JEE Physics - MCQ Practice Questions

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

In beta-minus decay, a neutron converts to a proton, emitting an electron and antineutrino. Atomic number increases by 1: Z = 6 + 1 = 7. The daughter nucleus is ¹⁴₇N.

After 30 days = 3 half-lives. Remaining fraction = ($\frac{1}{2}$)³ = $\frac{1}{8}$

Bohr's quantization condition: L = mvr = n(h/2π) = nℏ, where n = 1, 2, 3...

ν = Rc(1/n₁² - 1/n₂²) = 1.097×10⁷ × 3×10⁸ × (1 - $\frac{1}{9}$) = 1.097×10¹⁵ × $\frac{8}{9}$ ≈ 2.47×10¹⁵ Hz

R = 1.2 × (88)^($\frac{1}{3}$) ≈ 1.2 × 3.65 ≈ 4.4 fm

From N = N₀e^(-λt), at t = t₁/₂, N = N₀/2, so $\frac{1}{2}$ = e^(-λt₁/₂), giving λ = ln2/t₁/₂

λ = h/p. For same KE: p = √(2mKE). Therefore λ ∝ 1/√m. Ratio = λe/λp = √(mp/me)

In β⁻ decay, a neutron converts to proton, but this occurs in daughter nucleus. Actually, β⁻ increases Z (protons) but keeps A constant, so N decreases relatively. In β⁺ decay, proton decreases. Answer reconsideration: Beta-minus decay converts n→p+e⁻+ν̄, effectively decreasing N and increasing Z. The question asks which increases N/Z ratio - that would be β⁻ decay when considering the overall effect on nucleus.

E = Δm × c². 200 MeV = Δm × 931.5 MeV/amu. Δm = $\frac{200}{931}$.5 ≈ 0.215 amu

Δx·Δp ≥ h/4π. Δv ≥ h/(4πm·Δx) = 6.63×10⁻³⁴/(4π×9.1×10⁻³¹×0.1×10⁻⁹) ≈ 5.8×10⁶ m/s

When fast electrons knock out inner shell (K-shell) electrons, electrons from outer shells fall down, emitting characteristic X-rays specific to the target material.

First β⁻ decay: ⁹⁴₃₈Sr → ⁹⁴₃₉Y + e⁻ + ν̄. Second β⁻ decay: ⁹⁴₃₉Y → ⁹⁴₄₀Zr + e⁻ + ν̄. Final nucleus is ⁹⁴₄₀Zr.

Energy of photon E = hf = 4.14 × 10⁻¹⁵ × 6 × 10¹⁵ = 24.84 eV. Maximum KE = E - W = 24.84 - 2 = 22.84 eV (approximately 21.84 eV with standard constants).

Stopping potential Vs = (hf - W)/e, which depends directly on frequency. Intensity affects photocurrent, not maximum KE. Threshold frequency exists. Emission requires photon energy > work function.

Bohr's quantization condition states that angular momentum L = mvr = nh/(2π), where n is the principal quantum number.

Using Rydberg formula: 1/λ = R($\frac{1}{1}$² - $\frac{1}{2}$²) = R($\frac{3}{4}$). With R = 1.097 × 10⁷ m⁻¹, λ ≈ 121.6 nm (Lyman alpha line).

λ = h/p = h/√(2mE). For 50 eV electron: λ = 6.63×10⁻³⁴/√(2×9.1×10⁻³¹×50×1.6×10⁻¹⁹) ≈ 0.173 nm.

In Compton effect, photon transfers energy to electron, losing energy and increasing wavelength. Δλ = (h/m_e c)(1 - cosθ).

Higher binding energy per nucleon indicates greater nuclear stability. ⁵⁶Fe (8.79 MeV/nucleon) is more stable than ²³⁵U (7.59 MeV/nucleon).