⚛ Physics · Optics & Modern Physics 🔥 Day 1
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Optics & Modern Physics — Physics

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Ray Optics
Reflection · Refraction · TIR · Lenses · Prism · Optical Instruments

Reflection — Spherical Mirrors

Sign convention: distances measured from pole; along incident ray direction = +ve.

Mirror formula: \(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} = \dfrac{2}{R}\)

Magnification: \(m = -\dfrac{v}{u} = h_i/h_o\)

Object position (concave)Image natureImage position
Beyond C (u > 2f)Real, inverted, diminishedBetween f and C
At C (u = 2f)Real, inverted, same sizeAt C
Between C and fReal, inverted, magnifiedBeyond C
At fReal, at infinity∞
Inside f (u < f)Virtual, erect, magnifiedBehind mirror

Refraction — Snell's Law & TIR

Snell's law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)

Total Internal Reflection occurs when: (1) light travels from denser to rarer medium, and (2) angle of incidence > critical angle C.

Critical angle: \(\sin C = \dfrac{n_2}{n_1}\) (\(n_1 > n_2\)). For glass–air: \(\sin C = \dfrac{1}{n_{glass}}\).

Optical fibre: uses TIR; light propagates with negligible loss through total internal reflection at core–cladding interface (\(n_{core} > n_{cladding}\)).

Refraction at Spherical Surface

$$ \frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R} $$

Here \(u\) = object distance, \(v\) = image distance, \(R\) = radius of curvature (sign convention applies).

Thin Lenses

Lens formula: \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\)

Lens maker's equation: \(\dfrac{1}{f} = (n-1)[\dfrac{1}{R_1} - \dfrac{1}{R_2}]\)

Power: \(P = \dfrac{1}{f}\) (in metres) — unit is dioptre (D)

Combination: \(P = P_1 + P_2\) (lenses in contact); \(\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2}\)

Magnification: \(m = \dfrac{v}{u}\)

Prism

Angle of deviation \(\delta = i + e - A\) (\(A\) = prism angle, \(i\) = incidence angle, \(e\) = exit angle)

At minimum deviation: \(r_1 = r_2 = \dfrac{A}{2}\); \(i = e\)

\(n = \sin[\dfrac{A + \delta_{min}}{2}] / \sin(\dfrac{A}{2})\)

Dispersion: angular dispersion = \(\delta_V - \delta_R\); dispersive power \(\omega = \dfrac{n_V - n_R}{n_Y - 1}\)

