⚛ 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 nature	Image position
Beyond C (u > 2f)	Real, inverted, diminished	Between f and C
At C (u = 2f)	Real, inverted, same size	At C
Between C and f	Real, inverted, magnified	Beyond C
At f	Real, at infinity	∞
Inside f (u < f)	Virtual, erect, magnified	Behind 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

Instrument	Magnification
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 correction	Concave (diverging) lens, $P = -\dfrac{1}{d}$ (in m)
Far-sighted correction	Convex (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: 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.

Using sign convention: f = −15 cm, u = −10 cm
1/v = 1/f − 1/u = 1/(−15) − 1/(−10) = −1/15 + 1/10 = (−2+3)/30 = 1/30
v = +30 cm
Image is at +30 cm (behind mirror) → virtual, erect, magnified
m = −v/u = −(30)/(−10) = +3 (magnified 3×, erect)

Example 2: A ray of light travels from glass (n = 1.5) to air. Find the critical angle.

$\sin C = \dfrac{n_{air}}{n_{glass}} = \dfrac{1}{1.5} = \dfrac{2}{3}$
$C = \sin^{-1}(\dfrac{2}{3}) \approx$ 41.8°
For angles of incidence > 41.8°, TIR occurs and no refracted ray exits into air.

Example 3: Two thin lenses of focal lengths +20 cm and −30 cm are placed in contact. Find the power of the combination.

$P_1 = \dfrac{1}{0.20} = +5$ D; $P_2 = \dfrac{1}{-0.30} = -3.33$ D
$P = P_1 + P_2 = 5 - 3.33 =$ +1.67 D
Equivalent focal length $f = \dfrac{1}{P} = \dfrac{1}{1.67} \approx 60$ cm (converging)

Example 4: A glass prism (A = 60°) produces minimum deviation of 40°. Find the refractive index.

$n = \sin[\dfrac{A + \delta_{min}}{2}] / \sin(\dfrac{A}{2}) = \sin[\dfrac{60° + 40°}{2}] / \sin(30°)$
$= \sin(50°) / \sin(30°) = 0.766 / 0.500 =$ 1.532

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Ray Optics — Diagrams

Click any diagram to zoom

Concave mirror: object beyond C gives real, inverted, diminished image between F and C

TIR occurs when incidence angle exceeds critical angle $C = \sin^{-1}(\dfrac{n_2}{n_1})$ in denser medium

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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)

Condition	Fringe 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: In YDSE, slit separation $d = 0.5$ mm, screen distance $D = 1$ m, $\lambda = 600$ nm. Find the fringe width.

$$ \beta = \lambda \frac{D}{d} = (600\times10^{-9} \times 1) / (0.5\times10^{-3}) $$
$$ = 600\times10^{-9} / 5\times10^{-4} = $$ $$ 1.2 \times 10^{-3} m = 1.2 mm $$

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.

For nth bright fringe: $y_n = n\lambda \dfrac{D}{d}$
$$ \lambda = y_n\cdot d / (nD) = (2.4\times10^{-3} \times 1\times10^{-3}) / (4 \times 1) $$
$= 2.4\times10^{-6} / 4 =$ 600 nm

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.

Malus's law: $I = I_{0} \cos^{2}\theta = 80 \times \cos^{2}(60^{\circ})$
$= 80 \times (0.5)^{2} = 80 \times 0.25 =$ 20 W/m²

Example 4: A glass surface (n = 1.732) is illuminated. Find the Brewster's angle.

$$ \tan \theta_B = n = 1.732 $$
$\theta_B = \tan^{-1}(1.732) =$ 60°
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

YDSE: alternating bright and dark fringes; fringe width $\beta = \lambda \dfrac{D}{d}$ (uniform spacing)

Single slit: wide central maximum flanked by narrower, weaker secondary maxima; minima at $a\sin\theta = n\lambda$

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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 type	Change in A	Change in Z	Change in N
α-decay (⁴He)	−4	−2	−2
β⁻-decay (e⁻)	0	+1	−1
β⁺-decay (e⁺)	0	−1	+1
γ-decay (photon)	0	0	0

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.

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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)

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Modern Physics — Solved Examples

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)

$$ KE_{max} = h\nu - \phi = (6.63\times10^{-34} \times 8\times10^{14}) / (1.6\times10^{-19}) - 2 eV $$
$$ = (5.3\times10^{-19} / 1.6\times10^{-19}) - 2 = 3.31 - 2 = 1.31 eV $$
Stopping potential $V_0 = \dfrac{KE_{max}}{e} =$ 1.31 V

Example 2: Find the de Broglie wavelength of an electron accelerated through 100 V.

$\lambda = \dfrac{12.27}{\sqrt V} Å = \dfrac{12.27}{\sqrt{100}} = \dfrac{12.27}{10} =$ 1.227 Å
This is comparable to X-ray wavelengths and atomic spacings — hence electron diffraction is possible.

Example 3: A radioactive sample has T₁/₂ = 20 days. What fraction remains after 60 days?

Number of half-lives $n = \dfrac{60}{20} = 3$
Fraction remaining = $(\dfrac{1}{2})^3$ = 1/8 = 12.5%
Activity also reduces to $\dfrac{A_0}{8}$ after 3 half-lives.

