every question. answered.

λ₁ = 350 nm, λ₂ = 450 nm; maximum velocities differ by factor 2 (v₁ = 2v₂). Find work function φ.

Einstein’s equations:hc/λ₁ = φ + ½mᵉv₁² ...(1)hc/λ₂ = φ + ½mᵉv₂² ...(2)Substitute v₁ = 2v₂ ⇒ v₁² = 4v₂² and subtract (2) from (1):hc(1/λ₁ − 1/λ₂) = (3/2)mᵉv₂²Calculate left side:1/350nm − 1/450nm = 100/(350×450) nm⁻¹ = 6.349×10⁵ m⁻¹hc × 6.349×10⁵ = 1.262×10⁻¹⁹ JSolve for v₂²:v₂² = (2/3) × 1.262×10⁻¹⁹ / 9.1×10⁻³¹ = 9.245×10¹⁰ m²/s²Substitute into equation (2):hc/λ₂ = (6.626×10⁻³⁴×3×10⁸) / (450×10⁻⁹) = 4.417×10⁻¹⁹ J½mᵉv₂² = 0.5 × 9.1×10⁻³¹ × 9.245×10¹⁰ = 4.206×10⁻²⁰ Jφ = 4.417×10⁻¹⁹ − 0.4206×10⁻¹⁹ ≈ 3.996×10⁻¹⁹ J ≈ 2.5 eVAnswer: Work function φ ≈ 2.5 eV

Charges −q, +q, −q at corners of equilateral triangle (side a). Find net force on +q at centroid O.

Distance from centroid to vertex: r = a/√3Force magnitude from each corner on +q at O:F = kq²/r² = kq²/(a²/3) = 3kq²/a²Directions (place +q corner at top, −q corners at bottom-left and bottom-right):From +q at top: repels downward (0, −F)From −q at bottom-L: attracts toward B-L (−F√3/2, −F/2)From −q at bottom-R: attracts toward B-R (+F√3/2, −F/2)Sum of components:Ex = 0 − F√3/2 + F√3/2 = 0Ey = −F − F/2 − F/2 = −2FF_net = 2F = 6kq²/a²Answer: F_net = 6kq²/a², directed away from the +q corner (toward midpoint of opposite side)Symmetry only guarantees x-cancellation here. The y-components all add never assume the force is zero without resolving components.

Two identical charges q₁ = q₂ = 5 μC. Find largest charge q to transfer so force decreases to 1/2.5 of original.

Step-1: After transferring charge q: new charges = (5−q) and (5+q) μCStep-2: New force = F₀ / 2.5:k(5−q)(5+q)/r² = k×25/(2.5×r²) = 10k/r²(5−q)(5+q) = 1025 − q² = 10 ⇒ q² = 15 ⇒ q = √15 μCAnswer: q = √15 μC ≈ 3.87 μCNote: (5−q)(5+q) = 25−q² is the difference-of-squares identity. Transferring charge always decreases the product q₁q₂ for charges of the same sign.

Charges +Q, −Q, +Q, −Q at corners of a square (side 5 cm). Find field at centre. (Q = 1 μC)

With alternating charges, the two +Q charges are at diagonally opposite corners, and the two −Q charges are at diagonally opposite corners.Field at centre from each pair:Both +Q fields: exactly opposite directions ⇒ cancelBoth −Q fields: exactly opposite directions ⇒ cancelEvery field vector has an equal and opposite partner by symmetry.Answer: Electric field at centre = 0 (zero)

P°A = 7×10³ Pa, P°B = 12×10³ Pa at 350 K. Liquid contains 40 mol% A. Find vapour composition.