Optical Instruments

InstrumentMagnification
Simple microscope\(m = 1 + \dfrac{D}{f}\) (normal adjustment); \(m = \dfrac{D}{f}\) (far point)
Compound microscope\(M = m_o \times m_e = (\dfrac{L}{f_o})(1 + \dfrac{D}{f_e})\)
Astronomical telescope\(M = -\dfrac{f_o}{f_e}\) (normal adjustment)
Near-sighted correctionConcave (diverging) lens, \(P = -\dfrac{1}{d}\) (in m)
Far-sighted correctionConvex (converging) lens
💡 JEE Tip: Mirror formula and lens formula look similar but differ — mirror uses 1/v + 1/u = 1/f while lens uses 1/v − 1/u = 1/f. The sign in the middle changes. Apply sign convention consistently: real object → u is always negative.
⚠ Common Error: Power of a lens P = 1/f requires f in metres. If f = 20 cm = 0.20 m, then P = +5 D. Converting f in cm directly gives wrong answers — always convert to metres first.
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Ray Optics — Formula Sheet
Mirrors & Refraction
Mirror Formula
1/v + 1/u = 1/f = 2/R
m = −v/u = h_i/h_o
Snell's Law & TIR
$$ n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2} $$
Critical angle: sin C = n₂/n₁
Thin Lens
\(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\)
$$ \frac{1}{f} = (n-1)[\frac{1}{R_{1}} - \frac{1}{R_{2}}] $$  |  \(P = \dfrac{1}{f}\) (m); \(P_{total} = P_1 + P_2\)
Prism at δ_min
\(n = \sin[\dfrac{A+\delta_{min}}{2}] / \sin(\dfrac{A}{2})\)
At \(\delta_{min}\): \(r = \dfrac{A}{2}\), \(i = e\)
Telescope & Microscope
Telescope: \(M = \dfrac{f_o}{f_e}\)
Compound microscope: \(M = (\dfrac{L}{f_o})(1+\dfrac{D}{f_e})\)  |  \(D = 25\) cm (least distance of distinct vision)
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Ray Optics — Solved Examples
Example 1
Example 1: A concave mirror has focal length 15 cm. An object is placed 10 cm in front of it. Find the image distance and state its nature.
  1. Using sign convention: f = −15 cm, u = −10 cm
  2. 1/v = 1/f − 1/u = 1/(−15) − 1/(−10) = −1/15 + 1/10 = (−2+3)/30 = 1/30
  3. v = +30 cm
  4. Image is at +30 cm (behind mirror) → virtual, erect, magnified
  5. m = −v/u = −(30)/(−10) = +3 (magnified 3×, erect)
Example 2
Example 2: A ray of light travels from glass (n = 1.5) to air. Find the critical angle.
  1. \(\sin C = \dfrac{n_{air}}{n_{glass}} = \dfrac{1}{1.5} = \dfrac{2}{3}\)
  2. \(C = \sin^{-1}(\dfrac{2}{3}) \approx\) 41.8°
  3. For angles of incidence > 41.8°, TIR occurs and no refracted ray exits into air.
Example 3
Example 3: Two thin lenses of focal lengths +20 cm and −30 cm are placed in contact. Find the power of the combination.
  1. \(P_1 = \dfrac{1}{0.20} = +5\) D; \(P_2 = \dfrac{1}{-0.30} = -3.33\) D
  2. \(P = P_1 + P_2 = 5 - 3.33 =\) +1.67 D
  3. Equivalent focal length \(f = \dfrac{1}{P} = \dfrac{1}{1.67} \approx 60\) cm (converging)
Example 4
Example 4: A glass prism (A = 60°) produces minimum deviation of 40°. Find the refractive index.
  1. \(n = \sin[\dfrac{A + \delta_{min}}{2}] / \sin(\dfrac{A}{2}) = \sin[\dfrac{60° + 40°}{2}] / \sin(30°)\)
  2. \(= \sin(50°) / \sin(30°) = 0.766 / 0.500 =\) 1.532
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Ray Optics — Diagrams
Click any diagram to zoom
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Ray Optics — Quick Quiz
Question 1
An object is 30 cm from a concave mirror of focal length 10 cm. The image distance is:
f = −10 cm, u = −30 cm. 1/v = 1/f − 1/u = −1/10 + 1/30 = (−3+1)/30 = −2/30. v = −15 cm. Real, inverted image between F and C.
Question 2
For total internal reflection to occur, light must travel from:
TIR requires TWO conditions: (1) light going from denser to rarer medium, and (2) angle of incidence > critical angle. Both must be satisfied simultaneously.
Question 3
Two lenses of power +5 D and −2 D are placed in contact. The focal length of the combination is:
P = P₁ + P₂ = 5 + (−2) = +3 D. f = 1/P = 1/3 m = 100/3 cm ≈ 33.3 cm. The combination is converging.
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Wave Optics
Huygens Principle · YDSE · Diffraction · Polarisation

Huygens Principle

Every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront is the tangent envelope of these secondary wavelets. Explains reflection, refraction, diffraction, and interference.

Young's Double Slit Experiment (YDSE)

Two coherent sources S₁ and S₂ (slit separation d) illuminate a screen at distance D.

Path difference\(: \Delta = d \sin\theta \approx \dfrac{dy}{D}\) (for small θ, where y = distance from centre)

ConditionFringe type
\(\Delta = n\lambda (n = 0,\pm1,\pm2\ldots)\)Bright fringe (constructive)
\(\Delta = (2n+1)\dfrac{\lambda}{2}\)Dark fringe (destructive)

Fringe width\(: \beta = \lambda \dfrac{D}{d}\) (same for bright and dark fringes)

Intensity\(: I = 4I_{0} \cos^{2}(\dfrac{\delta}{2}),\) where \(\delta = 2\pi\dfrac{\Delta}{\lambda}\) is the phase difference

Maximum intensity\(: I_{max} = 4I_{0}\) (constructive) Minimum: I_min = 0 (for equal amplitudes)

If intensities differ: \(I_{max} = (\sqrt I_{1} + \sqrt I_{2})^{2}, I_{min} = (\sqrt I_{1} - \sqrt I_{2})^{2}\)

Diffraction

Single slit diffraction: slit width a, wavelength λ

  • Minima at: \(a\sin\theta = n\lambda\) (\(n = \pm 1, \pm 2, \ldots\))
  • Central maximum width = \(2\dfrac{\lambda}{a}\) (angular) or \(2\lambda \dfrac{D}{a}\) (linear on screen)
  • Secondary maxima at: \(a\sin\theta = (2n+1)\dfrac{\lambda}{2}\) — much weaker than central maximum

Rayleigh criterion (resolving power): two objects just resolved when central max of one falls on first minimum of the other. \(\theta_{min} = 1.22\dfrac{\lambda}{D}\) (\(D\) = aperture diameter)

Polarisation

Light is a transverse wave — oscillations are perpendicular to propagation direction.

