteristic $$K_\alpha$$: $$\nu = R_H c (Z-1)^2 \left(1/1^2 - 1/2^2\right)$$ | Moseley’s law for $$K_\alpha$$ frequency | High-Z atoms, screening constant 1 |
| Moseley general: $$\sqrt{\nu} = a(Z-b)$$ | Linear fit for experimental data | Series held fixed |
Worked Example
Electrons accelerated through $$30\,\text{kV}$$ strike a target. Find minimum X-ray wavelength.
- $$\lambda_{\min} = 1240\,\text{nm eV}/(30\times10^3) = 0.0413\,\text{nm} = 0.413\,\text{Å}$$.
Answer: 0.041 nm
Shortcuts and Special Cases
- Every time voltage doubles, $$\lambda_{\min}$$ halves.
- For $$K_\alpha$$ of Cu (Z=29), wavelength ≈ 1.54 Å—appears in diffraction set-ups.
Watch Out
- Using $$V$$ in volts but constant 1240 in eV·nm—units must match.
- Writing $$\lambda_{\min} = hc/e$$ instead of divided by $$eV$$; voltage missing.
Nuclear Binding Energy & Mass Defect Formulas
Energy released in nucleus formation or fission is straight $$\Delta m c^2$$. JEE flavours include binding energy per nucleon comparison between isotopes. This sub-topic connects mass–energy relation to nuclear data.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\Delta m = Z m_p + N m_n - m_{\text{nucleus}}$$ | Mass defect | Masses in same units |
| $$B.E.= \Delta m c^2$$ | Total binding energy | --- |
| Binding energy per nucleon $$= B.E./A$$ | Stability measure | --- |
| 1 u $$= 931.5$$ MeV/$$c^2$$ | Conversion constant | --- |
Worked Example
Calculate binding energy per nucleon of $$^{56}\text{Fe}$$ (mass $$55.9349\,\text{u}$$). Data: $$m_p=1.0073\,\text{u},\; m_n=1.0087\,\text{u}$$.
- $$Z=26,\; N=30,\; A=56$$.
- $$\Delta m = 26(1.0073)+30(1.0087)-55.9349 = 56.4678 - 55.9349 = 0.5329\,\text{u}$$.
- $$B.E. = 0.5329\times931.5 = 496\,\text{MeV}$$.
- Per nucleon $$= 496/56 = 8.86\,\text{MeV}$$.
Answer: 8.9 MeV per nucleon
Shortcuts and Special Cases
- Binding energy per nucleon peaks near $$A\approx60$$ (Fe–Ni region); mention if comparison asked.
- For small $$\Delta m$$ in atomic mass units, $$B.E.$$(MeV) ≈ $$931\Delta m$$—safe rounding.
Watch Out
- Using atomic masses without subtracting electron masses when problem gives nuclear masses—follow what is supplied.
- Mixing MeV and Joule in intermediate steps.
Radioactivity Formulas
Nuclei decay exponentially. JEE loves half-life chains and activity ratios. Builds on nuclear definitions above.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$N(t) = N_0 e^{-\lambda_r t}$$ | Nuclei remaining at time $$t$$ | Single decay mode |
| $$A = \lambda_r N$$ | Activity (decay rate) | --- |
| $$T_{1/2} = \ln2 /\lambda_r$$ | Half-life | --- |
| Mean life $$\tau = 1/\lambda_r = T_{1/2}/\ln2$$ | Average lifetime | --- |
| For successive decays: $$N_B(t)=\dfrac{\lambda_A N_{0A}}{\lambda_B-\lambda_A}\left(e^{-\lambda_A t}-e^{-\lambda_B t}\right)$$ | Daughter nuclide count | Chain $$A\to B\to$$… |
Worked Example
Sample initially has activity $$800\,\text{decays s}^{-1}$$. After 30 min it is $$200\,\text{decays s}^{-1}$$. Find half-life.
- $$A \propto N$$, so $$A(t)=A_0 e^{-\lambda t}$$.
- $$200=800 e^{-\lambda (1800)} \Rightarrow e^{-\lambda 1800}=0.25$$.
- Take ln: $$-\lambda 1800 = \ln0.25 = -1.386$$.
- $$\lambda = 1.386/1800 = 7.70\times10^{-4}\,\text{s}^{-1}$$.
- Half-life $$T_{1/2} = \ln2/\lambda = 0.693/7.70\times10^{-4} = 900\,\text{s}=15\,\text{min}$$.
Answer: 15 minutes
Shortcuts and Special Cases
- Activity falls to $$\dfrac1{2^n}$$ in $$n$$ half-lives—count fingers instead of logs.
- For $$t=T_{1/2}$$, $$N = N_0/2$$, $$A = A_0/2$$ simultaneously—obvious but commonly forgotten.
Watch Out
- Using minutes for $$t$$ but $$\lambda$$ in s-1.
- Confusing mean life and half-life; $$\tau = 1.44 T_{1/2}$$.
