Quantization of Energy: Photoelectric Effect, Bohr’s Model, X-rays

TAKEAWAYS:

• Energy is quantized (comes in fixed packets) unlike classical physics, which assumes continuous energy.
• The photoelectric effect proves light behaves as particles (photons) with energy $E=h\nu$.
• Bohr’s model explains atomic spectra by quantizing electron orbits with discrete energy levels.
• X-rays are produced by high-speed electrons hitting a metal target, used in medical imaging and industry.
• Planck’s constant ($h$) links energy and frequency: $E=h\nu$.
• These ideas revolutionized physics, leading to quantum mechanics and technologies like LEDs and lasers.

1. Planck’s Quantum Theory: The Birth of Quantization

Key Idea: Energy is not continuous

Classical physics (like Newton’s laws) assumed energy could be divided infinitely. But in 1900, Max Planck proposed that energy is emitted or absorbed in discrete packets called quanta. This was a radical idea—like saying money comes only in ₹5 notes, not in fractions!

• Planck’s Equation: The energy ($E$) of a quantum is proportional to its frequency ($\nu$): $$E=h\nu$$ where:
  • $h=6.626\times 10^{-34}$ Js (Planck’s constant).
  • $\nu$ = frequency of radiation (Hz).

Why was this revolutionary?

Before Planck, scientists couldn’t explain why hot objects (like a glowing filament) emit light at specific frequencies. Planck’s theory solved this by saying energy is quantized—only certain amounts are allowed.

pie
    title Energy Emission in Classical vs. Quantum View
    "Classical (Continuous)" : 30
    "Quantum (Discrete Packets)" : 70

Visual: This pie chart shows the shift from continuous energy (classical) to discrete packets (quantum). The quantum view matches real-world observations (e.g., atomic spectra).

2. The Photoelectric Effect: Light as Particles

Observation that Stumped Classical Physics

When light shines on a metal surface, electrons are ejected (photoelectrons). Classical physics predicted:

• Brighter light → more energy → more electrons ejected.
• Any frequency of light should work if it’s bright enough.

But experiments showed:

• Only light above a certain frequency (called the threshold frequency, ${\nu }_0$) ejects electrons.
• Below ${\nu }_0$, no electrons come out, no matter how bright the light!
• The energy of ejected electrons depends only on frequency, not brightness.

Einstein’s Explanation (1905): Photons!

Einstein said light behaves as particles of energy called photons. Each photon has energy: $$E=h\nu$$

• If $h\nu <h{\nu }_0$ (where ${\nu }_0$ is the threshold frequency), no electrons are ejected.
• If $h\nu >h{\nu }_0$, excess energy becomes the kinetic energy (KE) of the electron: $$K{E}_{\text{max}}=h\nu -h{\nu }_0=h(\nu -{\nu }_0)$$

Key Terms

Worked Example: Calculating Maximum KE

A metal has a work function $\varphi =4.2$ eV. Light of frequency $\nu =1.2\times 10^{15}$ Hz shines on it. Find the maximum KE of ejected electrons.

Solution:

1. Convert $\varphi$ to joules: $$\varphi =4.2\text{ eV}=4.2\times 1.6\times 10^{-19}\text{ J}=6.72\times 10^{-19}\text{ J}$$
2. Calculate photon energy: $$E=h\nu =(6.626\times 10^{-34})(1.2\times 10^{15})=7.95\times 10^{-19}\text{ J}$$
3. Use $K{E}_{\text{max}}=E-\varphi$: $$K{E}_{\text{max}}=7.95\times 10^{-19}-6.72\times 10^{-19}=1.23\times 10^{-19}\text{ J}$$
4. Convert to eV: $$K{E}_{\text{max}}=\frac{1.23\times 10^{-19}}{1.6\times 10^{-19}}=0.77\text{ eV}$$

Answer: The maximum KE of ejected electrons is 0.77 eV.

3. Bohr’s Model of the Hydrogen Atom: Quantized Orbits

Problem with Rutherford’s Model

Rutherford’s model (nucleus + electrons orbiting like planets) had a flaw:

• Electrons should lose energy and spiral into the nucleus (like a satellite crashing into Earth).
• But atoms are stable!

Bohr’s Solution (1913): Quantized Electron Orbits

Niels Bohr added quantization to Rutherford’s model:

1. Electrons orbit the nucleus in specific, fixed paths (orbits) with discrete energies.
2. Only certain orbits are allowed—no in-between orbits!
3. Electrons jump between orbits by absorbing or emitting energy in photons.

Key Equations

1. Energy of an Electron in the $n^{th}$ Orbit: $$E_n=-\frac{13.6}{n^2}\text{ eV}$$

   • $n=1,2,3,\dots$ (principal quantum number).
   • Negative sign: energy is bound (less than zero).

2. Frequency of Emitted/Absorbed Photon: When an electron jumps from orbit $n_i$ to $n_f$: $$\Delta E=E_f-E_i=h\nu$$ $$\nu =\frac{\Delta E}{h}=\frac{13.6}{h}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$$

Worked Example: Calculating Wavelength of Emitted Light

An electron in hydrogen jumps from $n=3$ to $n=2$. Find the wavelength of the emitted photon.