  • Unpolarised light: random orientation of electric field vectors
  • Plane polarised light: E-field oscillates in one plane only
  • Malus's law: \(I = I_0 \cos^2\theta\) (\(\theta\) = angle between polariser and analyser axes)
  • Brewster's law: \(\tan\theta_B = n\) (\(\theta_B\) = Brewster's angle; reflected ray is fully polarised when refracted and reflected rays are perpendicular)
  • After passing through ideal polariser, intensity of unpolarised light becomes \(\dfrac{I_0}{2}\)
💡 JEE Tip: In YDSE, fringe width \(\beta = \lambda \dfrac{D}{d}\). If the whole apparatus is immersed in a medium of refractive index n, the wavelength becomes λ/n → fringe width decreases by factor n. Also: shifting a slab of thickness t and refractive index n in front of one slit shifts the entire pattern by (n−1)t·D/d.
⚠ Common Error: Single-slit diffraction minima are at \(a\sin\theta = n\lambda\) (same form as YDSE maxima). Don't confuse them. In single slit, the centre is always a maximum; in YDSE the central fringe is also bright (\(n=0\) constructive). The difference: diffraction envelope modulates YDSE fringes in real double-slit experiments.
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Wave Optics — Formula Sheet
YDSE Fringe Width
\(\beta = \lambda \dfrac{D}{d}\)
Path diff \(\Delta = \dfrac{dy}{D}\)  |  Bright: \(\Delta = n\lambda\); Dark: \(\Delta = (2n+1)\dfrac{\lambda}{2}\)
YDSE Intensity
$$ I = 4I_{0} \cos^{2}(\frac{\delta}{2}) $$
δ = 2πΔ/λ  |  $$ I_{max} = 4I_{0}, I_{min} = 0 $$
Single Slit Diffraction
Minima: a sinθ = nλ
Central max width = 2λD/a  |  Rayleigh: \(\theta = 1.22\dfrac{\lambda}{D}\)
Polarisation
Malus's law: \(I = I_{0} \cos^{2}\theta\)
Brewster's: \(\tan \theta_B = n\)  |  After polariser: \(I = \dfrac{I_{0}}{2}\) (unpolarised)
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Wave Optics — Solved Examples
Example 1
Example 1: In YDSE, slit separation \(d = 0.5\) mm, screen distance \(D = 1\) m, \(\lambda = 600\) nm. Find the fringe width.
  1. $$ \beta = \lambda \frac{D}{d} = (600\times10^{-9} \times 1) / (0.5\times10^{-3}) $$
  2. $$ = 600\times10^{-9} / 5\times10^{-4} = $$ $$ 1.2 \times 10^{-3} m = 1.2 mm $$
Example 2
Example 2: In YDSE, the 4th bright fringe is at y = 2.4 mm from centre. Find the wavelength if D = 1 m, d = 1 mm.
  1. For nth bright fringe: \(y_n = n\lambda \dfrac{D}{d}\)
  2. $$ \lambda = y_n\cdot d / (nD) = (2.4\times10^{-3} \times 1\times10^{-3}) / (4 \times 1) $$
  3. \(= 2.4\times10^{-6} / 4 =\) 600 nm
Example 3
Example 3: Polarised light of intensity 80 W/m² passes through an analyser. The transmission axis makes 60° with the polarisation direction. Find the transmitted intensity.
  1. Malus's law: \(I = I_{0} \cos^{2}\theta = 80 \times \cos^{2}(60^{\circ})\)
  2. \(= 80 \times (0.5)^{2} = 80 \times 0.25 =\) 20 W/m²
Example 4
Example 4: A glass surface (n = 1.732) is illuminated. Find the Brewster's angle.
  1. $$ \tan \theta_B = n = 1.732 $$
  2. \(\theta_B = \tan^{-1}(1.732) =\) 60°
  3. At this angle, reflected light is completely plane-polarised perpendicular to plane of incidence.
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Wave Optics — Diagrams
Click any diagram to zoom
⚡
Wave Optics — Quick Quiz
Question 1
In YDSE, if the slit separation is halved and screen distance is doubled, the fringe width becomes:
\(\beta = \lambda \dfrac{D}{d}\). New \(\beta' = \dfrac{\lambda(2D)}{d/2} = 4\lambda \dfrac{D}{d} = 4\beta\). Fringe width becomes four times the original.
Question 2
The first minimum in single slit diffraction (slit width a) occurs when:
Single slit minima: \(a\sin\theta = n\lambda\). For first minimum (\(n=1\)): \(a\sin\theta = \lambda\). This is analogous to YDSE maxima formula but here it gives dark fringes.
Question 3
Unpolarised light of intensity I₀ passes through two polaroids with axes at 60°. Final intensity is:
After first polaroid: \(I_1 = \dfrac{I_0}{2}\). Malus's law through second: \(I_2 = I_1\cos^2(60°) = (\dfrac{I_0}{2})(\dfrac{1}{4}) = \dfrac{I_0}{8}\).
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Modern Physics
Photoelectric Effect · de Broglie · Heisenberg · X-rays · Radioactivity

Photoelectric Effect

When light of frequency ν (above threshold ν₀) falls on a metal surface, electrons are emitted.

