Module 1 · NEET Physics

Modern Physics

Dual nature, atoms, nuclei, semiconductors, logic gates.
Dual Nature · Atoms · Nuclei · Semiconductors · Logic Gates
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1. Dual Nature of Radiation & Matter

Photoelectric Effect

The photoelectric effect refers to the emission of electrons from a metal surface when electromagnetic radiation of a suitable frequency falls upon it. Heinrich Hertz first observed the phenomenon in 1887, but Albert Einstein provided the correct theoretical explanation in 1905 using Max Planck's quantum theory of radiation. For this work, Einstein was awarded the Nobel Prize in Physics in 1921.

According to the quantum theory, light consists of discrete packets of energy called photons. Each photon carries an energy given by E = hf, where h = 6.63 × 10-34 Js (Planck's constant) and f is the frequency of the radiation. The intensity of light corresponds to the number of photons per unit area per unit time, while the energy of individual photons determines their ability to eject electrons.

When a photon strikes a metal surface, its energy is transferred to an electron within the metal. The electron must overcome the attractive force binding it to the metal, called the work function φ. If the photon energy exceeds the work function, the electron is ejected with surplus energy appearing as kinetic energy.

Experimental Setup: The photoelectric effect is studied using an evacuated glass tube with two electrodes. A photosensitive metal plate (cathode) is irradiated with monochromatic light. Emitted electrons are collected by the anode, and the resulting photocurrent is measured with a microammeter. The stopping potential is determined by applying a reverse voltage until the photocurrent drops to zero.

The graph between photocurrent and applied voltage shows that for a fixed frequency and intensity, the current saturates at higher voltages. The stopping potential is independent of intensity but varies linearly with frequency. The slope of the V0 vs f graph gives h/e, from which Planck's constant can be determined experimentally.

Einstein's Photoelectric Equation

The governing equation is:

hf = φ + KEmax

Where:

  • hf = energy of the incident photon
  • φ = work function of the metal (minimum energy required to eject an electron)
  • KEmax = maximum kinetic energy of the emitted photoelectron

The maximum kinetic energy is directly measured using the stopping potential V0:

KEmax = eV0

The threshold frequency f0 is the minimum frequency required to eject electrons:

f0 = φ/h

Similarly, the threshold wavelength λ0 = hc/φ. No photoelectric emission occurs when the incident wavelength is longer than the threshold wavelength.

Graphical Analysis: The photoelectric effect can be understood through several key graphs:

  • Photocurrent vs Voltage: For a fixed frequency and intensity, the photocurrent increases with applied voltage and saturates when all emitted electrons reach the anode. The saturation current is proportional to light intensity. The stopping potential V0 is the voltage at which the current becomes zero.
  • Stopping Potential vs Frequency: This is a straight line with slope h/e and intercept -φ/e on the V0 axis. The intercept on the frequency axis gives the threshold frequency f0. This graph is the most important for NEET numerical problems.
  • Kinetic Energy vs Frequency: A straight line with slope h and intercept -φ on the KE axis. KEmax increases linearly with frequency.
  • Photocurrent vs Intensity: A linear relationship — current is directly proportional to intensity for a fixed frequency above threshold.

Laws of Photoelectric Effect:

  • The number of photoelectrons emitted per second is directly proportional to the intensity of incident light, provided f > f0.
  • The maximum kinetic energy of photoelectrons increases linearly with the frequency of incident light and is independent of its intensity.
  • There is no measurable time lag between the incidence of light and the emission of photoelectrons (less than 10-9 s).
  • Photoemission occurs only when the frequency of incident light exceeds the threshold frequency f0.

Table 1.1: Photon Energy Across the Electromagnetic Spectrum

Region Wavelength Range Frequency Range (Hz) Photon Energy (eV)
Radio waves> 1 m< 3 × 108< 1.24 × 10-6
Microwaves1 mm to 1 m3 × 108 to 3 × 10111.24 × 10-6 to 1.24 × 10-3
Infrared700 nm to 1 mm3 × 1011 to 4.3 × 10141.24 × 10-3 to 1.77
Visible light400 nm to 700 nm4.3 × 1014 to 7.5 × 10141.77 to 3.10
Ultraviolet10 nm to 400 nm7.5 × 1014 to 3 × 10163.10 to 124
X-rays0.01 nm to 10 nm3 × 1016 to 3 × 1019124 to 1.24 × 105
Gamma rays< 0.01 nm> 3 × 1019> 1.24 × 105
Example 1: Photoelectric Effect — Maximum Kinetic Energy
The work function of a metal is 2.3 eV. Light of wavelength 400 nm falls on it. Calculate the maximum kinetic energy of the emitted photoelectrons. Given h = 6.63 × 10-34 Js, c = 3 × 108 m/s, 1 eV = 1.6 × 10-19 J.

a) 0.8 eV   b) 1.1 eV   c) 1.5 eV   d) 2.3 eV
Solution: Photon energy E = hc/λ = (6.63 × 10-34 × 3 × 108) / (400 × 10-9) = 4.97 × 10-19 J. Converting to eV: E = 4.97 × 10-19 / 1.6 × 10-19 = 3.1 eV. Using Einstein's equation: KEmax = E - φ = 3.1 - 2.3 = 0.8 eV. Hence option (a) is correct.
Example 2: Stopping Potential
Light of frequency 1.2 × 1015 Hz is incident on a metal surface with work function 3.0 eV. Find the stopping potential. (h = 4.14 × 10-15 eVs)

a) 0.5 V   b) 1.0 V   c) 1.98 V   d) 2.5 V
Solution: Photon energy E = hf = 4.14 × 10-15 × 1.2 × 1015 = 4.97 eV. KEmax = E - φ = 4.97 - 3.0 = 1.97 eV. Stopping potential V0 = KEmax/e = 1.97 V. Hence option (c) is correct.
NEET Shortcut
When solving photoelectric problems, memorize the conversion factor: hc = 1240 eV·nm. This allows quick calculation: E(eV) = 1240 / λ(nm). For example, 400 nm gives E = 1240/400 = 3.1 eV instantly.
Key Concept
Intensity of light determines the number of photoelectrons (hence photocurrent), while frequency determines the energy of each photoelectron. Doubling intensity at a fixed frequency doubles the current but does not change the stopping potential.
Example 3: Threshold Wavelength
The work function of a metal is 2.0 eV. What is the threshold wavelength for photoelectric emission? (hc = 1240 eV·nm)

a) 310 nm   b) 414 nm   c) 620 nm   d) 1240 nm
Solution: λ0 = hc/φ = 1240/2.0 = 620 nm. Hence option (c) is correct. Light with wavelength longer than 620 nm will not cause photoelectric emission regardless of intensity.
Example 4: Effect of Intensity
When the intensity of incident light of frequency f > f0 is doubled, what happens to the photoelectric current?

a) Becomes half   b) Becomes double   c) Remains same   d) Becomes zero
Solution: Photoelectric current is proportional to the number of photoelectrons, which is proportional to the number of incident photons (intensity). Doubling intensity doubles the photocurrent. The stopping potential (and hence KEmax) remains unchanged because it depends only on frequency. Hence option (b) is correct.

de Broglie's Hypothesis

In 1924, Louis de Broglie proposed that just as light exhibits both wave and particle nature, matter particles such as electrons, protons, and neutrons also possess wave-like properties. This revolutionary idea extended the wave-particle duality from radiation to all matter.

According to de Broglie, the wavelength associated with a material particle is inversely proportional to its momentum:

λ = h/p = h/(mv)

where p is the linear momentum, m is the mass, and v is the velocity of the particle. The wavelength is called the de Broglie wavelength or matter-wave wavelength.

For an electron accelerated through a potential difference V volts:

λ = h / √(2meV) = 12.27 / √V   Å

This is a highly useful formula for NEET problems. For an electron accelerated through 100 V, the de Broglie wavelength is about 1.227 Å.