Solution:

1. Calculate $\Delta E$: $$E_3=-\frac{13.6}{3^2}=-1.51\text{ eV}$$ $$E_2=-\frac{13.6}{2^2}=-3.4\text{ eV}$$ $$\Delta E=E_2-E_3=-3.4-(-1.51)=-1.89\text{ eV}$$ (Negative sign means energy is emitted.)
2. Convert $\Delta E$ to joules: $$\Delta E=1.89\times 1.6\times 10^{-19}=3.024\times 10^{-19}\text{ J}$$
3. Find frequency: $$\nu =\frac{\Delta E}{h}=\frac{3.024\times 10^{-19}}{6.626\times 10^{-34}}=4.56\times 10^{14}\text{ Hz}$$
4. Find wavelength ($\lambda =\frac{c}{\nu }$): $$\lambda =\frac{3\times 10^8}{4.56\times 10^{14}}=6.58\times 10^{-7}\text{ m}=658\text{ nm}$$

Answer: The wavelength is 658 nm (red light, part of the Balmer series).

4. X-Rays: High-Energy Photons

How X-Rays Are Produced

When high-speed electrons (from a heated cathode) hit a metal target (anode), they:

1. Slow down abruptly → emit bremsstrahlung (braking radiation) as X-rays.
2. Knock out inner-shell electrons → outer electrons fill the gap, emitting X-rays of specific energies (characteristic X-rays).

Key Features of X-Rays

Worked Example: Minimum Wavelength of X-Rays

Electrons are accelerated through a potential difference of $V=30$ kV. Find the minimum wavelength of X-rays produced.

Solution:

1. KE of electrons = $eV$: $$KE=1.6\times 10^{-19}\times 30\times 10^3=4.8\times 10^{-15}\text{ J}$$
2. Maximum photon energy = KE (all energy converted to one photon): $$E=h\nu =4.8\times 10^{-15}\text{ J}$$
3. Minimum wavelength (${\lambda }_{\text{min}}$): $${\lambda }_{\text{min}}=\frac{hc}{E}=\frac{(6.626\times 10^{-34})(3\times 10^8)}{4.8\times 10^{-15}}=4.14\times 10^{-11}\text{ m}=0.0414\text{ nm}$$

Answer: The minimum wavelength is 0.0414 nm.

5. Applications of Quantization

6. Comparison: Wave vs. Particle Nature of Light

Exam Tip: How to Score Full Marks in NEB Questions

1. Photoelectric Effect Questions:

   • Always draw a graph of KE vs. frequency (linear with slope $h$ and intercept $-\varphi$).
   • Remember: Brightness affects photon number, not energy.
   • Use $K{E}_{\text{max}}=h\nu -\varphi$ or $e{V}_0=h\nu -\varphi$.

2. Bohr’s Model Questions:

   • For energy levels, use $E_n=-\frac{13.6}{n^2}$ eV.
   • For wavelengths, use $\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$, where $R=1.097\times 10^7$ m${}^{-1}$ (Rydberg constant).
   • Label transitions (e.g., Lyman series: $n_f=1$).

3. X-Ray Questions:

   • Minimum wavelength: ${\lambda }_{\text{min}}=\frac{hc}{eV}$.
   • Characteristic X-rays: depend on the target material (e.g., tungsten in X-ray tubes).

4. Common Mistakes to Avoid:

   • Forgetting the negative sign in Bohr’s energy formula.
   • Confusing frequency and wavelength in photoelectric equations.
   • Not converting units (eV to joules or vice versa).

NEB Board-Style Questions (Practice!)

Short Answer (5 marks)

1. Explain the photoelectric effect. Why does increasing the intensity of light not increase the maximum kinetic energy of photoelectrons?
2. Derive the expression for the energy levels of an electron in the $n^{th}$ orbit of a hydrogen atom according to Bohr’s model.
3. How are X-rays produced in a Coolidge tube? Mention two medical applications of X-rays.

Long Answer (10 marks)

4. (a) State Planck’s quantum theory. (2) (b) Explain the photoelectric effect with the help of Einstein’s equation. (4) (c) A metal surface has a work function of 2.2 eV. Calculate the threshold frequency and the maximum kinetic energy of photoelectrons when light of frequency $1.5\times 10^{15}$ Hz is incident on it. (4)
5. (a) Describe Bohr’s model of the hydrogen atom. (3) (b) Calculate the wavelength of the photon emitted when an electron in a hydrogen atom transitions from $n=4$ to $n=2$. (4) (c) Why does Bohr’s model fail for atoms with more than one electron? (3)

Summary Table: Key Formulas

Final Thought

Quantization changed physics forever! From explaining why metals emit electrons only above a certain frequency to predicting atomic spectra and enabling technologies like X-ray machines, these ideas are foundational. Practice the numericals—NEB loves them! 🚀