Core Concept

The de Broglie wavelength λ for an electron accelerated through potential V is λ = h / √(2m e V), where h is Planck’s constant, m is electron mass, and e is electron charge. Kinetic energy gained equals eV, so momentum p = √(2m e V) and λ ∝ 1/√V.

Detailed Derivation

Electron starts from rest, gains kinetic energy KE = eV = (1/2) m v², so velocity v = √(2 e V / m). Momentum p = m v = √(2 m e V), thus λ = h / p = h / √(2 m e V). Initially λ₁ ∝ V₁^{-1/2}; new λ₂ ∝ (4 V₁)^{-1/2} = (1/2) V₁^{-1/2}. Ratio λ₂ / λ₁ = 1/2.

Option Analysis

Typical CSIR NET-style options: (A) 4, (B) 2, (C) 1/2, (D) 1/4.

• (A) 4: Wrong—inverse relation, wavelength shrinks with higher speed/energy.
• (B) 2: Wrong—matches doubling of V (√2 ≈ 1.41, but here quadrupled).
• (C) 1/2: Correct—1/√4 = 1/2, wavelength halves.
• (D) 1/4: Wrong—would apply if λ ∝ 1/V, but it’s 1/√V.

Formula Breakdown

Start with de Broglie relation: λ = h / p. For acceleration through V, p = √(2 m e V), so λ = h / √(2 m e V). Quadrupling V (to 4V) makes new wavelength λ’ = h / √(2 m e (4 V)) = λ / 2.

CSIR NET Relevance

This matches exam patterns like IIT JAM BT 2022, emphasizing non-relativistic approximations (valid for typical voltages). Common pitfalls: confusing with kinetic energy doubling (1/√2) or linear 1/V scaling. Practice verifies electron wavelengths (~0.1 nm) suit diffraction experiments.