Engineering Physics: Comprehensive Bullet-Point Notes Compton Scattering / Compton Effect Physical processCollision between a high-energy photon (typically X-ray or γ-ray) and a loosely bound / free electron. Treated as an elastic collision obeying conservation of energy and momentum. Key relationsWavelength shift: Δ λ = λ ′ − λ = λ c ( 1 − cos θ ) \Delta \lambda = \lambda' - \lambda = \lambda_c (1 - \cos \theta) Δ λ = λ ′ − λ = λ c ( 1 − cos θ ) Compton wavelength of the electron: λ < e m > c = h m < / e m > e c = 0.00243 nm = 0.0243 A ˚ \lambda<em>c = \frac{h}{m</em>e c} = 0.00243\,\text{nm} = 0.0243\,\text{Å} λ < e m > c = m < / e m > ec h = 0.00243 nm = 0.0243 A ˚ Energy of scattered photon: E ′ = h c λ ′ E' = \frac{hc}{\lambda'} E ′ = λ ′ h c Recoil electron kinetic energy: K e = E − E ′ K_e = E - E' K e = E − E ′ Momentum relations (1-D form shown for clarity):p < e m > γ = h λ , p < / e m > e = 2 m < e m > e K < / e m > e p<em>\gamma = \frac{h}{\lambda}, \; p</em>{e} = \sqrt{2 m<em>e K</em>e} p < e m > γ = λ h , p < / e m > e = 2 m < e m > eK < / e m > e SignificanceDemonstrated particle-like properties of light; could not be explained by classical wave theory. Confirmed relativistic energy–momentum relation for electrons. Example calculation (asked in paper)Incident X-ray wavelength λ = 1.01 A ˚ \lambda = 1.01 \,\text{Å} λ = 1.01 A ˚ , scattering angle θ = 30 ∘ \theta = 30^{\circ} θ = 3 0 ∘ . Shift: Δ λ = 0.0243 ( 1 − cos 30 ∘ ) = 0.0243 ( 1 − 0.8660 ) ≈ 0.00326 A ˚ \Delta \lambda = 0.0243(1-\cos 30^{\circ}) = 0.0243(1-0.8660) \approx 0.00326\,\text{Å} Δ λ = 0.0243 ( 1 − cos 3 0 ∘ ) = 0.0243 ( 1 − 0.8660 ) ≈ 0.00326 A ˚ Scattered wavelength: λ ′ = 1.01 + 0.00326 ≈ 1.0133 A ˚ \lambda' = 1.01 + 0.00326 \approx 1.0133\,\text{Å} λ ′ = 1.01 + 0.00326 ≈ 1.0133 A ˚ Note the very small but measurable increase due to low angle. Experimental featuresUse of graphite / beryllium targets; detection via Bragg diffraction. Peak positions shift with angle, verifying the 1 − cos θ 1-\cos\theta 1 − cos θ dependence. Newton’s Rings (Interference by Reflected Light) Formation mechanismPlano-convex lens on flat glass plate ⇒ air film of varying thickness. Monochromatic light → partial reflection at lens–air and air–plate interfaces. Circular interference fringes (alternate bright/dark rings) centered at point of contact. Diameter of the n th n^{\text{th}} n th dark ringFor reflected light: D n 2 = 4 n λ R D_n^2 = 4 n \lambda R D n 2 = 4 nλ R D n D_n D n = diameter, n n n = ring order (starting at n = 1 n=1 n = 1 ), λ \lambda λ = wavelength, R R R = radius of curvature of lens. Determining wavelength or R R R From two diameters: λ R = D < e m > n < / e m > 2 2 − D < e m > n < / e m > 1 2 4 ( n < e m > 2 − n < / e m > 1 ) \lambda R = \frac{D<em>{n</em>2}^2 - D<em>{n</em>1}^2}{4(n<em>2-n</em>1)} λ R = 4 ( n < e m > 2 − n < / e m > 1 ) D < e m > n < / e m > 2 2 − D < e m > n < / e m > 1 2 . Sample numerical problem (paper data)D < e m > 4 = 0.400 cm , D < / e m > 12 = 0.700 cm D<em>4 = 0.400\,\text{cm},\; D</em>{12} = 0.700\,\text{cm} D < e m > 4 = 0.400 cm , D < / e m > 12 = 0.700 cm .Constant k = D < e m > n 2 n k = \frac{D<em>n^2}{n} k = n D < e m > n 2 ⇒
