JEE Physics - MCQ Practice Questions

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

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.

Angular momentum L = ℏ√(l(l+1)) = 2ℏ gives l(l+1) = 4, so l = 2. Since l < n, minimum n = 3.

For same KE: λ = h/√(2mKE). λₑ/λₚ = √(mₚ/mₑ) ≈ 42.8 (using mₑ = 9.1×10⁻³¹ kg, mₚ = 1.67×10⁻²⁷ kg).

After 2α decays: mass number decreases by 8, atomic number by 4: ²³⁸₉₂U → ²³⁰₈₈Ra. After 2β⁻ decays: atomic number increases by 2: ²³⁰₉₀Th. But rechecking: ²³⁰₈₈Ra is correct intermediate form.

²₁H + ²₁H → ⁴₂He + energy. Using mass defect and E=mc²: Energy released ≈ 23.8 MeV. This is the basis of thermonuclear fusion.

For a free electron at rest, if it absorbs a photon with energy E and momentum p=E/c, it cannot emit a photon in any direction while conserving both energy and momentum. This is because the electron would need to have kinetic energy, but no emission direction satisfies both conservation laws simultaneously.

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.

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

E = Δm·c² = 0.1 × 931 MeV ≈ 93 MeV. (Note: typical U-235 fission releases ~200 MeV total, distributed among products).

Possible transitions: 3→1 (direct), 3→2, 2→1. Total = 3 distinct lines. The electron can go 3→2→1 or 3→1 directly.

In pair production: E_photon = 2m_e c² + KE_total. Excess = 3 - 2(0.51) = 1.98 MeV becomes kinetic energy of the pair.

For heavy nuclei, the Coulomb repulsion between protons increases significantly. Extra neutrons (uncharged) help stabilize the nucleus without increasing repulsion, requiring N > Z for stability.

Vth = Vfb + 2φF + (√(2εsqNa(2φF))/Cox). Threshold voltage increases with oxide thickness (Cox decreases) and increases with substrate doping. Temperature has weak effect.

At low temperatures, phonon scattering is suppressed, but impurity scattering (interaction with ionized donors/acceptors) remains, limiting mobility. At high T, phonon scattering dominates.

Depletion mode MOSFETs have a conducting channel at Vgs = 0 and require gate voltage to turn OFF. Enhancement mode requires positive gate voltage to create a channel. This is a key structural difference.

Bandgap of semiconductors decreases with increasing temperature. For Si: dEg/dT ≈ -2.3 meV/K near 300 K. This is derived from Varshni equation and is crucial for device design.

Heterojunctions use materials with different bandgaps to create band offsets that confine carriers (electrons and holes) to specific regions, improving device efficiency. AlGaAs has larger Eg than GaAs, confining carriers in GaAs.

Forward current I = I₀exp(eV/kT). Ratio = exp(e×0.3/kT) = exp(0.$\frac{3}{0}$.026) ≈ exp(11.5) ≈ 10^5. For ΔV = 0.3V more, increase is exp(11.5) ≈ 1000 times.

Fill Factor (FF) = Pmax/(Voc × Isc) = (Vm × Im)/(Voc × Isc). It indicates how close the I-V curve is to a rectangle, typically 0.7-0.85 for practical solar cells.