Grade 12 Physics: Quantum and Nuclear Science Comprehensive Guide

Planck's Concept of Quantization of Energy

• Historical Background: In 1900, German physicist Max Planck discovered he could calculate the correct emission spectrum of solids if he assumed an atom could only emit or absorb specific, discrete amounts of energy.

• The Energy Hypothesis: Planck hypothesized that an atom's energy changes in a solid are proportional to the frequency of its vibration multiplied by an integer.

• Formula for Energy of Vibration: The energy ($E$) emitted or absorbed by a vibrating atom is the product of an integer, Planck’s constant, and the frequency of vibration:

  • $E = nhf$

  • $f$: frequency of vibration ($Hz$

  • $h$: Planck’s constant ($6.62 \times 10^{-34}\,J/Hz$ or $J \cdot s$)

  • $n$: an integer ($0, 1, 2, 3, \dots$)

• Quantization Concept: Energy exists only in specific "bundles" or quanta. It can have values like $hf$, $2hf$, or $3hf$, but never fractional values such as $\frac{2}{3} hf$ or $\frac{3}{4} hf$.

• Changes in Vibrations:

  • Atoms emit radiation only at specific moments when their vibrational energy state changes.

  • Example: If an atom moves from $3hf \rightarrow 2hf$, the energy radiated ($\Delta E$) is $3hf - 2hf = hf$.

  • Example: If an atom absorbs energy $hf$, it could transition from a lower state to a higher state (e.g., $2hf \rightarrow 3hf$).

• Macroscopic vs. Atomic Observation:

  • In everyday life, energy changes appear continuous because the value of Planck's constant ($h$) is extremely small, making the energy-changing steps nearly imperceptible.

  • At the atomic level, these discrete steps are significant and observable.

Photons and Energy-Wavelength Calculations

• Relationship Between Energy and Wavelength:

  • The energy of a photon can be calculated using the formula: $E = \frac{1240\,eV \cdot nm}{\lambda}$.

  • Alternative calculation: $E = hf = \frac{hc}{\lambda}$.

• Constants:

  • $c$ (speed of light): $3 \times 10^8\,m/s$

  • Conversion factor: $1\,eV = 1.6 \times 10^{-19}\,J$

• Practice Problems (Page 72):

  • Problem 1: Find energy for a wavelength of $515\,nm$.

    • Calculation 1: $E = \frac{1240\,eV \cdot nm}{515\,nm} = 2.4\,eV$

    • Calculation 2: $E = \frac{(6.63 \times 10^{-34}\cdot s)(3 \times 10^8\,m/s)}{515 \times 10^{-9}\,m} = 3.86 \times 10^{-19}\,J$

    • Conversion: $\frac{3.86 \times 10^{-19}\,J}{1.6 \times 10^{-19}\,J/eV} = 2.4\,eV$

  • Problem 2: Find wavelength if photon energy is $2.03\,eV$.

    • $\lambda = \frac{1240\,eV \cdot nm}{E} = \frac{1240}{2.03} = 610.8\,nm$

  • Problem 3: Rank photons from least to greatest energy: (A) $4.0\,eV$, (B) $320\,nm$, (C) $811\,nm$, (D) $2.1\,eV$.

    • (B) $320\,nm \rightarrow E = \frac{1240}{320} = 3.87\,eV$

    • (C) $811\,nm \rightarrow E = \frac{1240}{811} = 1.53\,eV$

    • Ranking: C (1.53\,eV) < D (2.1\,eV) < B (3.87\,eV) < A (4.0\,eV).

  • Problem 4 (Visible Light Range): Find the energy range for visible light ($400\,nm$ to $700\,nm$).

    • For $400\,nm$: $E = \frac{1240}{400} = 3.1\,eV$

    • For $700\,nm$: $E = \frac{1240}{700} = 1.77\,eV$

• Extra Question 14: Calculate frequency if transition energy change is $5.44 \times 10^{-19}\,J$ for $n=1$.