k < / e m > 4 = 0.400 2 4 = 0.0400 k</em>4 = \frac{0.400^2}{4}=0.0400 k < / e m > 4 = 4 0.40 0 2 = 0.0400 ,k 12 = 0.700 2 12 = 0.0408 k_{12}=\frac{0.700^2}{12}=0.0408 k 12 = 12 0.70 0 2 = 0.0408 ⇒ average k ≈ 0.0404 cm 2 k \approx 0.0404\,\text{cm}^2 k ≈ 0.0404 cm 2 . D < e m > 20 2 = 20 × 0.0404 = 0.808 D<em>{20}^2 = 20 \times 0.0404 = 0.808 D < e m > 20 2 = 20 × 0.0404 = 0.808 ⇒ D < / e m > 20 ≈ 0.90 cm D</em>{20} \approx 0.90\,\text{cm} D < / e m > 20 ≈ 0.90 cm . Applications & remarksPrecise measurement of λ \lambda λ , lens curvature, refractive index of liquids. Destructive vs constructive conditions differ for reflected vs transmitted light (phase reversal). Wave Function ((\psi)) & Its Properties Interpretation∣ ψ ( r , t ) ∣ 2 |\psi(\mathbf r, t)|^2 ∣ ψ ( r , t ) ∣ 2 gives probability density of finding the particle at position r \mathbf r r at time t t t . Essential propertiesSingle-valued everywhere: avoids ambiguous probabilities. Finite everywhere: probability cannot be infinite. Normalizable: ∫ all space ∣ ψ ∣ 2 d V = 1 \int_{\text{all space}} |\psi|^2 \, dV = 1 ∫ all space ∣ ψ ∣ 2 d V = 1 . Continuous along with its first spatial derivative (ensures finite kinetic energy). Must satisfy appropriate boundary conditions (e.g., zero at infinite potential walls). Physical observables derived through operators acting on ψ \psi ψ (e.g., p ^ = − i ℏ ∇ \hat p = -i\hbar \nabla p ^ = − i ℏ∇ ). Time-Independent Schrödinger Equation (Derivation Sketch) Start with classical energy conservation of a non-relativistic particle in potential V ( r ) V(\mathbf r) V ( r ) :E = p 2 2 m + V E = \frac{p^2}{2m} + V E = 2 m p 2 + V . Replace dynamical variables by quantum operators:E ^ → i ℏ ∂ ∂ t \hat E \to i\hbar \frac{\partial}{\partial t} E ^ → i ℏ ∂ t ∂ ,p ^ → − i ℏ ∇ \hat p \to -i\hbar \nabla p ^ → − i ℏ∇ . Apply to total wave function Ψ ( r , t ) = ψ ( r ) e − i E t / ℏ \Psi(\mathbf r, t)=\psi(\mathbf r) e^{-iEt/\hbar} Ψ ( r , t ) = ψ ( r ) e − i Et /ℏ (separation of variables) ⇒i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∇ 2 Ψ + V Ψ i\hbar \frac{\partial \Psi}{\partial t}= -\frac{\hbar^2}{2m} \nabla^2 \Psi + V\Psi i ℏ ∂ t ∂ Ψ = − 2 m ℏ 2 ∇ 2 Ψ + V Ψ . Cancelling the common exponential factor yields the time-independent form:− ℏ 2 2 m ∇ 2 ψ + V ψ = E ψ -\frac{\hbar^2}{2m} \nabla^2 \psi + V\psi = E\psi − 2 m ℏ 2 ∇ 2 ψ + V ψ = E ψ . ImportanceCentral equation for bound-state problems (atoms, wells, oscillators). Eigenvalue equation; discrete E E E for bound systems, continuous for free particles. Heisenberg Uncertainty Principle (HUP) Fundamental statementΔ x Δ p x ≥ ℏ 2 , Δ E Δ t ≥ ℏ 2 \Delta x \; \Delta p_x \ge \frac{\hbar}{2}, \qquad \Delta E \; \Delta t \ge \frac{\hbar}{2} Δ x Δ p x ≥ 2 ℏ , Δ E Δ t ≥ 2 ℏ . Elementary proof (Fourier approach)A wave packet requires spread of wave numbers Δ k \Delta k Δ k to localize in space Δ x \Delta x Δ x . Relation Δ x Δ k ≥ 1 2 \Delta x \; \Delta k \ge \frac{1}{2} Δ x Δ k ≥ 2 1 from Fourier transforms. With p = ℏ k p = \hbar k p = ℏ k ⇒ Δ p = ℏ Δ k \Delta p = \hbar \Delta k Δ p = ℏΔ k leads to HUP. Application: electron inside nucleusNuclear radius ≈ 1 fm = 10 − 15 m \approx 1\,\text{fm}=10^{-15}\,\text{m} ≈ 1 fm = 1 0 − 15 m ⇒ Δ x ≈ 10 − 15 m \Delta x \approx 10^{-15}\,\text{m} Δ x ≈ 1 0 − 15 m . \Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.055\times10^{-34}}{2\times10^{-15}} \approx 5.3\times10^{-20}\,\text{kg·m·s}^{-1}. Minimum kinetic energy: K < e m > min = ( Δ p ) 2 2 m < / e m > e ≈ ( 5.3 × 10 − 20 ) 2 2 × 9.11 × 10 − 31 ≈ 1.5 × 10 − 9 J ≈ 9.5 MeV K<em>{\min}=\frac{(\Delta p)^2}{2m</em>e} \approx \frac{(5.3\times10^{-20})^2}{2\times9.11\times10^{-31}} \approx 1.5\times10^{-9}\,\text{J} \approx 9.5\,\text{MeV} K < e m > min = 2 m < / e m > e ( Δ p ) 2 ≈ 2 × 9.11 × 1 0 − 31 ( 5.3 × 1 0 − 20 ) 2 ≈ 1.5 × 1 0 − 9 J ≈ 