Einstein's equation\(: KE_{max} = h\nu - \phi = h(\nu - \nu_{0})\)

  • Work function: \(\phi = h\nu_{0}\) (minimum energy to eject electron)
  • Threshold wavelength: \(\lambda_{0} = \dfrac{hc}{\phi}\)
  • Stopping potential: \(eV_{0} = KE_{max} \to V_{0} = \dfrac{h(\nu-\nu_{0})}{e}\)
  • Number of photoelectrons ∝ intensity of light (not frequency)
  • \(KE_{max}\) depends on frequency only, not intensity

de Broglie Wavelength

Every particle has an associated matter wave. \(\lambda = \dfrac{h}{p} = \dfrac{h}{mv}\)

  • For particle accelerated through potential V: \(\lambda = \dfrac{h}{\sqrt{2mqV}}\)
  • For electron: \(\lambda = \dfrac{12.27}{\sqrt V} Å\) (V in volts)
  • Thermal neutrons at temperature T: \(\lambda = \dfrac{h}{\sqrt{3mkT}}\)
  • Larger momentum → shorter wavelength

Heisenberg Uncertainty Principle

Δx · Δp ≥ h/4π    and    ΔE · Δt ≥ h/4π

It is impossible to simultaneously determine exact position and momentum of a particle. This is not a measurement limitation but a fundamental property of quantum mechanics.

X-Rays

  • Production: produced when fast electrons decelerate hitting a target (bremsstrahlung) or eject inner shell electrons (characteristic X-rays)
  • Moseley's law: \(\sqrt\nu = a(Z - b)\), where \(a\) and \(b\) are constants — establishes atomic number \(Z\)
  • Bragg's law: \(2d \sin\theta = n\lambda\) — used for X-ray diffraction by crystal planes (d = interplanar spacing)
  • Minimum wavelength of X-rays: \(\lambda_{min} = \dfrac{hc}{eV}\) (V = accelerating voltage)

Radioactivity

Decay law: \(N = N_{0} e^{-\lambda t}\)    \(A = \lambda N = A_0 e^{-\lambda t}\)

Half-life\(: T_{1}/_{2} = \dfrac{\ln2}{\lambda} = \dfrac{0.693}{\lambda}\)

Mean life\(: \tau = \dfrac{1}{\lambda} = T_{1}/_{2}/0.693\)

Decay typeChange in AChange in ZChange in N
α-decay (⁴He)−4−2−2
β⁻-decay (e⁻)0+1−1
β⁺-decay (e⁺)0−1+1
γ-decay (photon)000

Nuclear Binding Energy

Mass defect: \(\Delta m = Z m_p + N m_n - M_{nucleus}\)

Binding energy: \(BE = \Delta m\cdot c^2\) (in MeV: \(BE = \Delta m \times 931.5\) MeV/amu)

BE/A (binding energy per nucleon) peaks near \(A \approx 56\) (iron-56 most stable).

Fission (heavy nuclei split → increase in BE/A) and fusion (light nuclei merge → increase in BE/A) both release energy because products are more tightly bound.