  • $f = \frac{E}{nh} = \frac{5.44 \times 10^{-19}}{1 \times 6.63 \times 10^{-34}} = 8.21 \times 10^{14}\,Hz$

The Photoelectric Effect

• Definition: The emission of electrons from a surface (usually metal) when electromagnetic radiation (light) falls on it.

• Failure of Classical Wave Theory:

  • Classical predictions: The electric field of light should accelerate and eject electrons regardless of frequency, depending only on intensity. It predicted that low-frequency light could eventually eject electrons if given enough time.

  • Experimental reality: Electrons are only ejected if the light frequency is above a specific minimum, called the Threshold Frequency ($f_0$).

• Observation via Photocell:

  • Cathode: Large electrode, often coated with cesium or alkali metals, which electrons are ejected from.

  • Anode: Small positive electrode (to minimize radiation blockage) that attracts photoelectrons.

  • Current: Formed by the flow of photoelectrons, measured by an ammeter.

• Threshold Frequency and Intensity Conclusions:

  • If f < f_0: No electrons are ejected, no matter how intense the light is.

  • If $f \geq f_0$: Electrons are ejected immediately. Increased intensity increases the number of electrons (current) but does not increase their individual kinetic energy.

  • High-frequency light provides enough energy per photon to eject an electron in a single interaction.

Einstein’s Photon Theory and Work Function

• Photon Interaction: Einstein proposed that light consists of discrete particles (photons). One photon interacts with exactly one electron.

• Work Function ($W$): The minimum energy required to free the most weakly bound electron from the metal.

  • $W = hf_0 = \frac{hc}{\lambda_0}$ h

• Energy States of Photoemission:

  • No Ejection: Photon Energy ($hf$) < Work Function ($W$).

  • Ejection with Zero Kinetic Energy: Photon Energy ($hf$) = Work Function ($W$). $f = f_0$.

  • Ejection with Kinetic Energy: Photon Energy ($hf$) > Work Function ($W$). f > f_0.

• Maximum Kinetic Energy Equation:

  • $KE_{max} = hf - hf_0$

  • $KE_{max} = E_{photon} - W$

• Variable Kinetic Energy: Not all ejected electrons have the same $KE$. Those deeper in the metal lose energy through collisions while escaping, so $KE_{max}$ refers to the surface electrons.

• Stopping Potential ($\Delta V_0$): The potential difference required to stop the most energetic photoelectrons.

  • $KE_{max} = -e\Delta V_0$

  • $e$ (charge of electron) = $-1.602 \times 10^{-19}\,C$

Photoelectric Effect Practice Problems

• Example Problem 1: Stopping potential is $4.0\,V$. Find $KE_{max}$.

  • $KE = |e|\Delta V_0 = (1.602 \times 10^{-19}\,C)(4.0\,V) = 6.4 \times 10^{-19}\,J$

  • In $eV$: $4.0\,eV$

• Example Problem 2: Sodium has a threshold wavelength of $526\,nm$.

  • a. Find Work Function ($W$) in $eV$:

    • $W = \frac{1240\,eV \cdot nm}{526\,nm} = 2.36\,eV$

  • b. If UV radiation ($348\,nm$) hits it, what is $KE_{max}$?

    • $E_{photon} = \frac{1240}{348} = 3.56\,eV$

    • Since 3.56\,eV > 2.36\,eV, electrons are discharged.

    • $KE_{max} = 3.56 - 2.36 = 1.2\,eV$

• Practice Problem 12: Zinc threshold wavelength is $310\,nm$. Find $f_0$ and $W$.

  • $W = \frac{1240}{310} = 4.0\,eV$

  • $f_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{310 \times 10^{-9}} = 9.67 \times 10^{14}\,Hz$

• Practice Problem 13: Cesium $W = 1.95\,eV$. Find $KE_{max}$ for $425\,nm$ light.

  • $E = \frac{1240}{425} = 2.917\,eV$

  • $KE_{max} = 2.917 - 1.95 = 0.967\,eV$

Graphical Analysis of Photoelectric Data

• Linear Graph Properties:

  • Slope: The slope of the line in a graph of $KE_{max}$ vs. Frequency equals Planck’s constant ($h$).

