Class 11Chemistry · Physical ChemistryFull chapter

Structure of Atom

The whole chapter in one place — read it, then test yourself. Clear notes, key facts, a practice quiz, and worked NCERT solutions & PYQs.

Discovery of the Electron, Proton and Neutron

Quick answer Cathode ray, anode ray and alpha-particle bombardment experiments revealed that the atom is made of three fundamental subatomic particles — the electron, proton and neutron.

By the late nineteenth century, experiments with electric discharge through gases at very low pressure (in a discharge tube, also called a Crookes tube) revealed that atoms are not indivisible — they are built from smaller charged particles.

When a high voltage (of the order of 10,000 V) is applied across a discharge tube containing gas at very low pressure (below about 0.01 mm Hg), invisible rays are found to travel from the cathode to the anode; these are called cathode rays, and their passage can be observed because they cause certain materials, such as zinc sulphide, to fluoresce. Cathode rays: (i) start from the cathode and move towards the anode; (ii) travel in straight lines in the absence of an electric or magnetic field; (iii) are deflected by external electric and magnetic fields in the manner expected of negatively charged particles; and (iv) have characteristics that do not depend on the material of the electrodes or the nature of the gas present in the tube. These observations showed that cathode rays consist of negatively charged particles that are a universal constituent of all matter. J. J. Thomson named this particle the electron and, using the deflections produced by electric and magnetic fields, measured its charge-to-mass ratio, e/m.

R. A. Millikan's oil drop experiment independently measured the actual charge, e, carried by the electron. Combining Millikan's value of e with Thomson's e/m ratio gives the mass of the electron.

Using a discharge tube with a perforated cathode, Eugen Goldstein observed a new kind of rays travelling in a direction opposite to the cathode rays, called canal rays or anode rays, made up of positively charged particles. Unlike cathode rays, the charge-to-mass ratio of these particles depended on the gas taken in the tube. The smallest and lightest positively charged particle was obtained when the tube was filled with hydrogen gas; this particle was named the proton.

Because atoms are electrically neutral overall but the proton accounts for only part of the atomic mass in heavier atoms, the existence of a third, neutral particle was suspected. In 1932, James Chadwick bombarded a thin sheet of beryllium with fast alpha particles and obtained a new, highly penetrating radiation that was not deflected by electric or magnetic fields. This neutral particle, of mass almost equal to that of the proton, was named the neutron.

Worked example. Determine the number of protons, neutrons and electrons in the chloride ion, written as 3517Cl (mass number A = 35, atomic number Z = 17, ionic charge = −1).

  • Number of protons = atomic number, Z = 17.
  • Number of neutrons = A − Z = 35 − 17 = 18.
  • Number of electrons in the neutral atom would equal the number of protons, i.e. 17; since the species carries a charge of −1, it has one extra electron, so the number of electrons in Cl = 17 + 1 = 18.

So the chloride ion 35Cl contains 17 protons, 18 neutrons and 18 electrons.

Charge-to-mass ratio of electron e/m = 1.758820 × 10^11 C kg⁻¹
Charge of electron (Millikan) e = 1.602176 × 10⁻19 C
Mass of electron m_e = 9.10939 × 10⁻31 kg
Mass of proton m_p ≈ 1.67262 × 10⁻27 kg
Mass of neutron m_n ≈ 1.67493 × 10⁻27 kg
Number of neutrons n = A − Z A = mass number, Z = atomic number
Remember
  • Cathode rays are streams of electrons, discovered to be a universal, negatively charged constituent of all atoms.
  • Thomson measured the electron's e/m ratio; Millikan's oil drop experiment measured its charge, giving its mass.
  • Anode (canal) rays revealed positively charged protons; the lightest one came from hydrogen gas.
  • Chadwick discovered the neutral neutron by bombarding beryllium with alpha particles.
  • Relative charges are electron = −1, proton = +1, neutron = 0; number of neutrons = mass number − atomic number.

Atomic Models: Thomson and Rutherford

Quick answer Thomson's uniform sphere-of-charge model was replaced by Rutherford's nuclear model after alpha-particle scattering showed a small, dense, positively charged nucleus surrounded by mostly empty space.

J. J. Thomson proposed that an atom is a sphere of uniform positive charge with electrons embedded in it, much like seeds in a watermelon (often called the plum-pudding model). The positive and negative charges were assumed to be equal in magnitude, making the atom electrically neutral, and the mass was assumed to be evenly spread throughout the sphere. While this model could account for the overall neutrality of the atom, it could not explain later experimental observations on how particles scatter off atoms.