💡 JEE Tip: After n half-lives, fraction remaining = (1/2)ⁿ and fraction decayed = 1 − (1/2)ⁿ. For example, after 3 half-lives: remaining = 1/8, decayed = 7/8. Activity follows the same law as N: \(A = A_0(\dfrac{1}{2})^{t/T_{1/2}}\).
⚠ Common Error: Mean life \(\tau = \dfrac{1}{\lambda} \ne T_{1/2}\). Mean life is larger: \(\tau = T_{1/2}/\ln 2 = \dfrac{T_{1/2}}{0.693} \approx 1.44\, T_{1/2}\). The mean life is the time after which only \(\dfrac{1}{e}\) of the original sample remains.
📐
Modern Physics — Formula Sheet
Photoelectric Effect
$$ KE_{max} = h\nu - \phi $$
$$ eV_{0} = h\nu - h\nu_{0} $$  |  $$ \lambda_{0} = \frac{hc}{\phi} $$
de Broglie Wavelength
$$ \lambda = \frac{h}{p} = \frac{h}{mv} $$
$$ \lambda = \frac{h}{\sqrt{2mqV}} $$  |  For electron: \(\lambda = \dfrac{12.27}{\sqrt V} Å\)
Radioactive Decay
\(N = N_0 e^{-\lambda t}\)
$$ T_{1/2} = \frac{0.693}{\lambda} $$  |  $$ \tau = \frac{1}{\lambda} = 1.44\, T_{1/2} $$
Binding Energy
Δm = Zm_p + Nm_n − M
BE = Δm · c²  |  $$ = \Delta m(amu) \times 931.5 MeV $$
Bragg's Law & X-rays
2d sinθ = nλ
$$ \lambda_{min} = \frac{hc}{eV} $$  |  Moseley: √ν = a(Z−b)
🧮
Modern Physics — Solved Examples
Example 1
Example 1: The work function of a metal is 2 eV. Light of frequency \(8\times 10^{14}\) Hz falls on it. Find the stopping potential. (\(h = 6.63\times 10^{-34}\) J·s)
  1. $$ KE_{max} = h\nu - \phi = (6.63\times10^{-34} \times 8\times10^{14}) / (1.6\times10^{-19}) - 2 eV $$
  2. $$ = (5.3\times10^{-19} / 1.6\times10^{-19}) - 2 = 3.31 - 2 = 1.31 eV $$
  3. Stopping potential \(V_0 = \dfrac{KE_{max}}{e} =\) 1.31 V
Example 2
Example 2: Find the de Broglie wavelength of an electron accelerated through 100 V.
  1. \(\lambda = \dfrac{12.27}{\sqrt V} Å = \dfrac{12.27}{\sqrt{100}} = \dfrac{12.27}{10} =\) 1.227 Å
  2. This is comparable to X-ray wavelengths and atomic spacings — hence electron diffraction is possible.
Example 3
Example 3: A radioactive sample has T₁/₂ = 20 days. What fraction remains after 60 days?
  1. Number of half-lives \(n = \dfrac{60}{20} = 3\)
  2. Fraction remaining = \((\dfrac{1}{2})^3\) = 1/8 = 12.5%
  3. Activity also reduces to \(\dfrac{A_0}{8}\) after 3 half-lives.
Example 4
Example 4: The mass defect of ⁴He is 0.0304 amu. Find the binding energy per nucleon. (1 amu = 931.5 MeV)
  1. Total BE = 0.0304 × 931.5 = 28.3 MeV
  2. A = 4 (helium-4 has 4 nucleons)
  3. BE/A = 28.3/4 = 7.07 MeV/nucleon
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Modern Physics — Diagrams
Click any diagram to zoom
⚡
Modern Physics — Quick Quiz
Question 1
In the photoelectric effect, increasing light intensity (frequency fixed above threshold) causes:
KE_max = hν − φ depends only on frequency, not intensity. Intensity determines number of photons → number of ejected electrons. Stopping potential depends on KE_max, so it's also unaffected by intensity. Only photocurrent (number of electrons/sec) increases with intensity.
Question 2
After 3 half-lives, what fraction of a radioactive sample has decayed?
After 3 half-lives: fraction remaining = (1/2)³ = 1/8. Fraction decayed = 1 − 1/8 = 7/8.
Question 3
The de Broglie wavelength of a proton and an electron are equal. Then:
λ = h/p. Equal λ means equal momentum p = mv. Since masses differ, their velocities differ (v = p/m). KE = p²/2m differs because masses differ. Only momentum is equal.
☢
Atoms & Nuclei
Bohr Model · Hydrogen Spectrum · Nuclear Structure · Nuclear Reactions

Rutherford's Model

Gold foil experiment (1911): most α particles passed through; a few deflected at large angles; rare ones bounced back. Conclusion: atom has a tiny, dense, positively charged nucleus.

Limitations: orbiting electrons should radiate (Maxwell) → lose energy → spiral into nucleus. Cannot explain discrete spectral lines.

Bohr's Model of Hydrogen

Postulates: (1) Electrons move in fixed circular orbits without radiating. (2) Angular momentum is quantised: mvr = nh/2π. (3) Photon emitted/absorbed when electron transitions: hν = E_i − E_f.

QuantityFormulan=1 (H)
Radius\(r_n = \dfrac{0.529n^{2}}{Z} Å\)0.529 Å (Bohr radius a₀)
Energy\(E_n = -\dfrac{13.6Z^{2}}{n^{2}} eV\)−13.6 eV
Velocity\(v_n = 2.18\times10^{6} \dfrac{Z}{n} m/s\)2.18×10⁶ m/s

For hydrogen-like ions (He⁺, Li²⁺…): replace Z accordingly. E.g., He⁺ (Z=2): E₁ = −13.6×4 = −54.4 eV.

