Planck Quantum Theory | IIT JEE & NEET | Vineet Khatri | ATP STAR

Planck's Quantum Theory Fundamentals
📌 The theory, developed by Dr. Max Planck, explains that electromagnetic radiation energy travels in discrete packets, not a continuous form.
💡 These discrete energy packets are called Quanta; in the context of visible light, a quantum is specifically termed a Photon.
🚫 Energy transmission is discontinuous, meaning one cannot receive a fraction like $1.5$ quanta; energy arrives only in whole packets.

Energy of a Photon (Quantum)
⚛️ The energy ($E$) of a photon is directly proportional to the wave's frequency ($\nu$): $E \propto \nu$.
➗ Removing the proportionality sign introduces Planck's Constant ($h$): $E = h\nu$.
📏 Substituting frequency ($\nu = c/\lambda$, where $c$ is the speed of light and $\lambda$ is the wavelength) yields the key equation: $E = \frac{hc}{\lambda}$.

Constants and Unit Conversions
📏 Planck's Constant ($h$) is $6.6 \times 10^{-34} \text{ J}\cdot\text{s}$ and the Speed of Light ($c$) is $3 \times 10^8 \text{ m/s}$.
🛠️ A simplified relationship for energy in Joules using these constants is $E = \frac{2 \times 10^{-25}}{\lambda}$, where $\lambda$ is in meters.
⚡ For calculations using energy in electron volts (eV), the simplified formula is $E = \frac{12400}{\lambda}$, where $\lambda$ must be in Angstroms ($\text{\AA}$).

Calculations and Applications
🔢 If wavelength ($\lambda$) is given in nanometers ($\text{nm}$), use $E = \frac{1240}{\lambda_{\text{nm}}}$ for energy in electron volts.
📊 The ratio of energies ($E_1/E_2$) for two photons is inversely proportional to the ratio of their wavelengths ($\lambda_2/\lambda_1$): $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
📉 Inverse Relationship: Lower wavelength corresponds to higher energy photons (e.g., $\lambda_1 = 3000 \text{\AA}$ and $\lambda_2 = 4000 \text{\AA}$ results in an energy ratio of $4:3$).

Key Points & Insights
➡️ Energy transfer for electromagnetic radiation occurs in quantized packets (Quanta/Photons), overturning the classical view of continuous transfer.
➡️ Always use the appropriate energy equation based on the required unit: $E = h\nu$ (fundamental), $E = \frac{hc}{\lambda}$ (SI units), or $E = \frac{12400}{\lambda_{\text{\AA}}}$ (for $\text{eV}$).
➡️ In problem-solving, ensure consistency in units; if energy is in Joules, use $h$ and $c$ values that result in Joules, and convert $\lambda$ to meters; if energy is in $\text{eV}$, use the $\text{12400}$ constant with $\lambda$ in Angstroms.

📸 Video summarized with SummaryTube.com on Feb 24, 2026, 00:30 UTC