Competitive Level — 6 Marks Questions — Tap a question to reveal full solution

⚛️ Structure of Atom (Q21–Q40)

Competitive Level — 6 Marks Questions — Tap a question to reveal full solution

🔵 Step 1: Frequency of Incident Light

ν = c/λ = 3×10⁸ ÷ 200×10⁻⁹ = 1.5×10¹⁵ Hz

🟢 Step 2: Energy of Incident Photon

E = hν = 6.626×10⁻³⁴ × 1.5×10¹⁵ = 9.94×10⁻¹⁹ J = 6.21 eV

🟣 Step 3: Maximum KE of Photoelectron

KE_max = E − φ = 6.21 − 4.2 = 2.01 eV (= 3.22×10⁻¹⁹ J)

✅ Final Answer

Frequency = 1.5×10¹⁵ Hz  |  KE_max = 2.01 eV

🔵 Step 1: Einstein's Photoelectric Equation

KE_max = h(ν − ν₀)

🟢 Step 2: Substitute Values

KE_max = 6.626×10⁻³⁴ × (8×10¹⁴ − 5×10¹⁴)

= 6.626×10⁻³⁴ × 3×10¹⁴ = 1.99×10⁻¹⁹ J

✅ Final Answer

KE_max = 1.99×10⁻¹⁹ J = 1.24 eV

🔵 Step 1: Energy of One Photon

E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸) ÷ 589×10⁻⁹

= 3.375×10⁻¹⁹ J

🟢 Step 2: Number of Photons per Second

n = Power ÷ Energy per photon = 100 ÷ 3.375×10⁻¹⁹

✅ Final Answer

Photons emitted per second = 2.96×10²⁰

🔵 Step 1: Bohr Velocity Formula

v_n = (2.18×10⁶ × Z)/n  m s⁻¹

🟢 Step 2: Substitute Z=1 (H atom), n=1

v₁ = 2.18×10⁶ × 1 ÷ 1

✅ Final Answer

v₁ = 2.18×10⁶ m s⁻¹

🔵 Step 1: Energy Formula

ΔE = −2.18×10⁻¹⁸ (1/n_f² − 1/n_i²)

n_i=∞ (E=0), n_f=1

🟢 Step 2: Substitute

ΔE = −2.18×10⁻¹⁸ (1/1 − 0) = −2.18×10⁻¹⁸ J (energy released)

✅ Final Answer

Energy released = 2.18×10⁻¹⁸ J per atom (numerically equal to the ionization energy of H)

🔵 Step 1: Protons and Electrons

Protons = Atomic number Z = 24

Electrons (neutral atom) = 24

🟢 Step 2: Neutrons

Neutrons = Mass number − Atomic number = 52 − 24 = 28

🟣 Step 3: Electronic Configuration

This is Chromium (Cr): [Ar] 3d⁵ 4s¹ (half-filled d exception)

✅ Final Answer

Protons = 24, Neutrons = 28, Electrons = 24

Configuration = [Ar] 3d⁵ 4s¹

🔵 Step 1: Identify the Transition

Balmer series: transitions end at n=2

1st line: n=3→2  |  2nd line: n=4→2

🟢 Step 2: Apply Rydberg Equation

1/λ = 1.097×10⁷ × (1/4 − 1/16) = 1.097×10⁷ × 0.1875

🟣 Step 3: Calculate

1/λ = 2.057×10⁶ m⁻¹

✅ Final Answer

λ = 486 nm (H_β line)

🔵 Step 1: Formula

Number of lines = n(n−1)/2, where n = highest level involved

🟢 Step 2: Substitute n=4

= 4×3 ÷ 2 = 6

✅ Final Answer

Total spectral lines = 6

🔵 Step 1: Convert Wavelength

λ = 4000 Å = 4×10⁻⁷ m

🟢 Step 2: Apply E = hc/λ

E = (6.626×10⁻³⁴ × 3×10⁸) ÷ 4×10⁻⁷ = 4.97×10⁻¹⁹ J

✅ Final Answer

E = 4.97×10⁻¹⁹ J = 3.11 eV

🔵 Step 1: Convert KE to Joules

KE = 100 × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁷ J

🟢 Step 2: Calculate Momentum

p = √(2m_eKE) = √(2 × 9.11×10⁻³¹ × 1.6×10⁻¹⁷)

= 5.4×10⁻²⁴ kg m s⁻¹

🟣 Step 3: Calculate Wavelength

λ = h/p = 6.626×10⁻³⁴ ÷ 5.4×10⁻²⁴

✅ Final Answer

λ = 1.227×10⁻¹⁰ m = 1.227 Å

Heisenberg's Uncertainty Principle: It is impossible to simultaneously determine the exact position and exact momentum (or velocity) of a microscopic particle like an electron. The product of the uncertainties in position (Δx) and momentum (Δp) is always greater than or equal to h/4π.

🔵 Step 1: Formula

Δx · Δp ≥ h/4π  ⟹  Δv ≥ h/(4π m_e Δx)

🟢 Step 2: Substitute Values

Δv ≥ 6.626×10⁻³⁴ ÷ (4π × 9.11×10⁻³¹ × 1×10⁻¹²)

= 6.626×10⁻³⁴ ÷ 1.145×10⁻⁴¹

✅ Final Answer

Δv ≥ 5.79×10⁷ m s⁻¹

In Bohr's model, E_n = −13.6 Z²/n² eV. The negative sign indicates that the electron in an orbit is in a bound state — it has lower energy than a free electron at rest infinitely far from the nucleus (which is taken as the zero-energy reference, E = 0 at n = ∞).

