Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
The de Broglie wavelength of a neutron moving with kinetic energy 1 eV is approximately:
Answer: A
Using λ = h/√(2mKE), where m = 1.67×10⁻²⁷ kg, KE = 1.6×10⁻¹⁹ J, h = 6.63×10⁻³⁴ J·s. Calculation yields λ ≈ 0.286 nm.
Q.282Medium
Two radioactive nuclei A and B have decay constants λₐ and λᵦ respectively, where λₐ = 2λᵦ. Initially, both have the same number of nuclei. The ratio of their half-lives (t₁/₂ₐ : t₁/₂ᵦ) is:
Answer: A
Half-life t₁/₂ = ln(2)/λ. Since λₐ = 2λᵦ, we have t₁/₂ₐ/t₁/₂ᵦ = λᵦ/λₐ = 21. Therefore, t₁/₂ₐ : t₁/₂ᵦ = 1:2.
Q.283Medium
An electron transitions from n=3 to n=1 in a hydrogen atom. The ratio of wavelengths emitted to that expected for Lyman alpha (n=2 to n=1) is:
Answer: B
Using 1/λ = R(1/n₁² - 1/n₂²). For 3→1: 1/λ₃₋₁ = R(1 - 91) = 8R/9. For Lyman alpha 2→1: 1/λ₂₋₁ = R(1 - 41) = 3R/4. Ratio λ₃₋₁/λ₂₋₁ = (43)/(98) = 3227. So λ₂₋₁/λ₃₋₁ = 2732.
Q.284Medium
In Compton scattering, a photon of wavelength λ₀ collides with a stationary electron. After scattering at angle θ = 90°, the wavelength becomes λ. The relationship is:
A radioactive sample has a half-life of 10 days. After how many days will 93.75% of the sample decay?
Answer: B
If 93.75% decays, 6.25% remains. 6.25% = 6.10025 = 161 = (21)⁴. So 4 half-lives have passed. Time = 4 × 10 = 40 days. Correction: 6.25% = 161, which requires 4 half-lives = 40 days.
Q.286Medium
The frequency of K-alpha X-ray for a target material depends on:
Answer: A
Characteristic X-ray frequency (Moseley's law) depends on atomic number Z of the target. f = R(Z - σ)²(1/n₁² - 1/n₂²). It is independent of incident electron energy (which only affects intensity).
Q.287Medium
The cutoff wavelength (λ₀) in X-ray spectrum produced by deceleration of electrons is determined by:
Answer: A
Maximum photon energy = eV. E = hc/λ₀, so λ₀ = hc/eV. This is the minimum wavelength or cutoff wavelength.
Q.288Medium
In a cathode ray tube with accelerating potential V, electrons reach the anode with kinetic energy. If V is doubled, the maximum frequency of X-rays produced will:
Answer: B
Maximum X-ray frequency: fmax = eV/h. If V is doubled, fmax also doubles, since f ∝ V.
Q.289Medium
A hydrogen atom transitions from n=4 to n=2 state. The wavelength of emitted photon is approximately:
Answer: A
Using Rydberg formula: 1/λ = R(41 - 161) = 3R/16. With R = 1.097×10⁷ m⁻¹, we get λ ≈ 486 nm (H-beta line).
Q.290Medium
The binding energy per nucleon for ⁵⁶Fe is maximum among all nuclei. This suggests that:
Answer: D
Maximum binding energy per nucleon means Fe-56 has highest stability and also highest mass defect. This is the peak of the nuclear stability curve.
Q.291Medium
A radioactive element has 75% of its original mass remaining after 6 hours. What is its half-life?
Answer: C
Using N = N₀(21)^(t/T₁/₂): 0.75N₀ = N₀(21)^(6/T₁/₂). Taking log: ln(0.75) = (6/T₁/₂)ln(0.5), giving T₁/₂ = 6√3 hours ≈ 10.39 hours.
Q.292Medium
The de Broglie wavelength of an electron accelerated through a potential difference of 100 V is approximately:
Answer: A
λ = h/√(2meV). For V = 100V: λ = 12.27/√(2×9.1×10⁻³¹×1.6×10⁻¹⁹×100) ≈ 1.23 Å = 0.123 nm.
Q.293Medium
In a hydrogen atom, the ground state energy is -13.6 eV. What is the energy required to ionize from n=3?
Answer: A
Energy at n=3: E₃ = -13.96 = -1.51 eV. Ionization energy = 0 - (-1.51) = 1.51 eV.
Q.294Medium
When a moving electron is stopped completely in matter, the kinetic energy is converted to:
Answer: C
When high-speed electrons are stopped in matter, their kinetic energy produces both continuous X-ray spectrum (Bremsstrahlung radiation) and heat.
Q.295Medium
In X-ray production, the minimum wavelength of continuous X-ray spectrum depends on:
Answer: B
Minimum wavelength λ_min = hc/eV, depends only on accelerating voltage V. Higher voltage produces shorter wavelength X-rays.
Q.296Medium
In Compton effect, the wavelength shift depends on:
Answer: D
Compton shift: Δλ = (h/mₑc)(1 - cosθ). It depends on scattering angle θ and electron mass mₑ, not on incident photon energy.
Q.297Medium
The de Broglie wavelength of a neutron at room temperature (T = 300 K) is approximately:
Answer: A
Using λ = h/√(3mkT), where m = 1.67×10⁻²⁷ kg for neutron, h = 6.63×10⁻³⁴ J·s, k = 1.38×10⁻²³ J/K. λ ≈ 0.016 nm
Q.298Medium
An electron transitions from n=3 to n=1 in hydrogen atom. The frequency of emitted photon is (R = 1.097×10⁷ m⁻¹):