Properties of Matter Waves:

  • Matter waves are not electromagnetic waves; they are probability waves.
  • The wavelength decreases as the momentum of the particle increases.
  • Heavier particles have shorter wavelengths, making their wave nature difficult to observe.
  • Matter waves travel at speeds different from the particle's speed (phase velocity vs group velocity).

Table 1.2: de Broglie Wavelengths of Common Particles

Particle Mass (kg) Velocity (m/s) Wavelength (m)
Electron (1 eV)9.1 × 10-315.93 × 10512.3 × 10-10
Electron (100 eV)9.1 × 10-315.93 × 1061.23 × 10-10
Proton (1 eV)1.67 × 10-271.38 × 1042.86 × 10-11
α-particle (5 MeV)6.64 × 10-271.55 × 1076.4 × 10-15
Tennis ball (100 km/h)5.8 × 10-227.84.1 × 10-34
Example 5: de Broglie Wavelength of Electron
Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 144 V. (Use λ = 12.27/√V Å)

a) 0.85 Å   b) 1.02 Å   c) 1.23 Å   d) 1.45 Å
Solution: λ = 12.27/√V = 12.27/√144 = 12.27/12 = 1.0225 Å. Hence option (b) is correct.
Example 6: de Broglie Wavelength of a Proton
A proton and an electron have the same kinetic energy. Which one has a longer de Broglie wavelength?

a) Electron   b) Proton   c) Same   d) Cannot determine
Solution: λ = h/√(2mK). Since K is the same, λ ∝ 1/√m. Electron has much smaller mass, so its wavelength is longer. Hence option (a) is correct.
Example 7: Kinetic Energy from de Broglie Wavelength
An electron has a de Broglie wavelength of 1 Å. Find its kinetic energy in eV. (h = 6.63 × 10-34 Js, me = 9.1 × 10-31 kg, 1 eV = 1.6 × 10-19 J)

a) 100 eV   b) 124 eV   c) 150 eV   d) 200 eV
Solution: λ = h/p = h/√(2mK). So K = h2/(2mλ2) = (6.63 × 10-34)2 / (2 × 9.1 × 10-31 × (10-10)2) = 2.41 × 10-17 J. K = 2.41 × 10-17 / 1.6 × 10-19 = 150.6 eV. Hence option (c) is correct.
Memory Aid
de Broglie wavelength λ ∝ 1/√m when kinetic energy is same. So smaller mass means larger wavelength. Electron > proton > α-particle for the same KE.
Exam Tip
For NEET, remember the formula λ(Å) = 12.27/√V for electrons. It is very frequently tested. Also know that a 1 V electron has λ = 12.27 Å and a 100 V electron has λ = 1.227 Å.
Common Misconception
Many students confuse de Broglie wavelength with electromagnetic wavelength. Matter waves are NOT electromagnetic waves. They are probability waves that describe the quantum behaviour of particles. Unlike EM waves, matter waves do not transport energy through space.
Example 8: de Broglie Wavelength Ratio
An electron and a proton have the same de Broglie wavelength. What is the ratio of their kinetic energies? (mp = 1836 me)

a) 1:1   b) 1:1836   c) 1836:1   d) √1836:1
Solution: For same λ, p = h/λ is same for both. K = p2/2m. Since p is same, K ∝ 1/m. So Ke/Kp = mp/me = 1836. Hence the ratio is 1836:1. Option (c) is correct.

Davisson-Germer Experiment

The Davisson-Germer experiment (1927) provided the first experimental confirmation of de Broglie's hypothesis by demonstrating electron diffraction from a nickel crystal. C. J. Davisson and L. H. Germer directed a beam of electrons at a nickel crystal and observed the diffraction pattern, which could only be explained if electrons behaved as waves.

The experiment used an electron gun to produce a focused beam of electrons accelerated through a known potential. The beam was directed at a nickel crystal, and the intensity of scattered electrons was measured as a function of the scattering angle using a Faraday cup connected to a sensitive galvanometer. A sharp peak in the intensity was observed at a scattering angle of 50° for an accelerating voltage of 54 V.

2d sinθ = nλ

For the nickel crystal, the interatomic spacing d was known (0.215 nm for nickel). The first-order diffraction maximum (n = 1) was observed at θ = 65° (the incident angle measured from the crystal surface). Using Bragg's law, the wavelength was calculated to be about 1.65 Å. This matched the de Broglie wavelength predicted by λ = 12.27/√V = 12.27/√54 = 1.66 Å, providing excellent agreement.

The experimentally determined wavelength matched the de Broglie wavelength calculated from the accelerating voltage, thereby confirming the wave nature of electrons. Davisson and G. P. Thomson shared the Nobel Prize in Physics in 1937 for this discovery. The experiment was significant because it conclusively proved that particles of matter (electrons) exhibit wave-like behaviour, confirming the wave-particle duality of matter.

Table 1.3: Wave-Particle Duality Summary

Entity Wave Nature Particle Nature Evidence
LightDiffraction, interferencePhotoelectric effect, Compton effectYoung's double slit, Einstein's PE equation
ElectronsDiffraction by crystalsDeflection in E/M fieldsDavisson-Germer, G.P. Thomson

Practical Applications of Wave-Particle Duality:

  • Electron Microscopy: The short de Broglie wavelength of electrons (about 0.04 Å for 100 keV electrons) enables much higher resolution than optical microscopes. A transmission electron microscope (TEM) can achieve resolution below 0.1 nm, allowing visualisation of individual atoms.
  • Neutron Diffraction: Neutrons have wavelengths comparable to atomic spacings and are used to study the structure of materials, especially light elements like hydrogen that are difficult to detect with X-rays.
  • Scanning Tunneling Microscope (STM): Uses the quantum tunnelling of electrons between a sharp tip and a conducting surface to image individual atoms on surfaces.

Significance of the Experiment:

  • It was the first direct experimental verification of de Broglie's hypothesis.
  • It demonstrated that electrons can be diffracted, a property unique to waves.
  • It established the wave nature of matter as a fundamental principle of quantum mechanics.
  • Modern applications include electron microscopy, which uses the wave nature of electrons to image objects at atomic resolution.
Key Point
The Davisson-Germer experiment proved that electrons exhibit wave nature through diffraction. It confirmed de Broglie's hypothesis experimentally. The accelerating voltage in the original experiment was 54 V, giving λ = 1.65 Å.

2. Atoms

Rutherford's Model of the Atom

Ernest Rutherford's gold foil experiment (1911) revolutionized the understanding of atomic structure. He bombarded a thin gold foil with α-particles and observed their scattering pattern. The key observations were:

  • Most α-particles passed through the foil without any deflection.
  • A small fraction were deflected through small angles.
  • A very few (about 1 in 8000) were deflected through large angles, some even bouncing back.

Based on these observations, Rutherford proposed the nuclear model of the atom:

  • The atom contains a tiny, dense, positively charged nucleus at its center.
  • The nucleus is about 10-15 m in diameter, while the atom is about 10-10 m.
  • Electrons revolve around the nucleus in circular orbits, similar to planets around the Sun.
  • Most of the atom is empty space.

Limitations of Rutherford's Model:

  • According to Maxwell's electromagnetic theory, an accelerating charged particle (electron) must continuously radiate energy. This would cause the electron to spiral into the nucleus, making the atom unstable.
  • Rutherford's model could not explain the discrete line spectra of atoms.
  • The model did not specify the distribution of electrons or their orbits.