• As n increases, E_n becomes less negative (closer to zero), meaning the electron becomes progressively less tightly bound as it moves to higher orbits.

• At n = ∞, E = 0, corresponding to a free electron (atom is ionized).

• The magnitude of the negative energy represents the energy that must be supplied to remove the electron completely from that orbit (ionization energy), or equivalently the energy released when a free electron is captured into that orbit.

Orbit (Bohr's concept):
• A well-defined circular path around the nucleus in which the electron is assumed to revolve
• Has a fixed radius and fixed energy
• Both position and momentum of the electron are assumed known simultaneously — this violates Heisenberg's Uncertainty Principle
• Represents 2-dimensional planar motion

Orbital (Quantum mechanical concept):
• A three-dimensional region of space around the nucleus where the probability of finding an electron is maximum (usually ≥90%)
• Does not describe a fixed path; only gives a probability distribution
• Consistent with the wave nature of the electron and the Uncertainty Principle
• Obtained by solving the Schrödinger wave equation; has definite shape (s, p, d, f) and energy

s orbital (l=0): Spherically symmetric around the nucleus; probability of finding the electron is the same in all directions at a given distance. Number of angular nodal planes = l = 0.

p orbital (l=1): Dumbbell-shaped, with two lobes on either side of the nucleus along an axis (p_x, p_y, p_z). Number of angular nodal planes = l = 1 (one plane passing through the nucleus, perpendicular to the orbital's axis, where probability is zero).

d orbital (l=2): Mostly four-lobed "cloverleaf" shapes (d_xy, d_yz, d_xz, d_x²−y²), except d_z² which has two lobes along the z-axis plus a doughnut-shaped ring. Number of angular nodal planes = l = 2.

Aufbau Principle: Electrons are filled into orbitals in order of increasing energy (lower energy orbitals are filled first), following the (n+l) rule. Example: the 4s orbital (n+l = 4+0 = 4) is filled before the 3d orbital (n+l = 3+2 = 5).

Pauli's Exclusion Principle: No two electrons in the same atom can have identical values for all four quantum numbers; equivalently, an orbital can hold a maximum of two electrons, and they must have opposite spins. Example: in the 1s² configuration, one electron has m_s=+½ and the other has m_s=−½.

Hund's Rule of Maximum Multiplicity: Electrons are added singly to each degenerate orbital of a subshell (with parallel spins) before any orbital receives a second electron. Example: in nitrogen's 2p³ configuration, each of the three 2p orbitals gets one electron (all spins parallel) rather than pairing up in fewer orbitals.

🔵 Step 1: Configuration of Neutral Fe

Fe (Z=26): [Ar] 3d⁶ 4s²

🟢 Step 2: Remove 3 Electrons for Fe³⁺

Remove both 4s electrons first, then one 3d electron:

Fe³⁺ = [Ar] 3d⁵

🟣 Step 3: Apply Hund's Rule to d⁵

All 5 electrons occupy the 5 d-orbitals singly (half-filled, maximum unpaired)

✅ Final Answer

Unpaired electrons = 5

Magnetic behaviour = Paramagnetic (strongly, due to 5 unpaired electrons)

🔵 Element with Z=11 (Sodium, Na)

Configuration: 1s² 2s² 2p⁶ 3s¹

Group = 1 (one valence electron)  |  Period = 3 (outermost shell n=3)

🟢 Element with Z=17 (Chlorine, Cl)

Configuration: 1s² 2s² 2p⁶ 3s² 3p⁵

Group = 17 (7 valence electrons)  |  Period = 3 (outermost shell n=3)

✅ Final Answer

Z=11 (Na): Group 1, Period 3

Z=17 (Cl): Group 17, Period 3

🔵 Step 1: Series Limit Means n=∞ → n=1

1/λ = R_H (1/1² − 1/∞²) = R_H × 1

🟢 Step 2: Substitute R_H

1/λ = 1.097×10⁷ m⁻¹

✅ Final Answer

λ = 1 ÷ 1.097×10⁷ = 9.12×10⁻⁸ m = 91.2 nm

🔵 Step 1: Radius Formula

r_n = 0.529 × n²/Z  Å

🟢 Step 2: Radius of He⁺ (Z=2, n=2)

r = 0.529 × 4/2 = 1.058 Å

🟣 Step 3: Radius of H (Z=1, n=1)

r = 0.529 × 1/1 = 0.529 Å

✅ Final Answer

r(He⁺, n=2) = 1.058 Å = 2 × r(H, n=1)

🔵 Step 1: Identify n for "Second Excited State"

Ground state = n=1, 1st excited state = n=2, 2nd excited state = n=3

🟢 Step 2: Energy Formula (Z=3 for Li²⁺)

E_n = −13.6 × Z²/n²  eV

🟣 Step 3: Substitute Z=3, n=3

E₃ = −13.6 × 9/9 = −13.6 eV

✅ Final Answer

E₃ = −13.6 eV = −2.18×10⁻¹⁸ J

(Coincidentally equal in magnitude to the ground state energy of hydrogen, since Z² = n² = 9)