Ernest Rutherford tested the structure of the atom by directing a beam of fast-moving, positively charged alpha (α) particles from a radioactive source at a very thin sheet of gold foil, with a circular fluorescent zinc sulphide screen around the foil to detect the scattered particles. He observed that:

  • most of the α-particles passed straight through the gold foil without any deflection;
  • a small fraction of the α-particles were deflected through small angles;
  • a very small fraction (about 1 in 20,000) bounced back, i.e. were deflected through angles close to 180°.

From these observations, Rutherford proposed the nuclear model of the atom: most of the space inside an atom is empty (which is why most α-particles pass straight through); almost the entire mass of the atom and all of its positive charge are concentrated in an extremely small region at the centre, called the nucleus (which repels and strongly deflects the rare α-particle that travels close to or almost head-on into it); and the electrons revolve around this nucleus in circular paths at high speed. The radius of the nucleus (of the order of 10−15 m) is about 105 times smaller than the radius of the atom (of the order of 10−10 m).

Rutherford's model, however, had a serious limitation: it could not explain the stability of the atom. Classical electromagnetic theory requires that a charged particle undergoing acceleration (as an electron does while revolving in a circular orbit) must continuously emit radiation and lose energy. An electron losing energy in this way should spiral inward and fall into the nucleus almost instantly, which contradicts the observed stability of atoms. The model also said nothing about how electrons are arranged around the nucleus or what energies they possess, so it could not account for the line spectra of atoms.

Worked example. Using Rutherford's observations, explain why only a tiny fraction of α-particles were deflected through large angles while the vast majority passed straight through the foil.

  • Since the atom is mostly empty space, an α-particle travelling anywhere except very close to a nucleus experiences almost no force and continues in a straight line — this explains why the large majority pass through undeflected.
  • The nucleus occupies only an extremely small fraction of the atom's volume, so very few α-particles happen to travel close enough to it to be strongly repelled.
  • An α-particle that does travel close to, or almost directly at, a nucleus experiences a very large electrostatic repulsion from the concentrated positive charge and large mass of the nucleus, and can be deflected through a large angle or even bounced straight back — this explains the rare large-angle deflections.
Approx. atomic radius r_atom ≈ 10⁻10 m
Approx. nuclear radius r_nucleus ≈ 10⁻15 m
Atom-to-nucleus size ratio r_atom : r_nucleus ≈ 10^5 : 1
Remember
  • Thomson's plum-pudding model pictured the atom as a uniform sphere of positive charge with embedded electrons.
  • Rutherford's alpha-scattering experiment showed most alpha particles pass straight through gold foil, but a few are deflected sharply.
  • This led to the nuclear model: a tiny, dense, positively charged nucleus surrounded by mostly empty space, with electrons revolving around it.
  • The atom's radius (~10⁻10 m) is about 10⁵ times the nucleus's radius (~10⁻15 m).
  • Rutherford's model could not explain why atoms are stable or account for atomic line spectra.

Bohr's Model of the Hydrogen Atom

Quick answer Bohr's model places the electron in fixed, quantized circular orbits of definite energy, successfully explaining the hydrogen atom's line spectrum, though it fails for multi-electron atoms and finer spectral effects.

To overcome the failure of Rutherford's model to explain atomic stability and line spectra, Niels Bohr proposed a model for the hydrogen atom (and other one-electron/hydrogen-like species) based on three postulates:

  1. An electron in an atom moves around the nucleus only in certain fixed circular paths of definite energy, called stationary states or orbits. As long as an electron remains in a given stationary state, its energy does not change, i.e. it does not radiate energy.
  2. Only those orbits are permitted for which the angular momentum of the electron is quantized, i.e. an integral multiple of h/2π: mvr = nh/2π, where n = 1, 2, 3, … is called the principal quantum number.
  3. Energy is emitted or absorbed by the atom only when an electron moves (jumps) from one stationary state to another, and the energy of the radiation absorbed or emitted equals the difference in energy between the two states: ΔE = E2 − E1 = hν.

Solving the model for a hydrogen-like species of atomic number Z gives the radius, velocity and energy of the electron in its nth orbit:

  • radius, rn = 0.529 × (n²/Z) Å;
  • velocity, vn = 2.18 × 106 × (Z/n) m s−1;
  • energy, En = −2.18 × 10−18 × (Z²/n²) J = −13.6 (Z²/n²) eV.