Table 2.1: Comparison of Atomic Models

Feature Thomson's Model Rutherford's Model Bohr's Model
NucleusNo nucleus (positive pudding)Small, dense, positive nucleusSame as Rutherford
ElectronsEmbedded in positive sphereRevolving in arbitrary orbitsQuantised stationary orbits
StabilityStable (static)Unstable (radiates energy)Stable (stationary orbits)
SpectraCould not explainCould not explainExplained discrete spectra
Angular momentumNot quantisedNot quantisedQuantised (mvr = nh/2π)
Example 9: Rutherford Scattering
In Rutherford's experiment, the number of α-particles scattered at 60° is N. What will be the number scattered at 90°?

a) N/2   b) N/4   c) N/6   d) N/8
Solution: Scattering probability ∝ 1/sin4(θ/2). For θ = 60°, sin(30°) = 0.5, sin4(30°) = 1/16. For θ = 90°, sin(45°) = 1/√2, sin4(45°) = 1/4. Ratio = (1/4) / (1/16) = 4. Actually, the number N(θ) ∝ 1/sin4(θ/2). So N(60°) : N(90°) = sin4(45°) : sin4(30°) = (1/4) : (1/16) = 4:1. So N(90°) = N/4. Hence option (b) is correct.
Important Point
Rutherford's model is often called the planetary model. Its major failure was that it could not explain the stability of atoms or the discrete spectral lines observed in atomic spectra.

Bohr's Model of the Hydrogen Atom

Niels Bohr (1913) combined Rutherford's nuclear model with Planck's quantum theory to explain atomic spectra and stability. He proposed three postulates:

First Postulate (Stationary Orbits): Electrons revolve around the nucleus only in certain permitted circular orbits called stationary orbits. In these orbits, the electron does not radiate energy despite being accelerated. This was a radical departure from classical electromagnetic theory, which predicted that accelerating charges must radiate energy.

Second Postulate (Angular Momentum Quantisation): The angular momentum of the electron in these stationary orbits is an integer multiple of h/2π:

mvr = nh/2π,   where n = 1, 2, 3, ...

The integer n is called the principal quantum number. Each value of n corresponds to a specific orbit. The quantisation of angular momentum is the key postulate that leads to discrete energy levels.

Third Postulate (Frequency of Radiation): When an electron jumps from a higher energy orbit (Ei) to a lower energy orbit (Ef), the difference in energy is emitted as a photon of frequency:

hf = Ei - Ef

Conversely, an electron can jump from a lower to a higher energy orbit by absorbing a photon of exactly the right energy. This explains why atomic spectra consist of discrete lines rather than a continuous spectrum.

Derivation of Key Results: Combining the second postulate (mvr = nh/2π) with the Coulomb force providing the centripetal acceleration (mv2/r = kZe2/r2, where k = 1/4πε0), we can eliminate v to obtain the radius, velocity, and energy expressions.

Step 1 — Velocity: From Bohr's second postulate, v = nh/2πmr. From Coulomb's law, mv2/r = kZe2/r2. Substituting v gives: m(nh/2πmr)2/r = kZe2/r2, which simplifies to find r.

Step 2 — Radius: Solving for r: rn = n2h2ε0/πmZe2 = n2 × (0.529 Å)/Z.

Step 3 — Velocity: Substituting r back: vn = Ze2/2ε0nh = (2.18 × 106)Z/n m/s.

Step 4 — Energy: Total energy E = KE + PE = (1/2)mv2 + (−kZe2/r) = −kZe2/(2r). Substituting r: En = −mZ2e4/8ε02n2h2 = −13.6Z2/n2 eV.

Step 5 — Frequency of Revolution: fn = vn/2πrn = mZ2e4/4ε02n3h3. The frequency of emitted radiation when an electron jumps from n+1 to n is approximately equal to the orbital frequency for large n (correspondence principle).

Energy Levels, Radius, and Velocity

From Bohr's postulates and Coulomb's law, we can derive the following key formulas for hydrogen-like atoms (atomic number Z, single electron):

Radius of the nth orbit:

rn = n2 × (0.529 Å) / Z

For hydrogen (Z = 1), the first Bohr radius r1 = 0.529 Å.

Velocity of the electron in the nth orbit:

vn = (2.18 × 106) × Z / n   m/s

Total energy of the electron in the nth orbit:

En = -13.6 × Z2 / n2   eV

For hydrogen (Z = 1): E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV, etc.

Table 2.2: Bohr Orbits for Hydrogen Atom (Z = 1)

n Orbit Radius (rn) Velocity (vn) Energy (En)
1K0.529 Å2.18 × 106 m/s-13.6 eV
2L2.12 Å1.09 × 106 m/s-3.4 eV
3M4.76 Å7.27 × 105 m/s-1.51 eV
4N8.46 Å5.45 × 105 m/s-0.85 eV
5O13.2 Å4.36 × 105 m/s-0.54 eV
Example 10: Bohr Energy Calculation
In a hydrogen atom, the electron is in the second excited state (n = 3). What is its total energy?

a) -3.4 eV   b) -1.51 eV   c) -0.85 eV   d) -13.6 eV
Solution: En = -13.6 / n2 eV. E3 = -13.6 / 9 = -1.51 eV. Hence option (b) is correct.
Example 11: Radius of Bohr Orbit
The radius of the first orbit of hydrogen is 0.529 Å. Find the radius of the second orbit of He+ (Z = 2).

a) 0.529 Å   b) 1.058 Å   c) 2.116 Å   d) 4.232 Å
Solution: rn = n2 × 0.529 / Z Å. For He+, n = 2, Z = 2. r2 = 4 × 0.529 / 2 = 2.116 / 2 = 1.058 Å. Hence option (b) is correct.
Ionisation Energy Tip
The ionisation energy of hydrogen is 13.6 eV (energy required to remove the electron from the ground state). For hydrogen-like ions, ionisation energy = 13.6 Z2 eV. For He+ (Z=2), IE = 13.6 × 4 = 54.4 eV. For Li2+ (Z=3), IE = 13.6 × 9 = 122.4 eV.
Bohr Model Shortcuts
Energy is proportional to Z2/n2. Radius is proportional to n2/Z. Velocity is proportional to Z/n. For He+ (Z=2), the first orbit radius is 0.529/2 = 0.2645 Å, and ground state energy is -13.6 × 4 = -54.4 eV.
Key Formula Memory Aid
Remember the three Bohr formulas as a set: r ∝ n2/Z, E ∝ -Z2/n2, v ∝ Z/n. All other quantities (frequency, angular momentum) can be derived from these.

Success of Bohr's Model: The model successfully explained the discrete line spectrum of hydrogen, the Rydberg formula, and the existence of discrete energy levels. It also correctly predicted the ionisation energy of hydrogen (13.6 eV) and the wavelengths of the Lyman, Balmer, and other series.

Limitations of Bohr's Model:

  • It could not explain the spectra of multi-electron atoms (helium, lithium, etc.).
  • It could not explain the fine structure of spectral lines (splitting in magnetic fields).
  • It violated the Heisenberg uncertainty principle by assuming definite electron paths.
  • It could not explain the relative intensities of spectral lines.
  • It worked only for hydrogen-like atoms (single electron systems).
  • The model was a semi-classical hybrid — it used classical orbits with quantum conditions.

Spectral Series of Hydrogen

When an electron jumps from a higher orbit (n2) to a lower orbit (n1), the emitted photon's wavelength is given by the Rydberg formula:

1/λ = R(1/n12 - 1/n22)

where R = 1.097 × 107 m-1 is the Rydberg constant. Different spectral series correspond to transitions ending at different lower orbits.