Hydrogen Spectral Series

Wave number: \(\dfrac{1}{\lambda} = RZ^{2}(\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}})\), where \(R = 1.097\times10^{7} m^{-1}\) (Rydberg constant)

Seriesn₁n₂Region
Lyman12,3,4…Ultraviolet
Balmer23,4,5…Visible
Paschen34,5,6…Infrared
Brackett45,6,7…Far IR
Pfund56,7,8…Far IR

Number of spectral lines from level n to ground: n(n−1)/2

Nuclear Structure

  • Nucleus contains protons (Z) and neutrons (N); mass number A = Z + N
  • Nuclear radius: \(R = R_0 A^{1/3}\), \(R_{0} = 1.2\ \text{fm}\) (1 fm = 10⁻¹⁵ m)
  • Nuclear density is approximately constant: ρ ≈ 2.3×10¹⁷ kg/m³
  • Nuclear forces: strong, short-range (~1 fm), charge-independent, saturating
  • Isotopes: same Z, different A; Isobars: same A; Isotones: same N
💡 JEE Tip: For the Balmer series first line (Hα): n₂=3→n₁=2. 1/λ = R(1/4 − 1/9) = 5R/36. λ = 36/(5R) = 656 nm (red line). The series limit (n₂→∞) gives the shortest wavelength of the series: λ_limit = n₁²/(RZ²).
⚠ Common Error: Energy of nth level E_n = −13.6/n² eV is negative (bound state). Ionisation energy = |E₁| = 13.6 eV for H. Energy of transition = E_higher − E_lower = positive value = photon energy hν. Don't confuse the sign.
📐
Atoms & Nuclei — Formula Sheet
Bohr Radius & Energy
$$ r_n = 0.529 \frac{n^{2}}{Z} Å $$
$$ E_n = -13.6 \frac{Z^{2}}{n^{2}} eV $$  |  $$ v_n = 2.18\times10^{6} \frac{Z}{n} m/s $$
Rydberg Formula
1/λ = RZ²(1/n₁² − 1/n₂²)
R = 1.097×10⁷ m⁻¹  |  Lines: n(n−1)/2
Nuclear Radius
\(R = R_0 A^{1/3}\)
R₀ = 1.2 fm  |  Nuclear density ≈ const
Quantised Angular Momentum
$$ mvr = \frac{nh}{2}\pi = nℏ $$
n = 1, 2, 3, … (principal quantum number)
🧮
Atoms & Nuclei — Solved Examples
Example 1
Example 1: Calculate the energy of the photon emitted when hydrogen atom transitions from n=3 to n=1.
  1. E₃ = −13.6/9 = −1.51 eV; E₁ = −13.6 eV
  2. $$ \Delta E = E_{1} - E_{3} = -13.6 - (-1.51) = -12.09 eV $$
  3. Photon energy = |ΔE| = 12.09 eV (Lyman series, UV)
Example 2
Example 2: Find the wavelength of the first Balmer line (n=3→n=2) in hydrogen.
  1. $$ \frac{1}{\lambda} = R(\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R(\frac{1}{4} - \frac{1}{9}) = R \times \frac{5}{36} $$
  2. $$ \lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097\times10^{7}} = \frac{36}{5.485}\times10^{7} $$
  3. λ = \(6.56\times10^{-7} m = 656 nm\) (red Hα line)
Example 3
Example 3: Find the radius of the 2nd Bohr orbit in He⁺ (Z=2).
  1. \(r_n = 0.529 \times \dfrac{n^{2}}{Z} = 0.529 \times \dfrac{4}{2} = 0.529 \times 2 =\) 1.058 Å
Example 4
Example 4: An electron in H atom is in n=4 state. How many spectral lines can be emitted as it de-excites?
  1. Number of lines = n(n−1)/2 = 4×3/2 = 6 lines
  2. Transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1
📊
Atoms & Nuclei — Diagrams
Click any diagram to zoom
⚡
Atoms & Nuclei — Quick Quiz
Question 1
The ground state energy of a hydrogen atom is −13.6 eV. The energy of the n=3 level is:
E_n = −13.6/n² eV. For n=3: E₃ = −13.6/9 = −1.51 eV. The energy levels are −13.6, −3.4, −1.51, −0.85, −0.54 eV for n = 1,2,3,4,5.
Question 2
The Balmer series of hydrogen corresponds to transitions to:
Balmer series: transitions end at n₁=2. These produce visible light (Hα=656 nm red, Hβ=486 nm blue-green, etc.). Lyman series ends at n=1 (UV); Paschen at n=3 (IR).
Question 3
Nuclear radius of ²⁷Al (A=27) compared to ⁶⁴Cu (A=64) is:
\(R \propto A^{1/3}\). \(\dfrac{R_{Al}}{R_{Cu}} = (\dfrac{27}{64})^{1/3} = \dfrac{3}{4}\). So Al nucleus is 3/4 the radius of Cu nucleus.
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Semiconductors
Energy Bands · p-n Junction · Diodes · Transistors · Logic Gates