The negative sign shows that the electron is bound to the nucleus; as n increases, the energy becomes less negative (higher), and at n = ∞ the electron is completely free of the nucleus (E = 0, corresponding to ionization). When an electron falls from a higher orbit n2 to a lower orbit n1, radiation of a definite wavelength is emitted, given by the Rydberg formula, 1/λ = RH Z² (1/n1² − 1/n2²), where RH = 1.097 × 107 m−1 is the Rydberg constant. This correctly accounted for the observed line spectrum of the hydrogen atom, including the Lyman (n1 = 1), Balmer (n1 = 2), Paschen (n1 = 3), Brackett (n1 = 4) and Pfund (n1 = 5) series.

Despite this success, Bohr's model has several limitations: it fails to explain the line spectra of atoms or ions containing more than one electron; it cannot account for the splitting of spectral lines into finer lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect); it does not explain the ability of atoms to form molecules with covalent bonds; and it does not incorporate the wave nature of the electron or the uncertainty principle, since it assumes the electron follows a well-defined circular path.

Worked example. Calculate the energy needed to excite an electron in a hydrogen atom from n = 1 to n = 2, and the wavelength of radiation associated with this transition.

  • E1 = −2.18 × 10−18 × (1²/1²) = −2.18 × 10−18 J.
  • E2 = −2.18 × 10−18 × (1²/2²) = −5.45 × 10−19 J.
  • ΔE = E2 − E1 = (−5.45 × 10−19) − (−2.18 × 10−18) = 1.635 × 10−18 J, which must be absorbed.
  • Using ΔE = hc/λ: λ = hc/ΔE = (6.626 × 10−34 × 3 × 108) / (1.635 × 10−18) = 1.216 × 10−7 m = 121.6 nm.

This wavelength (121.6 nm, in the ultraviolet region) corresponds to the first line of the Lyman series.

Quantization of angular momentum mvr = nh/2π
Radius of nth Bohr orbit r_n = 0.529 × (n²/Z) Å
Velocity in nth orbit v_n = 2.18 × 10^6 × (Z/n) m s⁻¹
Energy of nth orbit E_n = −2.18 × 10⁻18 × (Z²/n²) J = −13.6 (Z²/n²) eV
Rydberg formula 1/λ = R_H Z² (1/n1² − 1/n2²) R_H = 1.097 × 10^7 m⁻¹
Energy of transition ΔE = hν = hc/λ = E_higher − E_lower
Remember
  • Bohr proposed that electrons move in fixed, non-radiating stationary orbits with quantized angular momentum, mvr = nh/2π.
  • Orbit radius grows as n², while orbit energy becomes less negative (higher) as n increases: E_n = −13.6(Z²/n²) eV.
  • Transitions between orbits emit or absorb radiation of a definite wavelength, correctly explaining the hydrogen spectral series.
  • Bohr's model fails for multi-electron atoms, cannot explain the Zeeman/Stark effects, and ignores the wave nature of the electron.

Dual Behaviour of Matter and Heisenberg's Uncertainty Principle

Quick answer Matter, like radiation, shows dual wave-particle behaviour (de Broglie's λ = h/mv), and Heisenberg's uncertainty principle shows that an electron's exact position and momentum can never both be known at once.

Experiments on black-body radiation and the photoelectric effect showed that light (and electromagnetic radiation in general) behaves as a stream of energy packets called photons, each of energy E = hν, in addition to its well-established wave nature (seen in diffraction and interference). Light therefore shows dual behaviour — it behaves as a wave in some experiments and as a stream of particles in others.

In 1924, Louis de Broglie proposed that this dual behaviour is not special to light but is a general property of all moving matter — electrons, protons, or even larger particles. He proposed that a wave, called a matter wave or de Broglie wave, of wavelength λ = h/mv (h = Planck's constant, m = mass, v = velocity) is associated with every moving particle. Because Planck's constant h is extremely small, the de Broglie wavelength is only significant (measurable) for particles of very small mass, such as electrons; for macroscopic objects (like a ball or a car), the associated wavelength is far too small to be observed.

Werner Heisenberg's uncertainty principle (1927) states that it is impossible to determine, simultaneously and with perfect accuracy, both the exact position and the exact momentum (or velocity) of a microscopic moving particle such as an electron. Mathematically, if Δx is the uncertainty in position and Δp is the uncertainty in momentum, then Δx · Δp ≥ h/4π. The more precisely the position of the electron is known, the less precisely its momentum can be known, and vice versa. This principle rules out the idea of an electron following a well-defined path or orbit (as Bohr had assumed), and instead leads to the idea that we can only speak of the probability of finding an electron at a given point in space — the foundation of the quantum mechanical model of the atom.