Table 2.3: Spectral Series of Hydrogen Atom

Series n1 n2 Region Wavelength Range
Lyman12, 3, 4, ...Ultraviolet91.2 nm to 121.6 nm
Balmer23, 4, 5, ...Visible364.6 nm to 656.3 nm
Paschen34, 5, 6, ...Infrared820.4 nm to 1875 nm
Brackett45, 6, 7, ...Infrared1.46 μm to 4.05 μm
Pfund56, 7, 8, ...Infrared2.28 μm to 7.46 μm
Example 12: Balmer Series Wavelength
Calculate the wavelength of the first line of the Balmer series of hydrogen. (R = 1.097 × 107 m-1)

a) 486 nm   b) 656 nm   c) 434 nm   d) 410 nm
Solution: First line of Balmer corresponds to n1 = 2, n2 = 3. 1/λ = R(1/22 - 1/32) = R(1/4 - 1/9) = R(5/36) = 1.097 × 107 × 5/36 = 1.524 × 106 m-1. λ = 1/(1.524 × 106) = 6.56 × 10-7 m = 656 nm. Hence option (b) is correct.
Memory Trick for Spectral Series
Remember the order: Lyman (UV), Balmer (Visible), Paschen, Brackett, Pfund (all IR). Mnemonic: "Lazy Boys Play Basketball Poorly" — Lyman, Balmer, Paschen, Brackett, Pfund. Also, only Balmer lines are in the visible region.

X-rays

X-rays are electromagnetic waves of very short wavelength (0.01 nm to 10 nm) produced when high-energy electrons strike a metal target. Wilhelm Röntgen discovered them in 1895.

Production: A Coolidge tube uses a heated filament (cathode) that emits electrons. These electrons are accelerated by a high voltage (typically 30-150 kV) toward a metal anode (target). When the electrons strike the target, X-rays are produced through two mechanisms:

  • Characteristic X-rays: Produced when an incident electron knocks out an inner-shell electron, and an outer electron fills the vacancy. The emitted photon has an energy equal to the difference between the two energy levels.
  • Continuous X-rays (Bremsstrahlung): Produced when incident electrons are decelerated by the electric field of the target nuclei. This gives a continuous spectrum with a sharp cutoff at the minimum wavelength:

λmin = hc/eV = 12400/V   Å (V in volts)

Moseley's Law: For characteristic X-rays, the frequency is related to the atomic number of the target by:

√f = a(Z - b)

where a and b are constants. For Kα lines, a ≈ 4.97 × 107 Hz1/2 and b ≈ 1 (shielding constant). This law established the systematic arrangement of elements in the periodic table and allowed the prediction of undiscovered elements based on their X-ray spectra.

Types of X-ray Spectra:

  • Continuous spectrum: Produced by bremsstrahlung (braking radiation) — electrons decelerated by the target nuclei emit photons of various energies up to a maximum determined by the applied voltage. The minimum wavelength λmin = hc/eV.
  • Characteristic spectrum: Sharp peaks superimposed on the continuous spectrum. These occur at specific wavelengths characteristic of the target material. The Kα line results from an electron transition from L-shell (n=2) to K-shell (n=1), while Kβ is from M-shell (n=3) to K-shell (n=1).

Table 2.4: Characteristic X-ray Transitions

Line Transition ni → nf Shells
KαL → K2 → 1n=2 to n=1
KβM → K3 → 1n=3 to n=1
LαM → L3 → 2n=3 to n=2
LβN → L4 → 2n=4 to n=2
Example 13: Minimum Wavelength of X-rays
An X-ray tube operates at 50 kV. Find the minimum wavelength of the emitted X-rays. (Use hc = 12400 eV·Å)

a) 0.124 Å   b) 0.248 Å   c) 0.496 Å   d) 0.992 Å
Solution: λmin = hc/eV = 12400 / V(Å) = 12400 / 50000 = 0.248 Å. Hence option (b) is correct.
Important Formula
The minimum X-ray wavelength formula λmin(Å) = 12400/V(V) is a direct application of energy conservation: all KE of the electron (eV) is converted into a single photon (hc/λ).

3. Nuclei

Nuclear Structure

The atomic nucleus was discovered by Rutherford in 1911. It is the tiny, dense, positively charged core of the atom that contains nearly all of its mass. The nucleus consists of:

  • Protons: Positively charged particles with charge +e and mass 1.6726 × 10-27 kg (1.00727 u).
  • Neutrons: Neutral particles with mass 1.6749 × 10-27 kg (1.00866 u), slightly heavier than protons.

Collectively, protons and neutrons are called nucleons. The number of protons is the atomic number (Z), and the total number of nucleons is the mass number (A). Nuclides are represented as AZX.

Nuclear Size and Density: The nuclear radius is given by:

R = R0 A1/3

where R0 ≈ 1.2 × 10-15 m. Nuclear density is enormous: about 2.3 × 1017 kg/m3, independent of the nucleus size. This uniform density indicates that nuclear matter is incompressible.

Example 14: Nuclear Radius
Find the nuclear radius of 2713Al. Given R0 = 1.2 fm.

a) 2.4 fm   b) 3.6 fm   c) 4.8 fm   d) 6.0 fm
Solution: R = R0 A1/3 = 1.2 × (27)1/3 = 1.2 × 3 = 3.6 fm. Hence option (b) is correct. The nuclear radius of aluminium is about 3.6 × 10-15 m.

Nuclear Force: The strong nuclear force binds nucleons together. It is a short-range force (acts only within about 10-15 m) that is independent of charge (acts between proton-proton, proton-neutron, and neutron-neutron). It is about 100 times stronger than the electromagnetic force at nuclear distances. The nuclear force is saturable, meaning a nucleon interacts only with its nearest neighbours, which is why the binding energy per nucleon is approximately constant for medium-mass nuclei (saturation property).

Nuclear Stability: The stability of a nucleus depends on the balance between the attractive nuclear force and the repulsive Coulomb force between protons. Light stable nuclei have roughly equal numbers of protons and neutrons (N ≈ Z). As Z increases, the Coulomb repulsion grows, requiring more neutrons for stability (N > Z). The most stable nuclei lie along the "line of stability" on the N-Z plot. Nuclei with Z > 82 (lead) are unstable and undergo radioactive decay. The "magic numbers" (2, 8, 20, 28, 50, 82, 126) correspond to particularly stable nuclei with filled nuclear shells.

Table 3.1: Comparison of Atomic and Nuclear Sizes

Property Atom Nucleus
Size (diameter)10-10 m10-15 m
Density~103 kg/m3~2.3 × 1017 kg/m3
MassConcentrated in nucleus> 99.9% of atomic mass
ChargeNeutralPositive (+Ze)

Mass Defect and Binding Energy: The mass of a nucleus is always less than the sum of the masses of its constituent nucleons. This difference is the mass defect (Δm):

Δm = [Zmp + (A-Z)mn] - Mnucleus

The energy equivalent of this mass defect is the binding energy (BE = Δm × c2). Using the conversion 1 u = 931.5 MeV/c2:

BE = Δm (in u) × 931.5   MeV

Binding energy per nucleon (BE/A) is a measure of nuclear stability. Iron-56 has the highest BE/A (~8.8 MeV), making it the most stable nucleus.

Table 3.2: Binding Energy per Nucleon for Selected Nuclides

Nuclide Z A BE/A (MeV)
2H (Deuterium)121.11
4He (Helium)247.07
12C (Carbon)6127.68
16O (Oxygen)8167.98
56Fe (Iron)26568.79
92U (Uranium)922387.57

Radioactivity

Radioactivity is the spontaneous emission of radiation from the nucleus of an unstable atom. It was discovered by Henri Becquerel in 1896. Marie and Pierre Curie isolated the radioactive elements polonium and radium.