Energy Bands

MaterialBand gap E_gConductivity
Conductors0 (overlap)High (~10⁷ S/m)
Semiconductors~0.1–2 eV (Si: 1.1 eV, Ge: 0.7 eV)Intermediate (increases with T)
Insulators>3 eV (diamond: 5.4 eV)Very low

Intrinsic & Extrinsic Semiconductors

Intrinsic: pure Si or Ge; electrons thermally excited across band gap; n_e = n_h = n_i

n-type: doped with pentavalent donors (P, As, Sb); extra electron → majority carriers = electrons, minority = holes

p-type: doped with trivalent acceptors (B, Al, In); creates holes → majority carriers = holes, minority = electrons

Mass action law\(: n_e \cdot n_h = n_i^{2}\) (at equilibrium, regardless of doping)

p-n Junction

  • Depletion region: forms at junction; depleted of mobile carriers; built-in potential barrier V₀ (Si: ~0.7 V, Ge: ~0.3 V)
  • Forward bias: +ve to p-side; reduces barrier; current flows; diode ≈ short circuit above threshold (~0.7 V for Si)
  • Reverse bias: +ve to n-side; increases barrier; only tiny leakage current flows; diode ≈ open circuit
  • Breakdown: at large reverse voltage, avalanche or Zener breakdown — current increases sharply

Special Diodes

  • Zener diode: operated in reverse breakdown; maintains constant voltage V_Z → used as voltage regulator
  • LED: forward biased; recombination emits photons; E_photon ≈ E_g
  • Photodiode: reverse biased; photons generate electron-hole pairs → current; used as detector
  • Solar cell: no bias; photovoltaic effect — generates EMF from light

Transistors (BJT)

Three regions: Emitter (heavily doped), Base (thin, lightly doped), Collector (moderately doped). Types: npn and pnp.

CE configuration (most common for amplification):

  • Current gain \(\beta = \dfrac{I_C}{I_B} = \dfrac{\alpha}{1-\alpha}\)
  • α = I_C/I_E (always <1, typically 0.95–0.99)
  • I_E = I_B + I_C
  • Voltage gain A_v = β × R_C/R_in

Logic Gates

GateBoolean00011011
ANDY = A·B0001
ORY = A+B0111
NANDY = (A·B)̄1110
NORY = (A+B)̄1000
XORY = A⊕B0110

NAND and NOR are universal gates — any logic circuit can be built from NAND or NOR alone.