Worked example. Calculate the de Broglie wavelength associated with an electron (mass = 9.11 × 10−31 kg) moving with a velocity of 2.05 × 107 m s−1.

  • λ = h/(mv) = (6.626 × 10−34) / (9.11 × 10−31 × 2.05 × 107).
  • mv = 9.11 × 10−31 × 2.05 × 107 = 1.868 × 10−23 kg m s−1.
  • λ = 6.626 × 10−34 / 1.868 × 10−23 = 3.548 × 10−11 m ≈ 35.5 pm.

This wavelength is comparable to atomic dimensions, which is why wave-like behaviour of electrons (such as diffraction) is observable, unlike for everyday macroscopic objects.

de Broglie wavelength λ = h/mv = h/p
Photon energy E = hν = hc/λ
Heisenberg uncertainty principle Δx · Δp ≥ h/4π equivalently Δx · Δv ≥ h/4πm
Remember
  • Radiation (light) shows dual behaviour: wave nature (diffraction/interference) and particle nature (photoelectric effect).
  • de Broglie extended this duality to all matter: every moving particle has an associated wavelength, λ = h/mv.
  • The de Broglie wavelength is significant only for particles of very small mass, such as electrons.
  • Heisenberg's uncertainty principle, Δx·Δp ≥ h/4π, shows position and momentum of a micro-particle cannot both be known exactly.
  • This principle ruled out Bohr's fixed orbits and led to a probability-based (orbital) description of the electron.

Quantum Mechanical Model of the Atom

Quick answer The quantum mechanical model describes an electron by a probability-based orbital rather than a fixed path, with four quantum numbers fixing its size, shape, orientation and spin.

Erwin Schrödinger developed a mathematical equation, based on the wave nature of the electron, whose solutions (called wave functions, ψ) describe the electron in an atom. The wave function ψ itself has no direct physical meaning, but ψ² gives the probability density of finding the electron at a point in space. An orbital is defined as the three-dimensional region of space around the nucleus within which the probability of finding an electron is maximum (typically taken as 90%) — this is quite different from Bohr's idea of a fixed, well-defined, two-dimensional circular orbit.

The size, shape, orientation and energy of orbitals, as well as the spin of the electron occupying them, are described by four quantum numbers:

  • The principal quantum number, n (n = 1, 2, 3, …) determines the size and energy of the shell; the number of subshells in a shell equals n, the number of orbitals in a shell equals n², and the maximum number of electrons a shell can hold is 2n².
  • The azimuthal (angular momentum) quantum number, l (l = 0 to n − 1, for a given n) determines the shape of the subshell/orbital: l = 0, 1, 2, 3 correspond to s, p, d, f subshells respectively. The number of orbitals in a subshell is 2l + 1, and the orbital angular momentum of the electron is √[l(l + 1)] · h/2π.
  • The magnetic quantum number, ml (ml = −l to +l, including 0) describes the orientation of the orbital in space; there are (2l + 1) values of ml for a given l.
  • The spin quantum number, ms (+1/2 or −1/2) describes the two possible spin orientations of an electron about its own axis.

The shapes of orbitals depend on l: s orbitals (l = 0) are spherically symmetric about the nucleus, with size increasing as n increases (1s < 2s < 3s); a 2s orbital, for example, has a spherical node in addition to being larger than 1s. p orbitals (l = 1) are dumbbell-shaped, with two lobes on either side of the nucleus separated by a nodal plane passing through it; the three p orbitals of a given shell (px, py, pz) are identical in shape and energy but point along the three different axes. d orbitals (l = 2) have five orbitals (dxy, dyz, dxz, dx²−y², d); four of these have a four-lobed, clover-leaf shape lying along or between the axes, while d has two lobes along the z-axis together with a doughnut-shaped ring of electron density around the centre in the xy-plane. The total number of nodes (regions of zero probability) in an orbital equals n − 1, of which l are angular nodes and (n − l − 1) are radial nodes.

Worked example. Identify which of the following sets of quantum numbers (n, l, ml, ms) are not permissible for an electron, giving reasons.