Semiconductors and Diodes Formulas
Modern physics ends in solid-state devices. JEE asks for intrinsic carrier concentration, diode current at forward bias, or logic gates. This ties back to energy quantisation but with macroscopic currents.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$n_i = A T^{3/2} e^{-E_g/(2kT)}$$ | Intrinsic carrier concentration | $$E_g \gt 3kT$$ |
| Mass-action law: $$np = n_i^2$$ | Electron–hole product in thermal equilibrium | --- |
| Diode current $$I = I_s\!\big(e^{V/(\eta V_T)} -1\big)$$ | PN-junction I-V relation | Low level injection, V forward |
| Reverse bias $$I \approx -I_s$$ | Saturation current magnitude | $$|V|\gg V_T$$ negative |
| Cut-in voltage Si ≈ 0.7 V, Ge ≈ 0.3 V | Approximate turn-on potential | Room temperature |
Worked Example
A silicon diode with $$I_s = 2\,\mu\text{A}$$ is forward-biased at $$0.7\,\text{V}$$, $$\eta=1$$, $$T=300\,\text{K}$$. Find current.
- Thermal voltage $$V_T = kT/e = 25.9\,\text{mV}$$.
- $$I = 2\times10^{-6} \big(e^{0.7/0.0259}-1\big)$$.
- Exponent $$0.7/0.0259 = 27.0, \; e^{27}=5.32\times10^{11}$$.
- Current $$I \approx 2\times10^{-6}\times5.32\times10^{11} = 1.06\times10^{6}\,\text{A}$$—clearly unrealistic because series resistance ignored, examiner usually caps voltage at 0.6 V or lower. So answer expected: saturation occurs, question likely asks for “ratio”, not absolute value. (Lesson: look out for context.)
Answer: Use formula carefully; unrealistic numbers flag simplified model.
Shortcuts and Special Cases
- At room temperature each additional 60 mV forward raises diode current tenfold (log-scale rule).
- For digital questions, treat diode as switch: ON if $$V_{forward}\gt 0.7\,\text{V}$$, OFF otherwise.
Watch Out
- Plugging $$I_s$$ in mA instead of A.
- For $$n\!p=n_i^2$$ confusion: in n-type large $$n$$, small $$p$$ but product fixed.
- Temperature dependence of $$V_T$$ ignored when question gives different T.
Commonly Confused Modern Physics Formulas for JEE
| Formula | Confused with | How to choose correctly |
|---|---|---|
| $$K_{\text{max}} = eV_s$$ | $$E = eV$$ of accelerated electron | Check if electron is emitted from metal (photoelectric) or accelerated in vacuum (cathode ray). |
| $$\lambda = h/p$$ | $$\lambda_{\min}=hc/eV$$ | Former is matter wave (depends on momentum), latter is X-ray cutoff (depends on voltage). |
| $$N(t) = N_0 e^{-\lambda t}$$ | $$Q(t)=Q_0 e^{-t/RC}$$ in capacitors | Look for “nuclei”, “activity”, “half-life” keywords. |
| Bohr $$E_n = -13.6/n^2$$ | Particle in box $$E_n \propto n^2$$ | Hydrogen energy decreases with $$n$$; infinite well energy increases with $$n^2$$. |
Common Mistakes to Avoid in JEE Modern Physics Formulas
- Switching J and eV mid-calculation then adding them.
- Forgetting $$Z^2$$ factor in hydrogen-like formulas.
- Using non-relativistic de Broglie shortcut above 100 keV.
- Dropping exponential minus one term in diode equation for small forward voltages.
- Half-life assumed as $$1/\lambda$$ instead of $$\ln2/\lambda$$.
- Cutoff wavelength formula used for characteristic lines; they are independent.
- Incorrect unit conversion: 1 Å = 10-10 m, not 10-8 m.
Quick Revision of Modern Physics Formulas for JEE
- $$h\nu = \phi + eV_s$$ — photoelectric equation.
- $$\lambda(\text{Å}) = 12.27/\sqrt{V}$$ — electron de Broglie shortcut.
- Bohr $$E_n = -13.6Z^2/n^2$$ eV; $$r_n = n^2 a_0/Z$$.
- $$1/\lambda = R_H Z^2 (1/n_1^2 - 1/n_2^2)$$ — spectral lines.
- X-ray cutoff $$\lambda_{\min}=1240\,\text{eV nm}/V$$.
- Binding energy $$B.E.= \Delta m c^2$$; 1 u = 931.5 MeV.
- Radioactive decay $$N=N_0/2^{t/T_{1/2}}$$ — mental math route.
- Diode I–V: $$I = I_s(e^{V/(\eta V_T)}-1)$$; each 60 mV → ×10 current.
Modern Physics Formulas for JEE: Conclusion
Modern Physics Formulas for JEE connect photon energy, matter waves, atomic structure, nuclear processes, and semiconductor behaviour. Effective revision involves understanding each relation, its assumptions, and the units used in calculations. Keep a JEE Mains Formula Sheet organised by topic so you can quickly review key concepts and distinguish between similar expressions.
Strengthen your understanding by solving Questions from JEE Mains previous papers . Then attempt a JEE Mains mock test to assess how accurately you apply these concepts under time pressure. Review mistakes involving unit conversions, energy levels, decay calculations, and mass definitions, and use them to guide your next revision session.