9.5 MeV . Far exceeds typical nuclear binding (~MeV scale) & β-decay energies ⇒ free electron cannot be confined in nucleus. Hall Effect & Hall Coefficient PhenomenonCharge carriers in a conductor/semiconductor carrying current I I I experience magnetic force in field B \mathbf B B , producing transverse electric field E y E_y E y (Hall voltage). Force balance for steady state: q v < e m > d B = q E < / e m > y q v<em>d B = q E</em>y q v < e m > d B = qE < / e m > y . Current density: J < e m > x = n q v < / e m > d J<em>x = n q v</em>d J < e m > x = n q v < / e m > d . Hall coefficient derivation
R < e m > H = E < / e m > y J x B = 1 n q R<em>H = \frac{E</em>y}{J_x B} = \frac{1}{n q} R < e m > H = J x B E < / e m > y = n q 1 .For electrons (negative carriers): q = − e q=-e q = − e ⇒ R H = − 1 n e R_H=-\frac{1}{n e} R H = − n e 1 . For holes (positive carriers): q = + e q=+e q = + e ⇒ R H = + 1 p e R_H=+\frac{1}{p e} R H = + p e 1 . UsesSign of R H R_H R H immediately distinguishes n-type (negative) vs p-type (positive) semiconductors. Determines carrier concentration, mobility (with conductivity data). Magnetic field sensing, Hall probes. Fresnel’s Biprism – Interference of Light (Short Note) Biprism: single prism with very small refracting angle (~1°) producing two virtual coherent sources from one slit. Fringe width: β = D λ 2 d \beta = \frac{D \lambda}{2d} β = 2 d D λ ((D) = distance to screen, (d) = source separation). High fringe contrast due to common origin; enables laboratory measurement of λ \lambda λ of sodium light etc. Nuclear Liquid Drop Model (Short Note) Treats nucleus as incompressible droplet of nuclear fluid with surface tension. Semi-empirical mass formula (Weizsäcker):B = a < e m > v A − a < / e m > s A 2 / 3 − a < e m > c Z ( Z − 1 ) A 1 / 3 − a < / e m > a ( A − 2 Z ) 2 A ± a p A − 1 / 2 B = a<em>v A - a</em>s A^{2/3} - a<em>c \frac{Z(Z-1)}{A^{1/3}} - a</em>a \frac{(A-2Z)^2}{A} \pm a_p A^{-1/2} B = a < e m > v A − a < / e m > s A 2/3 − a < e m > c A 1/3 Z ( Z − 1 ) − a < / e m > a A ( A − 2 Z ) 2 ± a p A − 1/2 . ExplainsBinding-energy systematics, fission (surface tension vs Coulomb repulsion), approximate constant density. Cannot account for magic numbers ⇒ supplemented by shell model. Geiger–Müller Counter (Short Note) Gas-filled radiation detector operating in Geiger region of ionization curve. Principals parts: cylindrical cathode, axial anode wire, fill gas + quench gas. Ionizing particle → avalanche → large output pulse independent of primary ionization ⇒ simple counting. Plateau characteristics, dead time, quenching (halogen or organic) to suppress continuous discharge. Meissner Effect (Short Note) Below critical temperature T c T_c T c , superconductors expel magnetic flux: B = 0 \mathbf B = 0 B = 0 inside bulk. Perfect diamagnetism ((\chi = -1)). Distinguishes superconductivity from mere perfect conductivity; requires modification of Maxwell equations (London equations). Practical applications: magnetic levitation, MRI stability. Kronig–Penney Model (Short Note) One-dimensional periodic square-well potential used to explain electronic band structure in crystals. Dispersion relation:cos k a = cos α a + P α a sin α a \cos k a = \cos \alpha a + \frac{P}{\alpha a} \sin \alpha a cos ka = cos α a + α a P sin α a , where α 2 = 2 m E ℏ 2 \alpha^2 = \frac{2mE}{\hbar^2} α 2 = ℏ 2 2 m E , P P P = potential strength. Yields allowed (|rhs|≤1) & forbidden energy bands; origin of band gaps. Introduces concept of Brillouin zones, effective mass.