Example 4: The mass defect of ⁴He is 0.0304 amu. Find the binding energy per nucleon. (1 amu = 931.5 MeV)

Total BE = 0.0304 × 931.5 = 28.3 MeV
A = 4 (helium-4 has 4 nucleons)
BE/A = 28.3/4 = 7.07 MeV/nucleon

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Modern Physics — Diagrams

Click any diagram to zoom

$KE_{max}$ vs $\nu$: straight lines for different metals (same slope $h$, different x-intercepts $\nu_0$)

BE/A peaks near Fe-56 (~8.8 MeV/nucleon). Fusion (light nuclei) and fission (heavy nuclei) both release energy

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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.

Quantity	Formula	n=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)

Series	n₁	n₂	Region
Lyman	1	2,3,4…	Ultraviolet
Balmer	2	3,4,5…	Visible
Paschen	3	4,5,6…	Infrared
Brackett	4	5,6,7…	Far IR
Pfund	5	6,7,8…	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.

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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)

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Atoms & Nuclei — Solved Examples

Example 1: Calculate the energy of the photon emitted when hydrogen atom transitions from n=3 to n=1.

E₃ = −13.6/9 = −1.51 eV; E₁ = −13.6 eV
$$ \Delta E = E_{1} - E_{3} = -13.6 - (-1.51) = -12.09 eV $$
Photon energy = |ΔE| = 12.09 eV (Lyman series, UV)

Example 2: Find the wavelength of the first Balmer line (n=3→n=2) in hydrogen.

$$ \frac{1}{\lambda} = R(\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R(\frac{1}{4} - \frac{1}{9}) = R \times \frac{5}{36} $$
$$ \lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097\times10^{7}} = \frac{36}{5.485}\times10^{7} $$
λ = $6.56\times10^{-7} m = 656 nm$ (red Hα line)

Example 3: Find the radius of the 2nd Bohr orbit in He⁺ (Z=2).

$r_n = 0.529 \times \dfrac{n^{2}}{Z} = 0.529 \times \dfrac{4}{2} = 0.529 \times 2 =$ 1.058 Å

Example 4: An electron in H atom is in n=4 state. How many spectral lines can be emitted as it de-excites?

Number of lines = n(n−1)/2 = 4×3/2 = 6 lines
Transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1

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Atoms & Nuclei — Diagrams

Click any diagram to zoom

Hydrogen energy levels; Lyman series (UV, to n=1), Balmer series (visible, to n=2), Paschen (IR, to n=3)

Bohr model: quantised orbits (mvr = nℏ); photon emitted when electron drops to lower orbit

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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

Material	Band gap E_g	Conductivity
Conductors	0 (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

Gate	Boolean	00	01	10	11
AND	Y = A·B	0	0	0	1
OR	Y = A+B	0	1	1	1
NAND	Y = (A·B)̄	1	1	1	0
NOR	Y = (A+B)̄	1	0	0	0
XOR	Y = A⊕B	0	1	1	0

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.

📐

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

🧮

Semiconductors — Solved Examples

Example 1: A silicon sample is doped with 10¹⁶ donor atoms/cm³. Intrinsic carrier density n_i = 1.5×10¹⁰/cm³. Find hole concentration.

In n-type: $n_e \approx N_D = \dfrac{10^{16}}{cm^{3}}$ (donor concentration dominates)
Mass action: $n_h = \dfrac{n_i^{2}}{n_e} = (1.5\times10^{10})\dfrac{^{2}}{10^{16}}$
$= 2.25\dfrac{\times10^{20}}{10^{16}} =$ 2.25×10⁴/cm³
Holes are minority carriers — much fewer than electrons.

Example 2: In a CE transistor circuit, I_B = 20 μA and β = 50. Find I_C and I_E.

$I_C = \beta \times I_B = 50 \times 20\times10^{-6} =$ 1 mA
$I_E = I_B + I_C = 20 \mu A + 1000 \mu A =$ 1.02 mA
$$ \alpha = \frac{I_C}{I_E} = \frac{1}{1.02} \approx 0.98 $$

Example 3: Find the output of NAND gate for inputs A=1, B=0.

NAND: Y = (A·B)̄ = (1·0)̄ = (0)̄ = 1
NAND output is 0 only when both inputs are 1. For any other input combination, output is 1.

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.

V_out = V_Z = 6 V (Zener regulates)
I_series = (V_in − V_Z)/R_s = (10 − 6)/200 = 4/200 = 20 mA
I_load = V_Z/R_L = 6/1000 = 6 mA
I_Z = I_series − I_load = 20 − 6 = 14 mA

📊

Semiconductors — Diagrams

Click any diagram to zoom

p-n junction: depletion region at interface; forward bias narrows it (current flows), reverse bias widens it (blocks current)

CE npn transistor: small base current I_B controls large collector current I_C = βI_B; β typically 50–300

⚡

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.

🔧

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

Experiment	Quantity 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 viscosity	Terminal velocity of a sphere (Stokes' law)
Resonance tube	Speed of sound in air
Metre bridge / Ohm's law	Resistance and resistivity of a wire
p-n junction / Zener diode	Forward & 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.

🔧

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

Total Reading

$$ R = \text{MSR} + (\text{VSR}\times\text{LC}) - (\text{zero error}) $$

Correct for the signed zero error

—

🔧

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.

Instrument	Magnification
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 correction	Concave (diverging) lens, \(P = -\dfrac{1}{d}\) (in m)
Far-sighted correction	Convex (converging) lens

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

Quantity	Formula	n=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

Experiment	Quantity 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 viscosity	Terminal velocity of a sphere (Stokes' law)
Resonance tube	Speed of sound in air
Metre bridge / Ohm's law	Resistance and resistivity of a wire
p-n junction / Zener diode	Forward & reverse characteristic curves