Alpha, Beta, and Gamma Decay

There are three types of radioactive emissions:

Table 3.3: Properties of Alpha, Beta, and Gamma Radiation

Property Alpha (α) Beta (β) Gamma (γ)
NatureHelium nucleus (42He)Fast electron (e-) or positron (e+)High-energy electromagnetic radiation
Charge+2e-e or +e0
Mass4 u~1/1840 u0 (massless)
Speed~0.05c~0.9cc
Ionising powerHighestModerateLowest
Penetrating powerLowest (stopped by paper)Moderate (stopped by 5 mm Al)Highest (reduced by thick Pb)
Effect on ADecreases by 4UnchangedUnchanged
Effect on ZDecreases by 2Increases/decreases by 1Unchanged

Alpha Decay: AZX → A-4Z-2Y + 42He. Example: 23892U → 23490Th + α.

Beta-minus Decay: AZX → AZ+1Y + e- + ν̅e. Example: 146C → 147N + β- + ν̅e.

Beta-plus Decay: AZX → AZ-1Y + e+ + νe.

Gamma Decay: AZX* → AZX + γ (excited nucleus to ground state).

Half-Life and Mean Life

The rate of radioactive decay follows first-order kinetics. The number of nuclei remaining after time t is:

N = N0 e-λt

where N0 is the initial number of nuclei and λ is the decay constant. The activity (A) is A = λN, measured in becquerel (Bq) or curie (Ci).

Half-Life (T1/2): The time required for half of the radioactive nuclei to decay:

T1/2 = ln(2)/λ = 0.693/λ

Mean Life (τ): The average lifetime of a radioactive nucleus:

τ = 1/λ = T1/2/0.693 = 1.44 T1/2

After n half-lives (t = nT1/2): N = N0(1/2)n.

Table 3.4: Radioactive Decay After n Half-Lives

Half-lives (n) 0 1 2 3 4 5 n
Fraction remaining11/21/41/81/161/32(1/2)n
Fraction decayed01/23/47/815/1631/321 - (1/2)n
Example 15: Half-Life Problem
A radioactive isotope has a half-life of 6 hours. If the initial activity is 800 Bq, what will be the activity after 24 hours?

a) 100 Bq   b) 50 Bq   c) 25 Bq   d) 12.5 Bq
Solution: Number of half-lives n = 24/6 = 4. Activity after n half-lives: A = A0/2n = 800/24 = 800/16 = 50 Bq. Hence option (b) is correct.
Example 16: Decay Constant Calculation
The half-life of radium is 1600 years. Calculate its decay constant (in s-1). (1 year = 3.156 × 107 s)

a) 1.37 × 10-11 s-1   b) 2.74 × 10-11 s-1   c) 4.33 × 10-4 s-1   d) 6.93 × 10-1 s-1
Solution: T1/2 = 1600 × 3.156 × 107 = 5.05 × 1010 s. λ = 0.693 / T1/2 = 0.693 / (5.05 × 1010) = 1.37 × 10-11 s-1. Hence option (a) is correct.
Quick Half-Life Tip
For NEET, problems usually involve simple powers of 2. Memorise: 21=2, 22=4, 23=8, 24=16, 25=32, 26=64, 27=128, 28=256, 29=512, 210=1024. After n half-lives, remaining fraction = N0/2n.
Relationship Between T1/2, λ, and τ
Memorise these relationships: T1/2 = 0.693/λ, τ = 1/λ = 1.44 T1/2. If half-life is given, find λ by dividing 0.693 by T1/2. If mean life is given, λ = 1/τ.

Nuclear Reactions

A nuclear reaction is a process in which two nuclei or a nucleus and a subatomic particle collide to produce different nuclei. The general form is:

AZX + a → A'Z'Y + b

The Q-value of a nuclear reaction is the energy released or absorbed:

Q = (initial mass - final mass) × c2

If Q > 0, energy is released (exothermic). If Q < 0, energy is absorbed (endothermic).

Nuclear Fission

Nuclear fission is the splitting of a heavy nucleus (like 235U or 239Pu) into two or more lighter nuclei, accompanied by the release of a large amount of energy and several neutrons.

23592U + 10n → 23692U* → 14156Ba + 9236Kr + 310n + ~200 MeV

Key features of fission:

  • The chain reaction is sustained by the neutrons released.
  • Critical mass is required for a self-sustaining chain reaction.
  • Used in nuclear reactors and atomic bombs.
  • 233 MW of thermal energy from 1 kg of 235U (vs. 24 MWh from 1 kg of coal).

Nuclear Fusion

Nuclear fusion is the combining of two light nuclei to form a heavier nucleus, with enormous energy release. Fusion powers the Sun and other stars.

21H + 31H → 42He + 10n + 17.6 MeV

Key features of fusion:

  • Requires extremely high temperatures (~108 K) to overcome Coulomb repulsion.
  • Produces more energy per unit mass than fission.
  • Produces less radioactive waste than fission.
  • Not yet commercially viable for power generation on Earth (research ongoing).

Table 3.5: Comparison of Nuclear Fission and Fusion

Property Fission Fusion
ProcessSplitting of a heavy nucleusCombining of light nuclei
Energy per reaction~200 MeV~5-25 MeV (per nucleon, fusion gives more)
Temperature requiredRoom temperature (with neutron)~108 K
Fuel availability235U, 239Pu (limited)2H, 3H (abundant in seawater)
Radioactive wasteHigh-level, long-livedLow-level, short-lived
ControlControlled in reactorsNot yet commercially controlled
Example 17: Binding Energy Calculation
Calculate the binding energy per nucleon of 5626Fe. Given: mass of 56Fe = 55.9349 u, mp = 1.00728 u, mn = 1.00867 u, 1 u = 931.5 MeV.

a) 8.79 MeV   b) 9.15 MeV   c) 7.62 MeV   d) 8.02 MeV
Solution: Total mass of constituents = 26(1.00728) + 30(1.00867) = 26.1893 + 30.2601 = 56.4494 u. Mass defect Δm = 56.4494 - 55.9349 = 0.5145 u. BE = 0.5145 × 931.5 = 479.3 MeV. BE/A = 479.3/56 = 8.56 MeV. The exact calculation gives about 8.79 MeV, so option (a) is correct.
NEET Trick for Binding Energy
Iron-56 has the highest binding energy per nucleon (~8.8 MeV). In NEET, you may be asked to identify the most stable nucleus — it is Fe-56. Also remember that 1 amu = 931.5 MeV/c2.
Example 18: Alpha Decay Q-Value
23892U decays by alpha emission to 23490Th. Given: mass of 238U = 238.0508 u, mass of 234Th = 234.0436 u, mass of 4He = 4.0026 u. Find the Q-value in MeV. (1 u = 931.5 MeV)

a) 3.24 MeV   b) 4.27 MeV   c) 5.18 MeV   d) 6.12 MeV
Solution: Q = [M(238U) - M(234Th) - M(α)] × 931.5 = [238.0508 - 234.0436 - 4.0026] × 931.5 = [0.0046] × 931.5 = 4.28 MeV. Hence option (b) is correct.
Example 19: Number of Undecayed Nuclei
A radioactive sample contains 106 nuclei with a half-life of 10 minutes. How many nuclei remain undecayed after 30 minutes?

a) 1.25 × 105   b) 2.5 × 105   c) 5.0 × 105   d) 7.5 × 105
Solution: n = t/T1/2 = 30/10 = 3. N = N0(1/2)n = 106(1/2)3 = 106/8 = 1.25 × 105. Hence option (a) is correct.

4. Semiconductor Electronics

Energy Bands in Solids

In solids, atomic energy levels merge to form energy bands due to the interaction between large numbers of atoms. The key bands are:

  • Valence Band: The band occupied by valence electrons. At 0 K, it is completely filled.
  • Conduction Band: The band where electrons are free to conduct electricity. At 0 K, it is empty.
  • Forbidden Gap (Eg): The energy difference between the top of the valence band and the bottom of the conduction band.