Step-1: Partial pressures (Raoult’s Law):PA = 0.40 × 7000 = 2800 PaPB = 0.60 × 12000 = 7200 PaStep-2: Total pressure:P_total = 2800 + 7200 = 10000 PaStep-3: Vapour mole fractions (Dalton’s Law):yA = 2800 / 10000 = 0.28yB = 7200 / 10000 = 0.72Answer: Mole fraction of A in vapour = 0.28, B in vapour = 0.72The more volatile component (B, higher P°) is enriched in the vapour. This is the principle behind fractional distillation.

O₂ at 293 K, partial pressure = 0.98 bar, KH = 34 kbar. Find solubility in g/L.

Step-1: Mole fraction:x(O₂) = 0.98 / 34000 = 2.882×10⁻⁵Step-2: Moles of O₂ in 1 L:n(O₂) ≈ 2.882×10⁻⁵ × 55.5 = 1.60×10⁻³ molStep-3: Mass (Mᵣ of O₂ = 32 g/mol):mass = 1.60×10⁻³ × 32 = 0.0512 gAnswer: Solubility of O₂ ≈ 0.0512 g/LO₂ has a smaller KH than N₂ (34 vs 76.48 kbar) meaning O₂ is more soluble, which is why aquatic life can survive using dissolved O₂.

Find density (g/mL) of 3.60 M H₂SO₄ that is 29% by mass. (Mᵣ = 98 g/mol)

Formula linking molarity, density, and mass percent:M = (10 × d × w%) / Molar massSubstituting values:3.60 = (10 × d × 29) / 98d = (3.60 × 98) / (10 × 29) = 352.8 / 290 = 1.216 g/mLAnswer: Density = 1.216 g/mLNote: Memorise M = (10×d×w%)/Mᵣ. Always keep w% as a plain number (29, not 0.29) and density in g/mL.

N₂ gas at 293 K, partial pressure = 0.987 bar, KH = 76.48 kbar. Find millimoles dissolved in 1 L water.

Step 1: Apply Henry’s Law (p = KH × x):x(N₂) = 0.987 / 76480 = 1.290×10⁻⁵Step 2: Moles of water in 1 L = 1000/18 = 55.5 molStep 3: Moles of N₂ (since x << 1):n(N₂) ≈ x × n(H₂O) = 1.290×10⁻⁵ × 55.5 = 7.16×10⁻⁴ mol = 0.716 mmolAnswer: ≈0.716 millimoles of N₂ dissolveKey Point: Henry’s Law: gas solubility ∝ partial pressure. The large KH for N₂ means it is only sparingly soluble a key concept in JEE solutions.

50 g of ethylene glycol (CH₂OH)₂ dissolved in 200 g water; cooled to −9.3°C. Find mass of ice that separates. (Kf = 1.86 K·kg/mol)

Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/molStep 1 — Moles of solute:n = 50 / 62 = 0.8065 molStep 2 — Freezing point depression formula: ΔTf = Kf × m9.3 = 1.86 × (0.8065 / W_solvent_kg)Step 3 — Mass of liquid water remaining:W_solvent = (1.86 × 0.8065) / 9.3 = 0.1613 kg = 161.3 gStep 4 — Ice separated = total water − liquid water:Ice = 200 − 161.3 = 38.7 gAnswer: ≈38.71 g of ice separates outWhen a solution cools, only pure water freezes. As ice forms, the remaining solution becomes more concentrated, lowering its freezing point further.

2.8×10⁻³ mol of CO₂ is left after removing 10²¹ molecules from its ‘x’ mg sample. Find x. (Nₐ = 6.02×10²³ mol⁻¹)

Molar mass of CO₂ = 44 g/molStep 1 — Convert removed molecules to moles:Moles removed = 10²¹ / (6.02×10²³) = 1.661×10⁻³ molStep 2 — Initial moles = remaining + removed:n_initial = 2.8×10⁻³ + 1.661×10⁻³ = 4.461×10⁻³ molStep 3 — Initial mass:mass = 4.461×10⁻³ × 44 = 0.19629 g = 196.3 mgAnswer: x ≈ 196.3 mgDividing molecule count by Nₐ gives moles; multiplying by molar mass gives mass the two-step core of all JEE mole-concept problems.