  • Slope (Stopping Potential): The slope of the line in a graph of Stopping Potential ($\Delta V_0$) vs. Frequency equals $\frac{h}{e}$.

  • X-Intercept: The point where the line intersects the x-axis represents the Threshold Frequency ($f_0$).

  • Metal Comparisons: All metals produce graphs with the same slope ($h$) because Planck's constant is universal; they only differ by their x-intercept ($f_0$), determined by the identity of the metal.

De Broglie Waves

• Matter Wave Proposal (1923): Louis de Broglie proposed that moving particles have wavelike properties.

• De Broglie Wavelength Formula:

  • $\lambda = \frac{h}{P} = \frac{h}{mv}$

  • $m$: mass; $v$: velocity; $P$: momentum.

• Experimental Evidence: Electron diffraction was confirmed in 1927, proving and validating de Broglie's theory.

• Visibility of Effects: Wavelike properties are only observable for very small particles (electrons/protons) because macroscopic objects (e.g., bowling balls) have masses so large that their wavelengths are too small to be detected.

• Practice Problems (Page 80):

  • Electron accelerated by $75\,V$:

    • $v = \sqrt{\frac{-2e\Delta V}{m}} = \sqrt{\frac{-2(-1.602 \times 10^{-19})(75)}{9.11 \times 10^{-31}}} = 5.1 \times 10^6\,m/s$

    • $\lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(5.1 \times 10^6)} = 1.4 \times 10^{-10}\,m$

  • Bowling Ball ($7.0\,kg$, $8.5\,m/s$):

    • $\lambda = \frac{6.63 \times 10^{-34}}{(7)(8.5)} = 1.1 \times 10^{-35}\,m$ (Too small to observe).

  • Condition for Stable Orbits (Bohr Model Integration): The circumference ($2\pi r$) of a stable electron orbit must be an integer multiple of the de Broglie wavelength: $2\pi r = n\lambda$.

Heisenberg Uncertainty Principle

• Core Concept: It is fundamentally impossible to measure both the position and momentum of a particle simultaneously with infinite precision.

• Mechanism of Uncertainty:

  • To locate a particle precisely, one must use shorter-wavelength radiation to reduce diffraction.

  • Shorter-wavelength radiation (high energy photons) causes the Compton effect, changing the particle's momentum upon impact.

  • Actively measuring position disturbs momentum, and vice versa.

• Implications: Newton’s and Maxwell's classical models work for everyday objects, but Quantum Theory is required for atomic-scale descriptions.

• Double Slit Interference: Explain patterns even when particles pass one by one because the uncertainty in momentum as it passes the slit makes it impossible to define which slit was traversed, allowing for wave interference distribution.

Atomic Models and Nuclear Structure

• Thomson’s Model: The "Raisin in a Muffin" model. The atom is a massive, positively charged substance with negative electrons distributed throughout.

• Rutherford’s Scattering Experiment (1911):

  • A beam of massive, high-speed alpha ($\alpha$) particles was directed at thin gold foil.

  • Expected: Minor deflections.

  • Observed: Most passed through undeflected; some were scattered at large angles; a few rebounded (>90^{\circ}).

• Rutherford’s Nuclear Model:

  • Concentrated positive charge and nearly all mass (99.9%) are in a tiny core called the nucleus.

  • Electrons orbit the nucleus (planetary model).

  • The atom is mostly empty space (diameter is $50,000$ times larger than the nucleus).

• Flaws in Planetary Model:

  1. Stability: Accelerating electrons should radiate energy and spiral into the nucleus within a fraction of a second.

  2. Discrete Spectra: It could not explain why atoms emit specific wavelengths rather than a continuous spectrum.

Atomic Spectra and Bohr's Energy Levels

• Emission Spectrum: A set of distinct, colored lines emitted by a gas in a discharge tube. Used as a "fingerprint" for elements.

• Absorption Spectrum: Dark lines appearing in a continuous spectrum when white light passes through a cool gas sample. A gas absorbs the same wavelengths it emits.

