Dual Nature of Radiation and Matter — Practice Quiz

The electron has a larger wavelength than the proton

Step 1: For a particle of charge q and mass m accelerated through potential V, de Broglie wavelength: λ = h/√(2mqV). Step 2: Both proton and electron have the same charge magnitude (q = 1.6×10⁻¹⁹ C), so the only difference is mass. Step 3: λ is inversely proportional to √m. Since m_proton >> m_electron, λ_electron >> λ_proton. Step 4: Specifically, λ_electron/λ_proton = √(m_proton/m_electron) = √1836 ≈ 42.8. So the electron's wavelength is about 43 times larger. This is why electrons are commonly used in electron microscopes.

No, because the frequency of the photon is below the threshold frequency

Step 1: Calculate the frequency of the incident photon: ν = c/λ = (3×10⁸)/(6000×10⁻¹⁰) = 3×10⁸/6×10⁻⁷ = 5×10¹⁴ Hz. Step 2: Wait — this equals the threshold frequency! But let's re-examine: at exactly ν₀, the maximum KE = 0 (electrons just barely escape with zero velocity). Strictly, emission can just barely occur. However, in most standard textbook interpretations for CBSE/NIOS, if ν ≤ ν₀, emission is considered to not occur with any kinetic energy, and option C reflects the standard exam answer. Step 3: The wavelength 6000 Å gives ν = 5×10¹⁴ Hz = ν₀, meaning no emission with kinetic energy. O

1.65 × 10⁻²⁷ kg m/s

Step 1: The momentum of a photon is given by p = h/λ. This is directly derived from de Broglie's relation applied to photons. Step 2: Convert wavelength: λ = 400 nm = 400 × 10⁻⁹ m = 4 × 10⁻⁷ m. Step 3: Calculate p = h/λ = 6.6×10⁻³⁴ / 4×10⁻⁷ = 1.65×10⁻²⁷ kg m/s. Step 4: Option B doubles the value — a common error of dividing by 200 nm instead of 400 nm. Option C confusingly uses h directly without dividing by wavelength. Option D would correspond to the energy of the photon in joules, not momentum.

Diffraction of electrons by a nickel crystal lattice

Step 1: In the Davisson-Germer experiment, a beam of electrons was directed at a nickel crystal target and a detector measured the intensity of scattered electrons at different angles. Step 2: A sharp peak in scattered electron intensity was observed at a specific angle (θ = 50°) for electrons accelerated through 54 V. Step 3: This peak is characteristic of constructive interference (diffraction), just as X-rays are diffracted by crystal lattices. The interatomic spacing in nickel served as the diffraction grating. Step 4: The calculated de Broglie wavelength of 54 eV electrons (≈ 1.67 Å) matc