JEE Physics - MCQ Practice Questions

Using E=mc², the rest mass energy of electron (m=9.11×10⁻³¹ kg) is 0.511 MeV, a fundamental constant in modern physics.

de Broglie proposed that λ = h/p, where h is Planck's constant and p is momentum. This is fundamental to wave-particle duality.

Iron-56 has the maximum binding energy per nucleon (~8.8 MeV/nucleon), making it the most stable nucleus. This is a key concept in nuclear physics.

Number of half-lives = $\frac{30}{10}$ = 3. Remaining fraction = ($\frac{1}{2}$)³ = $\frac{1}{8}$

Compton scattering shows that photons behave as particles with momentum, providing evidence for particle nature of light. Wavelength shift Δλ = (h/mₑc)(1 - cosθ).

Since c = fλ, we have f = c/λ. Therefore E = hf = hc/λ

rₙ = n²a₀, where a₀ is Bohr radius (0.53 Å). Orbital radius is proportional to n².

α-particle: charge = 2e, mass = 4u. Proton: charge = e, mass = u. (e/m)ₐ/(e/m)ₚ = (2e/4u)/(e/u) = $\frac{1}{2}$

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.

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.

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

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.

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).

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

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.

The radius of nth orbit in hydrogen atom is given by rₙ = n²a₀/Z, where a₀ = 0.53 Å is the Bohr radius and Z is atomic number.

In β⁻ decay, a neutron converts to proton, emitting electron and antineutrino: n → p + e⁻ + ν̄ₑ

Alpha particles (⁴₂He) are helium nuclei containing 2 protons and 2 neutrons. They have low penetrating power but high ionizing power.