  1. n = 1, l = 1, ml = 0, ms = +1/2
  2. n = 2, l = 1, ml = 0, ms = −1/2
  3. n = 3, l = 0, ml = 0, ms = +1/2
  4. n = 2, l = 2, ml = −1, ms = +1/2

For a given n, l can only range from 0 to n − 1. In set (1), n = 1 allows only l = 0, but l = 1 is given, so this set is not permissible. Set (2) is valid: for n = 2, l can be 0 or 1, and for l = 1, ml can be −1, 0 or +1, so ml = 0 is allowed. Set (3) is valid: for n = 3, l can be 0, 1 or 2, and l = 0 requires ml = 0. In set (4), n = 2 allows only l = 0 or 1, but l = 2 is given, so this set is also not permissible. So sets (1) and (4) are not permissible; sets (2) and (3) are permissible.

Orbitals per shell Number of orbitals = n²
Max electrons per shell Maximum electrons = 2n²
Orbitals per subshell Number of orbitals = 2l + 1
Orbital angular momentum √[l(l+1)] · h/2π
Nodes in an orbital Total nodes = n − 1; radial nodes = n − l − 1; angular nodes = l
Remember
  • An orbital is the 3-D region around the nucleus where the probability of finding an electron is maximum (usually 90%).
  • Four quantum numbers — n, l, ml, ms — together specify the size, shape, orientation and spin of an electron.
  • n fixes shell size/energy; l fixes subshell shape (s, p, d, f); ml fixes orbital orientation; ms fixes electron spin.
  • s orbitals are spherical; p orbitals are dumbbell-shaped (px, py, pz); d orbitals are mostly four-lobed, except dz².
  • Total nodes in an orbital = n − 1, split into l angular nodes and (n − l − 1) radial nodes.

Aufbau Principle, Pauli Exclusion, Hund's Rule and Electronic Configurations

Quick answer The Aufbau principle, Pauli exclusion principle and Hund's rule together determine how electrons fill orbitals to give an atom's ground-state electronic configuration.

Once the shapes and relative energies of orbitals are known, the ground-state electronic configuration of an atom — how its electrons are distributed among the various orbitals — is built up using three rules.

The Aufbau principle states that, in the ground state of an atom, orbitals are filled in order of increasing energy. The order of filling is obtained from the (n + l) rule: an orbital with a lower value of (n + l) is filled before one with a higher value; if two orbitals have the same (n + l) value, the orbital with the lower value of n is filled first. This gives the familiar filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, ….

The Pauli exclusion principle states that no two electrons in an atom can have the same set of all four quantum numbers. Equivalently, an orbital can hold a maximum of two electrons, and these two electrons must have opposite (paired) spins.

Hund's rule of maximum multiplicity states that electron pairing in the orbitals of a subshell (such as the three p orbitals or the five d orbitals, which all have the same energy) does not begin until each orbital of that subshell contains one electron each, and all these singly occupied electrons have parallel spins.

Applying these rules generally gives the correct configuration, but for a few elements the configuration predicted by the simple Aufbau filling order is not the one actually observed, because completely filled (s², p⁶, d¹⁰, f¹⁴) and exactly half-filled (s¹, p³, d⁵, f⁷) sets of degenerate orbitals have extra stability. This extra stability arises from (i) the symmetrical distribution of electrons among the orbitals of a subshell, and (ii) a larger exchange energy released when electrons of parallel spin occupy such a symmetrical arrangement.

Worked example. Write the ground-state electronic configuration of chromium (Cr, Z = 24), and explain why it differs from the configuration predicted by simply filling orbitals in order of increasing energy.

  • Simple Aufbau filling for Z = 24 would predict: [Ar] 3d4 4s2.
  • The actual, experimentally observed configuration is: [Ar] 3d5 4s1.
  • One electron shifts from the 4s orbital into the 3d orbitals so that the 3d subshell becomes exactly half-filled (3d5), with all five d orbitals singly occupied and parallel spins (following Hund's rule). This symmetrical, half-filled arrangement, together with the larger exchange energy it allows, makes 3d54s1 more stable than 3d44s2.

A similar shift occurs for copper (Cu, Z = 29): instead of the expected [Ar] 3d9 4s2, the observed configuration is [Ar] 3d10 4s1, because a completely filled 3d10 subshell is more stable.