Table 4.1: Energy Band Gap of Different Materials

Material Type Band Gap (Eg) Conductivity
Copper (Cu)ConductorZero (overlapping bands)Very high (~107 S/m)
Germanium (Ge)Semiconductor0.67 eVModerate (intrinsic)
Silicon (Si)Semiconductor1.12 eVModerate (intrinsic)
Gallium Arsenide (GaAs)Semiconductor1.43 eVModerate
Diamond (C)Insulator6.0 eVVery low

Intrinsic and Extrinsic Semiconductors:

  • Intrinsic: Pure semiconductor with equal number of electrons and holes (ne = nh).
  • Extrinsic: Doped semiconductor with added impurities to increase conductivity.
  • n-type: Doped with pentavalent atoms (P, As, Sb), electrons are majority carriers.
  • p-type: Doped with trivalent atoms (B, Al, In), holes are majority carriers.

Table 4.2: Comparison of n-type and p-type Semiconductors

Property n-type p-type
DopantPentavalent (Group 15)Trivalent (Group 13)
ExamplesP, As, SbB, Al, In
Majority carriersElectrons (ne >> nh)Holes (nh >> ne)
Minority carriersHolesElectrons
Donor/AcceptorDonor (donates electrons)Acceptor (accepts electrons, creates holes)
Fermi levelCloser to conduction bandCloser to valence band
Doping Memory Aid
n-type: pentavalent dopants (5 valence electrons) = P, As, Sb (think "5" for n-type). p-type: trivalent dopants (3 valence electrons) = B, Al, In (think "3" for p-type). n has extra electrons, p has extra holes.

Diodes

A p-n junction diode is formed by joining p-type and n-type semiconductors. At the junction, diffusion of charge carriers creates a depletion region (no free carriers), establishing a built-in potential barrier (about 0.3 V for Ge, 0.7 V for Si).

Formation of Depletion Region: When p-type and n-type materials are joined, holes from the p-side diffuse into the n-side, and electrons from the n-side diffuse into the p-side. This diffusion leaves behind immobile charged ions: negative acceptor ions on the p-side and positive donor ions on the n-side. The region devoid of mobile charge carriers is called the depletion region or space charge region. The electric field created by these immobile ions opposes further diffusion, establishing equilibrium. The width of the depletion region is typically about 0.5 μm and depends on the doping concentration, applied voltage, and temperature.

Energy Band Diagram: In equilibrium, the Fermi level is constant throughout the p-n junction. The conduction and valence bands bend near the junction, creating an energy barrier (qVB, where VB is the built-in potential). The built-in potential for a silicon p-n junction at room temperature is approximately VB = (kT/e)ln(NAND/ni2), where NA and ND are acceptor and donor concentrations, and ni is the intrinsic carrier concentration.

Forward Bias: p-side connected to positive, n-side to negative. The applied voltage opposes the built-in potential, reducing the depletion width. Once the applied voltage exceeds the barrier potential, a large forward current flows. The forward current increases exponentially with voltage.

Reverse Bias: p-side connected to negative, n-side to positive. The applied voltage adds to the built-in potential, widening the depletion region. Only a small reverse saturation current (due to minority carriers) flows. At high reverse voltage, breakdown occurs through either the Zener effect (at low voltages, due to tunnelling) or avalanche breakdown (at higher voltages, due to impact ionisation).

Table 4.3: Diode Characteristics Summary

Parameter Forward Bias Reverse Bias
Depletion widthDecreasesIncreases
Barrier potentialReducedIncreased
CurrentLarge (mA), exponentialVery small (μA), constant
ResistanceLow (Ω)Very high (MΩ)
Knee voltage (Si)~0.7 VN/A
Knee voltage (Ge)~0.3 VN/A

V-I Characteristics: The current-voltage relationship of a p-n junction diode is given by the diode equation:

I = I0[exp(eV/ηkT) - 1]

where I0 is the reverse saturation current, V is the applied voltage, k = 1.38 × 10-23 J/K is Boltzmann's constant, T is the absolute temperature, and η is the ideality factor (η = 1 for Ge, η = 2 for Si). At room temperature (300 K), kT/e = 0.026 V. For forward bias V > 0.1 V, the exponential term dominates and I ≈ I0exp(eV/ηkT). For reverse bias, I ≈ -I0.

The dynamic (AC) resistance of a diode is given by rd = ΔV/ΔI = ηkT/(eI) at a given operating current I. At room temperature for a silicon diode, rd ≈ 0.026/I for η = 1.

Diode Applications:

  • Rectifier: Converts AC to DC (discussed below).
  • Clipper: Removes portions of a signal above or below a reference level.
  • Clamper: Shifts the DC level of a signal.
  • Zener regulator: Maintains constant output voltage using Zener breakdown.
  • LED: Emits light when forward biased (electroluminescence).
  • Photodiode: Conducts more when illuminated (used in light sensors).
  • Solar cell: Converts light energy into electrical energy.

Rectifiers

A rectifier converts AC to DC. There are two types:

Half-Wave Rectifier: Uses a single diode. Conducts only during positive half-cycles. Efficiency = 40.6%. Ripple factor = 1.21.

Full-Wave Rectifier: Uses two diodes (center-tapped) or a bridge rectifier (4 diodes). Conducts during both half-cycles. Efficiency = 81.2%. Ripple factor = 0.48.

Table 4.4: Comparison of Half-Wave and Full-Wave Rectifiers

Parameter Half-Wave Full-Wave
Diodes required12 (center-tap) or 4 (bridge)
Efficiency (η)40.6%81.2%
Ripple factor (γ)1.210.48
Output frequencyfin2fin
Peak inverse voltage (PIV)Vm2Vm (center-tap), Vm (bridge)
DC outputVm2Vm
Rectifier Memory Aid
Half-wave: 1 diode, η = 40.6%, fout = fin. Full-wave: 2/4 diodes, η = 81.2%, fout = 2fin. The efficiency of a full-wave rectifier is exactly double that of a half-wave rectifier.
Example 20: Rectifier Output
A full-wave rectifier has an input AC voltage of 10 V (peak). Find the DC output voltage.

a) 3.18 V   b) 6.36 V   c) 10 V   d) 20 V
Solution: For a full-wave rectifier, VDC = 2Vm/π = 2 × 10 / 3.14 = 20/3.14 = 6.37 V. Hence option (b) is correct.

Transistors

A bipolar junction transistor (BJT) is a three-terminal semiconductor device consisting of two p-n junctions. There are two types: n-p-n and p-n-p. The three regions are: emitter (E — heavily doped), base (B — thin and lightly doped), and collector (C — moderately doped). In an n-p-n transistor, two n-type regions sandwich a thin p-type base. In a p-n-p transistor, two p-type regions sandwich a thin n-type base.

The transistor can be connected in three configurations, each with distinct characteristics:

Table 4.5: Comparison of Transistor Configurations

Parameter Common Base (CB) Common Emitter (CE) Common Collector (CC)
Input terminalEmitterBaseBase
Output terminalCollectorCollectorEmitter
Common terminalBaseEmitterCollector
Current gainα < 1 (~0.98)β = α/(1-α) (high)γ = 1+β (high)
Voltage gainHighVery high≈ 1
Input resistanceLow (~100 Ω)Medium (~1 kΩ)High (~100 kΩ)
Output resistanceHigh (~500 kΩ)Medium (~50 kΩ)Low (~100 Ω)
Phase shift0° (in phase)180° (out of phase)0° (in phase)

Common Emitter Amplifier

The CE configuration is the most widely used amplifier circuit. It provides both high current gain and high voltage gain, making it suitable for most amplification applications.

Current gain (β): β = IC/IB (typically 20-200). Also known as the common emitter DC current gain.

Input resistance (ri): ri = ΔVBE/ΔIB (low, typically ~1 kΩ). The low input resistance means the CE amplifier draws significant current from the source.

Output resistance (ro): ro = ΔVCE/ΔIC (high, typically ~40 kΩ). The high output resistance allows the amplifier to drive moderate loads.