Total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C₄H₈O is

Degree of unsaturation = (2×4 + 2 − 8) / 2 = 1 → One ring, no double bonds.A cyclic ether contains one O atom inside the ring. Possible ring sizes: 3-membered (oxirane), 4-membered (oxetane), 5-membered (tetrahydrofuran).StructureIsomer Count2,2-dimethyloxirane1(R)-ethyloxirane1(S)-ethyloxirane1cis-2,3-dimethyloxirane (meso)1(R,R)-2,3-dimethyloxirane1(S,S)-2,3-dimethyloxirane13-methyloxetane1Tetrahydrofuran (THF)1Answer: 8 isomersAlways check for stereocentres in epoxide rings. The meso compound of 2, 3-dimethyloxirane is a frequent JEE trap.

What does 1s, 2s, 2p, 3s, 3p mean?

NotationShellSubshellMax Electrons1s1sts (spherical)22s2nds22p2ndp (dumbbell)63s3rds23p3rdp6Number = energy level (shell) | Letter = subshell type.

Why is it incorrect to say that an electron is 'stationary' in an atom?

It is incorrect because electrons are constantly in motion, orbiting or existing in a probability cloud around the nucleus. Their movement is a fundamental aspect of atomic structure and behavior.Electrons are not static. Their continuous motion is responsible for phenomena like chemical bonding and the atom's magnetic properties. Quantum mechanics describes their location as a probability distribution rather than a fixed orbit, but it always implies motion.

Imagine you have an unknown element. What single piece of information about its subatomic particles would immediately allow you to identify the element?

The number of protons (or its atomic number) would immediately allow you to identify the element.As stated, the atomic number (number of protons) is unique to each element. Knowing this value is sufficient to identify the element on the periodic table.

How does the concept of 'electron shells' or 'energy levels' further refine our understanding of electron 'orbiting'?

Electron shells or energy levels suggest that electrons don't orbit randomly, but occupy specific regions at discrete energy states around the nucleus. This explains the quantized nature of electron energies and how they absorb and emit light.The Bohr model and later quantum mechanical models introduced the idea that electrons exist in specific energy levels or shells. This isn't a simple planetary orbit but a probabilistic distribution of electrons within these defined energy regions, explaining atomic stability and spectral lines.

What is the primary force that holds the protons and neutrons together within the nucleus, despite the electrostatic repulsion between protons?

The primary force is the strong nuclear force (or strong interaction).Protons, being positively charged, naturally repel each other. The strong nuclear force is an extremely powerful attractive force that acts over very short distances, overcoming this electrostatic repulsion to bind the nucleons (protons and neutrons) together in the nucleus.

Explain why the mass number is always a whole number, even though atomic masses often have decimal values.

The mass number represents the count of whole protons and neutrons in a single atom's nucleus, which are discrete particles. Atomic masses, however, are weighted averages of the masses of all naturally occurring isotopes of an element, which often results in decimal values.

If an atom has an atomic number of 20 and is electrically neutral, how many protons, neutrons, and electrons could it potentially have (give one possible combination)?

Protons = 20, Electrons = 20. For neutrons, it could vary, for example, 20 neutrons, making the mass number 40.Atomic number 20 means 20 protons. For neutrality, it must have 20 electrons. The number of neutrons can vary, creating isotopes. A common isotope for an element with 20 protons (Calcium) often has 20 neutrons.

How does the concept of 'orbiting' electrons relate to the overall size and volume of an atom?

The 'orbiting' electrons occupy a vast region of space around the nucleus, defining the overall size and volume of the atom. The nucleus itself is extremely small in comparison.Despite the nucleus containing almost all the mass, the electrons, due to their rapid motion and the probabilistic nature of their location, determine the atomic radius and, consequently, the atom's volume. The atom is mostly empty space.