(n + l) rule Lower (n+l) fills first; equal (n+l) → lower n fills first
Pauli exclusion principle Max. 2 electrons per orbital, with opposite spins
Chromium exception Cr (Z = 24): [Ar] 3d⁵ 4s¹ not [Ar] 3d⁴4s²
Copper exception Cu (Z = 29): [Ar] 3d¹⁰ 4s¹ not [Ar] 3d⁹4s²
Remember
  • The Aufbau principle fills orbitals in order of increasing (n + l) value, with ties broken by the lower n.
  • The Pauli exclusion principle limits every orbital to a maximum of two electrons, which must have opposite spins.
  • Hund's rule requires degenerate orbitals to be singly occupied (with parallel spins) before any pairing occurs.
  • Exactly half-filled and completely filled subshells have extra stability from symmetry and exchange energy.
  • This extra stability explains exceptions to simple Aufbau filling, such as Cr ([Ar]3d⁵4s¹) and Cu ([Ar]3d¹⁰4s¹).

Key facts & terms

Every formula in this chapter, in one place — screenshot it before your exam.

e/m = 1.758820 × 10^11 C kg⁻¹
Charge-to-mass ratio of electron
e = 1.602176 × 10⁻19 C
Charge of electron (Millikan)
m_e = 9.10939 × 10⁻31 kg
Mass of electron
m_p ≈ 1.67262 × 10⁻27 kg
Mass of proton
m_n ≈ 1.67493 × 10⁻27 kg
Mass of neutron
n = A − Z
Number of neutrons
r_atom ≈ 10⁻10 m
Approx. atomic radius
r_nucleus ≈ 10⁻15 m
Approx. nuclear radius
r_atom : r_nucleus ≈ 10^5 : 1
Atom-to-nucleus size ratio
mvr = nh/2π
Quantization of angular momentum
r_n = 0.529 × (n²/Z) Å
Radius of nth Bohr orbit
v_n = 2.18 × 10^6 × (Z/n) m s⁻¹
Velocity in nth orbit
E_n = −2.18 × 10⁻18 × (Z²/n²) J = −13.6 (Z²/n²) eV
Energy of nth orbit
1/λ = R_H Z² (1/n1² − 1/n2²)
Rydberg formula
ΔE = hν = hc/λ = E_higher − E_lower
Energy of transition
λ = h/mv = h/p
de Broglie wavelength
E = hν = hc/λ
Photon energy
Δx · Δp ≥ h/4π
Heisenberg uncertainty principle
Number of orbitals = n²
Orbitals per shell
Maximum electrons = 2n²
Max electrons per shell
Number of orbitals = 2l + 1
Orbitals per subshell
√[l(l+1)] · h/2π
Orbital angular momentum
Total nodes = n − 1; radial nodes = n − l − 1; angular nodes = l
Nodes in an orbital
Lower (n+l) fills first; equal (n+l) → lower n fills first
(n + l) rule
Max. 2 electrons per orbital, with opposite spins
Pauli exclusion principle
Cr (Z = 24): [Ar] 3d⁵ 4s¹
Chromium exception
Cu (Z = 29): [Ar] 3d¹⁰ 4s¹
Copper exception

Test yourself

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0 correct · 0/12 answered
Q1 Discovery of the electron easy

Cathode rays, produced in a discharge tube at low pressure, were found to consist of streams of negatively charged particles regardless of the gas taken or the material of the electrodes. This universal negatively charged particle is called the:

Q2 Discovery of the electron easy

R. A. Millikan's oil drop experiment was used to determine the:

Q3 Discovery of the neutron medium

The neutron was discovered by James Chadwick by bombarding a thin sheet of which element with alpha particles?

Q4 Rutherford's nuclear model medium

In Rutherford's alpha-particle scattering experiment, the observation that only a very small fraction of alpha particles were deflected by large angles (close to 180°) led to the conclusion that:

Q5 Limitations of Rutherford's model medium

The major drawback of Rutherford's nuclear model of the atom was that it could not explain:

Q6 Bohr's postulates medium

According to Bohr's model, the angular momentum of an electron revolving in its nth stationary orbit is quantized according to the relation:

Q7 Bohr model — orbit radius hard

Using r_n = 0.529 (n²/Z) Å, calculate the radius of the second orbit (n = 2) of a hydrogen atom (Z = 1).

Q8 Limitations of Bohr's model medium

Which of the following is NOT a limitation of Bohr's model of the hydrogen atom?

Q9 de Broglie relation medium

The de Broglie wavelength (λ) associated with a particle of mass m moving with velocity v is given by:

Q10 Heisenberg's uncertainty principle medium

Heisenberg's uncertainty principle is mathematically expressed as:

Q11 Quantum numbers hard

Which one of the following sets of quantum numbers (n, l, ml, ms) is permissible for an electron in an atom?