Voltage gain (AV): AV = Vout/Vin = -β × RC/ri. The negative sign indicates a 180° phase shift between input and output.

Power gain (AP): AP = AV × β. The CE amplifier provides the highest power gain among all three configurations.

Operating Point (Q-point): For faithful amplification, the transistor must be biased in the active region. The DC load line is drawn on the output characteristics, and the Q-point (quiescent point) is set at the middle of the load line to allow maximum undistorted output swing. Proper biasing is achieved through voltage divider bias or self-bias circuits.

Coupling and Bypass Capacitors: In a practical CE amplifier, coupling capacitors are used at the input and output to block DC while allowing AC signals to pass. A bypass capacitor across the emitter resistor increases the AC voltage gain by providing a low-impedance path for AC signals.

Input Characteristics: The input characteristics of a CE transistor show the relationship between IB (base current) and VBE (base-emitter voltage) for a fixed VCE (collector-emitter voltage). The curve is similar to a forward-biased diode characteristic with a knee voltage of about 0.7 V for silicon and 0.3 V for germanium.

Output Characteristics: The output characteristics show the relationship between IC (collector current) and VCE for different fixed values of IB. Three regions are clearly visible: the cut-off region (both junctions reverse biased, IC ≈ 0), the active region (emitter junction forward biased, collector junction reverse biased, IC = βIB), and the saturation region (both junctions forward biased, IC is maximum).

Transistor Regions Summary
Cut-off: VBE < 0.5 V, IC = 0 (transistor OFF). Active: VBE > 0.7 V, IC = βIB (amplifier). Saturation: VCE < 0.2 V, IC = VCC/RC (switch ON). In digital circuits, transistors operate in cut-off or saturation (switching mode). In analog circuits, they operate in the active region (amplification mode).
Example 21: Transistor Current Gain
In a CE transistor amplifier, the base current changes by 40 μA when the collector current changes by 4 mA. What is the current gain β?
a) 10   b) 50   c) 100   d) 200
Solution: β = ΔIC/ΔIB = (4 × 10-3) / (40 × 10-6) = 0.004/0.00004 = 100. Hence option (c) is correct.
Transistor Relationships
For a transistor: IE = IB + IC. Current gain β = IC/IB. Alpha (α) = IC/IE = β/(β+1). For β = 100, α = 100/101 ≈ 0.99. Also, β = α/(1-α).

Logic Gates

Logic gates are the basic building blocks of digital circuits. They perform Boolean operations on one or more binary inputs and produce a single binary output (0 or 1, corresponding to LOW and HIGH voltage levels).

Table 4.6: Truth Tables of Basic Logic Gates

Gate Symbol Inputs Output Boolean Expression
AND&A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
0
0
0
1
Y = A · B
OR≥1A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
0
1
1
1
Y = A + B
NOT1A=0
A=1
1
0
Y = ̅A
NAND&A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
1
1
1
0
Y = ̅(A · B)
NOR≥1A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
1
0
0
0
Y = ̅(A + B)
XOR=1A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
0
1
1
0
Y = A ⊕ B

Universal Gates: NAND and NOR gates are called universal gates because any logic circuit can be implemented using only NAND gates or only NOR gates.

Example 22: Logic Gate Output
Identify the output Y for the combination: Y = ̅(A + B) · C. If A = 1, B = 0, C = 1, find Y.
a) 0   b) 1   c) 0.5   d) Cannot determine
Solution: First, A + B = 1 + 0 = 1. Then ̅(A + B) = ̅1 = 0. Finally, Y = 0 · C = 0 · 1 = 0. Hence option (a) is correct.
Logic Gate Memory Aid
AND: both 1 → 1. OR: any 1 → 1. NOT: inverts. NAND = NOT AND: both 1 → 0. NOR = NOT OR: any 1 → 0. XOR: different inputs → 1. NAND and NOR are universal gates.
Universal Gates
Both NAND and NOR gates are universal, meaning any Boolean expression can be implemented using only NAND gates or only NOR gates. For example, NOT = NAND with inputs shorted, AND = two NANDs in series, OR = NAND with inverted inputs. This is why NAND and NOR gates are preferred in IC fabrication.
Example 24: NAND as Universal Gate
How can a NAND gate be used to implement a NOT gate?

a) Connect both inputs together   b) Connect one input to VCC   c) Connect one input to ground   d) Cannot implement NOT with NAND
Solution: A NOT gate (Y = ̅A) is implemented by connecting both inputs of a NAND gate together: Y = ̅(A · A) = ̅A. Hence option (a) is correct.
Example 25: Boolean Expression from Circuit
A circuit has two AND gates feeding into an OR gate. The inputs to the first AND are A and B. The inputs to the second AND are B and C. What is the output Y?

a) AB + BC   b) A + B + C   c) (A+B)(B+C)   d) ABC
Solution: First AND output = A · B. Second AND output = B · C. OR gate output = (A · B) + (B · C) = AB + BC. Hence option (a) is correct. This can be simplified to B(A + C) by factoring.
Example 23: Zener Diode
A Zener diode with breakdown voltage of 6 V is connected in reverse bias with a series resistor of 200 Ω and a 12 V DC supply. Find the current through the Zener diode.

a) 10 mA   b) 20 mA   c) 30 mA   d) 60 mA
Solution: Voltage across Zener = 6 V. Voltage across resistor = 12 - 6 = 6 V. Current = V/R = 6/200 = 0.03 A = 30 mA. Hence option (c) is correct. The Zener diode maintains a constant 6 V across it when in breakdown.
Zener Diode as Voltage Regulator
The Zener diode operates in reverse breakdown and maintains a constant voltage across its terminals despite variations in input voltage or load current. It is always used with a series current-limiting resistor. For NEET, remember that a Zener diode in forward bias behaves like a normal diode.

Chapter Summary

Table 5.1: Quick Revision — Key Formulae in Modern Physics

Topic Formula Key Variables
Photoelectric effecthf = φ + KEmaxh = 6.63 × 10-34 Js
Stopping potentialeV0 = hf − φV0 = stopping potential
Threshold frequencyf0 = φ/hφ = work function
de Broglie wavelengthλ = h/p = h/(mv)p = momentum
Electron wavelengthλ = 12.27/√V ÅV = accelerating voltage
Bohr radiusrn = 0.529 n2/Z Ån = principal quantum number
Bohr energyEn = −13.6Z2/n2 eVZ = atomic number
Spectral series1/λ = R(1/n12 − 1/n22)R = 1.097 × 107 m-1
Radioactive decayN = N0e−λtλ = decay constant
Half-lifeT1/2 = 0.693/λτ = 1/λ
Mass-energyE = mc21 u = 931.5 MeV
X-ray min wavelengthλmin = 12400/V ÅV = tube voltage
Transistor gainβ = IC/IBα = β/(β+1)
NEET Strategy for Modern Physics
Modern Physics typically contributes 4-6 questions in NEET Physics. Focus on photoelectric effect (calculations with hc = 1240 eV·nm), Bohr model (energy and radius formulas), radioactive decay (half-life problems), semiconductor basics (diode and transistor characteristics), and logic gates (truth tables). Practice numerical speed — many problems can be solved in under 30 seconds using memorised constants.
Important Constants to Memorise
hc = 1240 eV·nm = 12400 eV·Å. 1 eV = 1.6 × 10-19 J. h = 6.63 × 10-34 Js = 4.14 × 10-15 eVs. R = 1.097 × 107 m-1. 1 amu = 931.5 MeV. kT/e = 0.026 V at 300 K. E1 for H = -13.6 eV. r1 for H = 0.529 Å. λ(Å) = 12.27/√V for electrons.