• Energy Level Transitions:

  • $\Delta E_{atom} = E_f - E_i = -E_{photon}$

  • $E_{photon} = hf = |\Delta E_{atom}|$

• Hydrogen Energy Levels (Bohr):

  • $E_n = -(13.6\,eV) \cdot \frac{1}{n^2}$

  • Orbital Radius: $r_n = r_1 \cdot n^2$ (where $r_1 = 5.3 \times 10^{-11}\,m$).

• Spectral Series of Hydrogen:

  • Lyman Series: Transitions into $n=1$; produces Ultraviolet light.

  • Balmer Series: Transitions into $n=2$; produces Visible light (4 lines).

  • Paschen Series: Transitions into $n=3$; produces Infrared radiation.

Nuclear Properties and Stability

• Nuclear Notation: ${}_{Z}^{A} X$

  • $Z$: Atomic Number (Protons).

  • $A$: Mass Number/Nucleons (Protons + Neutrons).

  • $N$: Number of Neutrons ($N = A - Z$).

• Charge of Nucleus: $Q = +Ze$.

• Strong Nuclear Force:

  • Very short-range ($1.4 \times 10^{-15}\,m$ - about a proton's radius).

  • Over $100$ times stronger than the electromagnetic repulsion at short range.

  • Attractive between all nucleons (P-P, N-N, P-N).

• Mass Defect ($\Delta m$): The difference between the actual mass of the nucleus and the sum of the masses of its individual nucleons.

  • $\Delta m = m_{nucleus} - [ (Z \times m_p) + (N \times m_n) ]$

• Binding Energy ($E$): Energy converted from mass that holds the nucleus together.

  • $E = \Delta m c^2$

  • Conversion: $1\,u = 931.49\,MeV$.

Radioactive Decay and Nuclear Reactions

• Alpha Decay ($\alpha$): Emission of a Helium nucleus.

  • ${}_{Z}^{A} X \rightarrow {}_{Z-2}^{A-4} Y + {}_{2}^{4} He$

  • Least penetration; stopped by paper.

• Beta Decay ($\beta$):

  • $\beta^-$: Neutron turns to Proton; emits electron (${}_{-1}^{0} e$) and antineutrino ($\bar{\nu}$).

  • $\beta^+$: Proton turns to Neutron; emits positron (${}_{+1}^{0} e$) and neutrino ($\nu$).

  • Medium penetration; stopped by $6\,mm$ aluminum.

• Gamma Decay ($\gamma$): Emission of high-energy photons.

  • ${}_{Z}^{A} X^* \rightarrow {}_{Z}^{A} Y + {}_{0}^{0} \gamma$

  • Highest penetration; stopped by several cm of lead.

• Conservation Laws: In any nuclear equation, Atomic Number ($Z$) and Mass Number ($A$) must be conserved.

Half-Life and Nuclear Energy

• Half-Life ($T_{1/2}$): Time required for half of the atoms in a sample to decay.

  • $Remaining = Original \times (\frac{1}{2})^t$

  • $t =$ number of half-lives passed.

• Nuclear Fission: The division of a heavy nucleus (like $U-235$) into two or more smaller fragments plus neutrons and energy ($e.g., \approx 173\,MeV$).

• Nuclear Fusion: The combination of small masses to form a larger nucleus (e.g., Proton-Proton chain in stars). Releases energy due to mass loss as the new nucleus is more tightly bound.

Questions & Discussion

• Q: Why don't the protons cause the nucleus to fly apart?

  • A: The strong nuclear force provides a powerful attraction that overcomes the electric repulsion between protons at very short ranges.

• Q: Why is the case $n=2.9$ unstable in the Bohr orbit?

  • A: The wave does not form a closed standing wave; it interferes destructively and becomes unstable. The circumference must fit exactly a whole number multiple of wavelengths ($2\pi r = n\lambda$).

• Q: How is the composition of stars determined?

  • A: By comparing the absorption spectra of stars with known emission spectra of chemical elements.

• Q: What quantities are conserved in a nuclear equation?

  • A: Atomic number (to conserve charge) and Mass number (to conserve number of nucleons).