Q12 Electronic configuration medium

The ground-state electronic configuration of copper (Cu, Z = 29) is:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Calculate the number of protons, neutrons and electrons in the chloride ion, ³⁵Cl⁻ (mass number = 35, atomic number = 17).Atomic number, mass number and isotopes

For any atom or ion, the number of protons is always equal to the atomic number (Z), and the number of neutrons is (mass number A − atomic number Z).

  • Atomic number, Z = 17, so number of protons = 17.
  • Number of neutrons = A − Z = 35 − 17 = 18.
  • In a neutral chlorine atom, number of electrons = number of protons = 17. Since the ion carries a charge of −1 (Cl), it has gained one extra electron.

Therefore, in Cl: protons = 17, neutrons = 18, and electrons = 18.

2 Calculate the energy associated with the first orbit (n = 1) and the second orbit (n = 2) of the hydrogen atom. Hence calculate the energy required to excite the electron from n = 1 to n = 2.Bohr's model — orbit energies

The energy of the electron in the nth orbit of a hydrogen atom (Z = 1) is given by En = −2.18 × 10−18 (Z²/n²) J.

  • For n = 1: E1 = −2.18 × 10−18 × (1²/1²) = −2.18 × 10−18 J.
  • For n = 2: E2 = −2.18 × 10−18 × (1²/2²) = −2.18 × 10−18/4 = −5.45 × 10−19 J.
  • Energy absorbed on excitation, ΔE = E2 − E1 = (−5.45 × 10−19) − (−2.18 × 10−18) = 1.635 × 10−18 J (absorbed, since E2 > E1).

So the electron must absorb 1.635 × 10−18 J (about 10.2 eV) of energy to jump from the n = 1 to the n = 2 orbit.

3 Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s⁻¹ (mass of electron = 9.11 × 10⁻31 kg, h = 6.626 × 10⁻34 J s).de Broglie wavelength

The de Broglie wavelength is given by λ = h/(mv).

  • Here, h = 6.626 × 10−34 J s, m = 9.11 × 10−31 kg, v = 2.05 × 107 m s−1.
  • mv = 9.11 × 10−31 × 2.05 × 107 = 1.868 × 10−23 kg m s−1.
  • λ = 6.626 × 10−34 / 1.868 × 10−23 = 3.548 × 10−11 m (that is, about 35.5 pm).
4 A microscope is used to locate an electron in an atom to within a distance of 0.1 Å (1 × 10⁻11 m). What is the uncertainty in the measurement of its velocity? (h = 6.626 × 10⁻34 J s, mass of electron = 9.11 × 10⁻31 kg)Heisenberg's uncertainty principle

By Heisenberg's uncertainty principle, Δx · Δp ≥ h/4π, and since Δp = m·Δv, this gives Δv ≥ h/(4π m Δx).

  • Δx = 1 × 10−11 m, m = 9.11 × 10−31 kg, h = 6.626 × 10−34 J s.
  • 4π m Δx = 4 × 3.1416 × 9.11 × 10−31 × 1 × 10−11 = 1.145 × 10−40.
  • Δv ≥ 6.626 × 10−34 / 1.145 × 10−40 = ≈ 5.79 × 106 m s−1.

The uncertainty in velocity is extremely large compared to typical electron speeds, showing that position and velocity of an electron cannot both be known precisely at the same time.

5 Using the Aufbau principle and Hund's rule, write the ground-state electronic configurations of chromium (Cr, Z = 24) and copper (Cu, Z = 29), and explain why they differ from the configuration expected by simply filling orbitals in order of increasing energy.Aufbau exceptions / extra stability

Filling orbitals strictly in order of increasing energy (Aufbau order) would predict:

  • Cr (Z = 24): expected [Ar] 3d4 4s2; actual configuration is [Ar] 3d5 4s1.
  • Cu (Z = 29): expected [Ar] 3d9 4s2; actual configuration is [Ar] 3d10 4s1.

In both cases, one electron shifts from the 4s orbital into the 3d orbitals so that the d subshell becomes exactly half-filled (d5, as in Cr) or completely filled (d10, as in Cu). Half-filled and fully-filled degenerate sets of orbitals are more stable because of (i) the symmetrical distribution of electrons among the orbitals and (ii) a larger exchange energy released when electrons with parallel spins occupy a completely half-filled or filled subshell. This extra stability makes 3d54s1 and 3d104s1 more favourable than the naively expected configurations.

6 Which of the following sets of quantum numbers is not permissible for an electron? State the reason in each case. (a) n = 1, l = 1, ml = 0, ms = +1/2 (b) n = 3, l = 2, ml = −2, ms = +1/2 (c) n = 2, l = 0, ml = 1, ms = −1/2Permissible quantum numbers

For a given n, l can only take integer values from 0 to (n−1), and for a given l, ml can only take integer values from −l to +l (including 0).

