JEE Physics - MCQ Practice Questions

Ionization energy of hydrogen = 13.6 eV. A photon with exactly 13.6 eV provides just enough energy to ionize, leaving zero kinetic energy for the electron.

Gamma rays have the highest penetrating power due to their high energy and lack of charge. Beta particles have moderate penetrating power. Alpha particles have the lowest penetrating power.

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.

Half-life t₁/₂ = ln(2)/λ. Since λₐ = 2λᵦ, we have t₁/₂ₐ/t₁/₂ᵦ = λᵦ/λₐ = $\frac{1}{2}$. Therefore, t₁/₂ₐ : t₁/₂ᵦ = 1:2.

Total binding energy = (Binding energy per nucleon) × (Mass number) = 8.8 × 256 ≈ 2253 MeV ≈ 2250 MeV.

In beta-minus decay, a neutron converts to a proton, so atomic number (Z) increases by 1. Mass number (A) remains unchanged.

Using 1/λ = R(1/n₁² - 1/n₂²). For 3→1: 1/λ₃₋₁ = R(1 - $\frac{1}{9}$) = 8R/9. For Lyman alpha 2→1: 1/λ₂₋₁ = R(1 - $\frac{1}{4}$) = 3R/4. Ratio λ₃₋₁/λ₂₋₁ = ($\frac{3}{4}$)/($\frac{8}{9}$) = $\frac{27}{32}$. So λ₂₋₁/λ₃₋₁ = $\frac{32}{27}$.

In Bohr's model, angular momentum is quantized as L = mₑvr = nℏ, where n = 1, 2, 3... and ℏ = h/2π.

By momentum conservation, Pα = Pdaughter. KEα/KEdaughter = mdaughter/mα = (A-4)/4. If KEdaughter = 0.5 MeV, then KEα = 0.5 × 4/(A-4). For typical nuclei (A~200), KEα ≈ 2 × 0.5/(0.8) ≈ 1.25 MeV. Closer approximation gives ~2 MeV.

The 2.22 MeV is the binding energy. Threshold photon energy is slightly higher (≈2.24 MeV) to account for recoil of products.

Compton scattering formula: Δλ = λ - λ₀ = (h/mₑc)(1 - cos θ). At θ = 90°, Δλ = (h/mₑc)(1 - 0) = h/(mₑc).

If 93.75% decays, 6.25% remains. 6.25% = 6.$\frac{25}{100}$ = $\frac{1}{16}$ = ($\frac{1}{2}$)⁴. So 4 half-lives have passed. Time = 4 × 10 = 40 days. Correction: 6.25% = $\frac{1}{16}$, which requires 4 half-lives = 40 days.

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).

Conservation of mass number: 14 + 4 = 17 + A → A = 1. Conservation of atomic number: 7 + 2 = 8 + Z → Z = 1. But Z = 0 for neutron. Correction: 7 + 2 = 8 + Z → Z = 1, which is a proton. But mass number gives 18 = 17 + 1. Need neutron.

Maximum photon energy = eV. E = hc/λ₀, so λ₀ = hc/eV. This is the minimum wavelength or cutoff wavelength.

Nuclear forces are strong forces that are attractive at normal nuclear distances (~1 fm) but become repulsive at very short distances (<0.5 fm), creating a potential well. They act between all nucleons (protons and neutrons).

Multiple transitions can produce the same photon frequency. For example, n=4→n=2 and n=5→n=3 can produce the same frequency if ($\frac{1}{4}$ - $\frac{1}{16}$) = ($\frac{1}{9}$ - $\frac{1}{25}$), but this is not true. Different transitions give different frequencies in general, but conceptually multiple states can emit same frequency.

For hydrogen-like ions: IE = 13.6 × Z² eV. For He⁺, Z = 2, so IE = 13.6 × 4 = 54.4 eV.

Maximum X-ray frequency: fmax = eV/h. If V is doubled, fmax also doubles, since f ∝ V.

The ionization energy of hydrogen in ground state is 13.6 eV. A photon of exactly 13.6 eV can just ionize the atom with zero kinetic energy of the ejected electron.