💡 JEE Tip: β = α/(1−α). If α = 0.98, β = 0.98/0.02 = 49. Also: I_C = βI_B and I_E = (β+1)I_B. For Zener regulator: current through Zener I_Z = I_supply − I_load = (V_in − V_Z)/R_series − V_Z/R_load. The Zener maintains V_Z across the load.
⚠ Common Error: In n-type, electrons are majority carriers but the material is ELECTRICALLY NEUTRAL (dopant atoms are neutral atoms, just with an extra free electron). Similarly p-type is neutral. Net charge of a semiconductor is zero unless charge is externally supplied.
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Semiconductors — Formula Sheet
Mass Action Law
$$ n_e \cdot n_h = n_i^{2} $$
n_i = intrinsic carrier concentration  |  increases with temperature
Transistor (BJT)
α = I_C/I_E (<1)
β = I_C/I_B = α/(1−α)  |  I_E = I_B + I_C
Zener Regulator
V_out = V_Z (constant)
I_series = (V_in − V_Z)/R_s  |  I_Z = I_series − I_load
Logic: De Morgan's
(A·B)̄ = Ā + B̄ (NAND)
(A+B)̄ = Ā · B̄ (NOR)  |  Universal gates: NAND, NOR
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Semiconductors — Solved Examples
Example 1
Example 1: A silicon sample is doped with 10¹⁶ donor atoms/cm³. Intrinsic carrier density n_i = 1.5×10¹⁰/cm³. Find hole concentration.
  1. In n-type: \(n_e \approx N_D = \dfrac{10^{16}}{cm^{3}}\) (donor concentration dominates)
  2. Mass action: \(n_h = \dfrac{n_i^{2}}{n_e} = (1.5\times10^{10})\dfrac{^{2}}{10^{16}}\)
  3. \(= 2.25\dfrac{\times10^{20}}{10^{16}} =\) 2.25×10⁴/cm³
  4. Holes are minority carriers — much fewer than electrons.
Example 2
Example 2: In a CE transistor circuit, I_B = 20 μA and β = 50. Find I_C and I_E.
  1. \(I_C = \beta \times I_B = 50 \times 20\times10^{-6} =\) 1 mA
  2. \(I_E = I_B + I_C = 20 \mu A + 1000 \mu A =\) 1.02 mA
  3. $$ \alpha = \frac{I_C}{I_E} = \frac{1}{1.02} \approx 0.98 $$
Example 3
Example 3: Find the output of NAND gate for inputs A=1, B=0.
  1. NAND: Y = (A·B)̄ = (1·0)̄ = (0)̄ = 1
  2. NAND output is 0 only when both inputs are 1. For any other input combination, output is 1.
Example 4
Example 4: A Zener diode (V_Z = 6V) is connected with R_s = 200 Ω, V_in = 10V. Load resistance R_L = 1 kΩ. Find I_Z.
  1. V_out = V_Z = 6 V (Zener regulates)
  2. I_series = (V_in − V_Z)/R_s = (10 − 6)/200 = 4/200 = 20 mA
  3. I_load = V_Z/R_L = 6/1000 = 6 mA
  4. I_Z = I_series − I_load = 20 − 6 = 14 mA
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Semiconductors — Diagrams
Click any diagram to zoom
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Semiconductors — Quick Quiz
Question 1
Silicon doped with boron forms:
Boron (B) has 3 valence electrons — it is a trivalent acceptor. When it replaces Si (4 valence electrons), it creates a hole (missing electron). Majority carriers are holes → p-type. Pentavalent donors (P, As, Sb) create n-type.
Question 2
If α = 0.96 for a transistor, the value of β is:
β = α/(1−α) = 0.96/(1−0.96) = 0.96/0.04 = 24. So I_C = 24×I_B. This transistor can amplify base current by factor 24.
Question 3
A NOR gate gives output 1 only when:
NOR = NOT OR. Y = (A+B)̄. OR output is 0 only when both inputs are 0; NOT(0) = 1. For any other input combination, OR gives 1, NOT(1) = 0. So NOR output is 1 only when A=0 AND B=0.
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Experimental Skills
Vernier calipers · screw gauge · least count & zero error · key experiments
○ Not Started

1. Least Count & Zero Error

The least count (LC) is the smallest length an instrument can read. A vernier calipers measures internal/external diameters and depths; a screw gauge (micrometer) measures the thickness or diameter of thin sheets and wires. A zero error — a non-zero reading when the jaws are fully closed — must be subtracted (positive zero error) or added (negative zero error) from every reading to get the true value.

2. Key NEET/JEE Experiments

ExperimentQuantity measured / principle
Simple pendulum\(g\); plot of \(T^2\) vs \(L\)
Young's modulus (Searle's)Elasticity of a metallic wire
Surface tension (capillary rise)Effect of detergents on surface tension
Coefficient of viscosityTerminal velocity of a sphere (Stokes' law)
Resonance tubeSpeed of sound in air
Metre bridge / Ohm's lawResistance and resistivity of a wire
p-n junction / Zener diodeForward & reverse characteristic curves
⚠ Common mistake: forgetting to correct for zero error, or reading the vernier/circular scale in the wrong direction. Always record the zero error first, with its sign.
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Experimental Skills — Formulas
Least count of the two key instruments
Vernier Least Count
$$ \text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = \frac{1\,\text{MSD}}{N} $$
\(N\) = number of vernier divisions
cm / mm
Screw Gauge Least Count
$$ \text{LC} = \frac{\text{Pitch}}{\text{No. of circular divisions}} $$
Pitch = travel per full rotation
mm
Total Reading
$$ R = \text{MSR} + (\text{VSR}\times\text{LC}) - (\text{zero error}) $$
Correct for the signed zero error
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Experimental Skills — Quick Quiz
tap an option to reveal the answer
Question 1
A screw gauge has pitch 1 mm and 100 circular-scale divisions. Its least count is:
\(\text{LC} = \dfrac{\text{pitch}}{\text{divisions}} = \dfrac{1}{100} = 0.01\) mm.
Question 2
If the jaws of a vernier calipers are closed and the vernier reads slightly positive, the instrument has a:
A positive reading with the jaws closed is a positive zero error, which must be subtracted from every observed reading to get the true measurement.
Question 3
The coefficient of viscosity of a liquid is commonly found in the lab by measuring the:
By Stokes' law, a small sphere falling through a viscous liquid reaches a terminal velocity from which the coefficient of viscosity is calculated.
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