  • (a) n = 1 allows only l = 0, but l = 1 is given — not permissible.
  • (b) n = 3 allows l = 0, 1, 2; for l = 2, ml can be −2, −1, 0, +1, +2, so ml = −2 is allowed, and ms = +1/2 is valid — this set is permissible.
  • (c) l = 0 allows only ml = 0, but ml = 1 is given — not permissible.

So sets (a) and (c) are not permissible; set (b) is permissible.

Previous-year board questions 4

Q1 Which quantum number determines the shape of an orbital? Name the shapes of the s and p orbitals. CBSE 2023 1 mark

The azimuthal (angular momentum) quantum number, l, determines the shape of an orbital.

An s orbital (l = 0) is spherical in shape, while a p orbital (l = 1) has a dumbbell shape, consisting of two lobes on either side of the nucleus.

Q2 What was the main drawback of Rutherford's nuclear model of the atom, and how did Bohr's model overcome it? CBSE 2022 2 marks

According to Rutherford's model, electrons revolve around the nucleus in circular orbits. However, classical electromagnetic theory predicts that an accelerating (revolving) charged particle must continuously radiate energy. An electron losing energy in this way would gradually spiral inward and fall into the nucleus, making the atom unstable — but real atoms are known to be stable. This was the main drawback of Rutherford's model.

Bohr overcame this by postulating that an electron can revolve only in certain fixed, allowed orbits (stationary states) for which its angular momentum is quantized (mvr = nh/2π), and that the electron does not radiate energy as long as it remains in one of these stationary orbits. Energy is emitted or absorbed only when the electron jumps between two such orbits. This assumption accounted for the observed stability of the atom, which Rutherford's model could not explain.

Q3 An electron in a hydrogen atom undergoes a transition from n = 4 to n = 2. Calculate the wavelength of the spectral line emitted. (R_H = 1.097 × 10⁷ m⁻¹) CBSE 2020 3 marks

The wavelength of a spectral line is given by the Rydberg formula:

1/λ = RH (1/n1² − 1/n2²), where n1 is the lower energy level and n2 is the higher energy level.

  • Here n1 = 2, n2 = 4, RH = 1.097 × 107 m−1.
  • 1/n1² − 1/n2² = 1/4 − 1/16 = 4/16 − 1/16 = 3/16 = 0.1875.
  • 1/λ = 1.097 × 107 × 0.1875 = 2.057 × 106 m−1.
  • λ = 1 / (2.057 × 106) = 4.86 × 10−7 m (486 nm).

This falls in the visible region of the spectrum and corresponds to a line of the Balmer series (since the electron falls to n = 2).

Q4 State the postulates of Bohr's model of the atom. Derive an expression for the radius of the nth orbit of the hydrogen atom. CBSE 2019 5 marks

Postulates of Bohr's model of the hydrogen atom:

  1. The electron revolves around the nucleus only in certain fixed circular paths of definite energy, called stationary states or orbits; as long as the electron stays in a particular orbit, it does not lose or radiate energy.
  2. The angular momentum of the electron in an orbit is quantized: mvr = nh/2π, where n = 1, 2, 3 … is the principal quantum number.
  3. Energy is emitted or absorbed by the atom, in the form of radiation, only when an electron jumps from one stationary state to another, with ΔE = hν equal to the difference in energy of the two states.

Derivation of the radius of the nth orbit:

Consider an electron of mass m and charge −e revolving with velocity v in a circular orbit of radius r around a nucleus of charge +Ze. The electrostatic force of attraction between the nucleus and electron supplies the centripetal force needed for circular motion:

Ze²/(4πε₀r²) = mv²/r, which gives mv² = Ze²/(4πε₀r) ... (1)

From Bohr's quantization postulate, mvr = nh/2π, so v = nh/(2πmr) ... (2)

Squaring (2): v² = n²h²/(4π²m²r²). Substituting this into (1):

m × n²h²/(4π²m²r²) = Ze²/(4πε₀r)

On simplifying and solving for r:

rn = n²h²ε₀ / (π m Z e²)

Putting in the values of h, m, e and ε₀ for the hydrogen atom (Z = 1), this reduces to the familiar result rn = 0.529 × (n²/Z) Å, so the first orbit (n = 1) of hydrogen has a radius of 0.529 Å.

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