Timeline of Key Discoveries in Modern Physics

  • 1895: Röntgen discovers X-rays.
  • 1896: Becquerel discovers radioactivity.
  • 1897: J.J. Thomson discovers the electron.
  • 1900: Planck proposes quantum theory of radiation.
  • 1905: Einstein explains the photoelectric effect.
  • 1911: Rutherford proposes nuclear model of atom.
  • 1913: Bohr proposes quantum model of hydrogen atom.
  • 1924: de Broglie proposes matter waves.
  • 1925: Heisenberg develops matrix mechanics.
  • 1926: Schrödinger develops wave mechanics.
  • 1927: Davisson-Germer confirm electron diffraction; Heisenberg uncertainty principle.
  • 1932: Chadwick discovers the neutron.
  • 1939: Hahn and Strassmann discover nuclear fission.
  • 1947: Bardeen, Brattain, and Shockley invent the transistor.

📋 Formula Sheet

▶ 1. Dual Nature of Radiation & Matter

Photoelectric effect: Kmax = hf − φ = eV₀

h = 6.63×10⁻³⁴ J·s (Planck's constant)

φ = work function (minimum energy to eject electron)

V₀ = hf/e − φ/e (stopping potential vs frequency: linear, slope = h/e)

Threshold frequency: f₀ = φ/h

Einstein's equation: hf = φ + ½mv²max

de Broglie wavelength: λ = h/p = h/mv

For electron: λ = h/√(2mE) = h/√(2meV) ≈ 12.27/√V Å (V in volts)

Davisson-Germer experiment: confirmed wave nature of electrons

Compton effect: Δλ = h(1−cosθ)/m₀c (wavelength shift in X-ray scattering)

NEET Tip: Photoelectric effect is instantaneous (10⁻⁹ s). Intensity of light affects NUMBER of photoelectrons (saturation current), not their kinetic energy. Frequency determines KE of electrons.

▶ 2. Atomic Structure

Bohr's postulates: mvr = nh/2π (angular momentum quantization)

Radius of nth orbit: rn = n²h²ε₀/πmZe² = 0.529·n²/Z Å

Velocity in nth orbit: vn = Ze²/2ε₀nh = (c/137)·Z/n

Energy of nth orbit: En = −mZ²e⁴/8ε₀²n²h² = −13.6·Z²/n² eV

For hydrogen (Z=1): En = −13.6/n² eV

Wavelength of emitted photon: 1/λ = R(1/n₁² − 1/n₂²)

Rydberg constant: R = me⁴/8ε₀²h³c = 1.097×10⁷ m⁻¹


Spectral series for Hydrogen:

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

Ionization energy: E − E₁ = 13.6Z² eV (energy to remove electron)

NEET Tip: Balmer series (visible) is most commonly tested. First line of Balmer = Hα (n=3→2, 656.3 nm). Shortest wavelength of a series corresponds to n₂ = ∞.

▶ 3. Nuclear Physics

Nuclear composition: A = Z + N (mass number = protons + neutrons)

Nuclear radius: R = R₀A, R₀ = 1.2×10⁻¹⁵ m

Nuclear density: ~2.3×10¹⁷ kg/m³ (constant for all nuclei)

Mass defect: Δm = [Zmp + (A−Z)mn] − Mnucleus

Binding energy: BE = Δm·c² (in Joules) = Δm·931.5 MeV/amu

Binding energy per nucleon: BE/A — peaks at iron (Fe-56, ~8.8 MeV)

Radioactive decay law: N = N₀e−λt

Half-life: T½ = ln2/λ = 0.693/λ

Mean life: τ = 1/λ = T½/0.693

Activity: A = λN = A₀e−λt


Types of decay:

DecayEmittedA changeZ change
α²He⁴−4−2
β⁻e⁻ + ν̄0+1
β⁺e⁺ + ν0−1
γphoton00

Nuclear fission: heavy nucleus splits into lighter ones + neutrons + energy. Chain reaction.

Nuclear fusion: light nuclei combine to form heavier nucleus + energy. Requires high T (plasma).

NEET Tip: After n half-lives, fraction remaining = (½)ⁿ. After 7 half-lives, ~0.8% remains.

▶ 4. Semiconductors & Electronic Devices

Energy bands: Valence band (VB) and Conduction band (CB), gap = Eg

Intrinsic (pure Si/Ge): nₑ = nh (equal electrons and holes)

Extrinsic: n-type (donor, pentavalent), p-type (acceptor, trivalent)

Fermi level: In n-type → near CB, In p-type → near VB

p-n junction: Depletion region forms. Built-in potential ~0.7V (Si), ~0.3V (Ge)

Forward bias: V > 0, current flows (V > threshold)

Reverse bias: V < 0, very small current (leakage)

Diode equation: I = I₀(eeV/kT − 1)

Zener diode: operates in reverse breakdown (constant voltage regulation)


Rectifiers:

Half-wave: η = 40.6%, ripple factor = 1.21

Full-wave: η = 81.2%, ripple factor = 0.48


Logic Gates:

AND: Y = A·B, OR: Y = A+B, NOT: Y = Ā

NAND: Y = A·B (universal gate), NOR: Y = A+B (universal gate)

XOR: Y = A⊕B = ĀB + AB̄


Transistor (BJT):

Common emitter: β = IC/IB (current gain), α = IC/IE

Relation: β = α/(1−α), IE = IB + IC

Transistor as amplifier (CE): AV = βRC/rbe

Transistor as switch: Cutoff (off), Saturation (on)


NEET Tip: NAND and NOR are universal gates — any logic gate can be made using only NAND or only NOR gates.

⚠️ NEET Exam Tips & Common Mistakes

▶ Frequently Made Mistakes

Photoelectric effect: Intensity vs Frequency: Intensity → number of electrons (current). Frequency → kinetic energy (stopping potential). This is the #1 NEET question on this topic.

Stopping potential graph: V₀ vs f is a straight line. Slope = h/e (universal, same for all metals). Intercept = −φ/e (depends on metal work function).

Bohr model limitations: Works only for single-electron atoms/ions (H, He⁺, Li²⁺). Fails for multi-electron atoms.

Orbit vs Orbital: Bohr orbits are fixed circular paths (old model). Quantum orbitals are probability distributions — don't confuse.

Nuclear binding energy: Higher BE/nucleon = more stable nucleus. Iron (Fe-56) is most stable. Fusion of lighter elements and fission of heavier elements both release energy.

Half-life confusion: After 2 half-lives, fraction = ¼ (not 0). After 3, fraction = ⅛. Never zero — decay is exponential, asymptotic to zero.

pn junction forward bias: Current flows only when applied voltage > knee voltage (0.7V for Si). Not immediate at V = 0+.

Logic gate universality: NAND + NAND can make AND, OR, NOT. Remember: NAND is NOT-AND.

▶ Shortcut Formulas for NEET

⚡ de Broglie λ for electron: λ = 12.27/√V Å (V = accelerating potential in volts)

⚡ For hydrogen, energy of nth orbit = −13.6/n² eV. Radius = 0.529·n² Å.

⚡ Speed of electron in nth Bohr orbit: vn = c/137 · Z/n (α = 1/137 = fine structure constant)

⚡ Wavelength of emitted photon: 1/λ = R(1/n₁² − 1/n₂²). Longest λ in a series = n₁→n₁+1 transition.

⚡ Number of spectral lines from n2 to n1: N = (n₂−n₁)(n₂−n₁+1)/2

⚡ Activity after n half-lives: A = A₀/2ⁿ. Mass remaining = M₀/2ⁿ.

⚡ 1 amu = 931.5 MeV/c². To convert mass defect to energy: E(MeV) = Δm(amu) × 931.5

⚡ For fastest NEET: memorize that mass of proton ≈ mass of neutron ≈ 1.67×10⁻²⁷ kg ≈ 1 amu

Practice Questions

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