A spectral line with reference frequency 5.00 × 10¹⁴ Hz is observed at 5.02 × 10¹⁴ Hz. Define Δf = f_obs − f₀ and take radial velocity positive for recession. Calculate Δf and estimate v using c = 3.00 × 10⁸ m s⁻¹. State whether the wavelength increases or decreases and explain how Cambridge's positive redshift magnitude convention relates to your signed calculation. For a different line, the reference wavelength is 500.0 nm and the observed wavelength is 502.0 nm. State the signed low-speed wavelength-shift relation for radial velocity, taking recession as positive, and estimate the radial velocity.

Question

Accepted Answer

Δf = +2.0 × 10¹² Hz; v ≈ −1.2 × 10⁶ m s⁻¹, so the source approaches and wavelength decreases. A redshift frequency magnitude is f₀ − f_obs for recession, opposite to signed Δf. v/c ≈ (λ_obs − λ₀)/λ₀. Therefore v ≈ (3.00 × 10⁸)(502.0 − 500.0)/500.0 = +1.20 × 10⁶ m s⁻¹, meaning recession. Δf = 5.02 × 10¹⁴ − 5.00 × 10¹⁴ = +2.0 × 10¹² Hz. v ≈ −cΔf/f₀ = −(3.00 × 10⁸)(2.0 × 10¹²)/(5.00 × 10¹⁴) = −1.2 × 10⁶ m s⁻¹. Negative v means approach; frequency increases and wavelength decreases, a blueshift. For a receding source the signed Δf is negative, while its frequency-decrease magnitude f₀ − f_obs is positive; the missing sign must be supplied by the definition, not ignored. The fractional-shift relation is appropriate here because |v|/c ≈ 0.004 ≪ 1. State v/c ≈ (λ_obs − λ₀)/λ₀ with the specified recession-positive convention (1). Substitute the wavelength ratio 0.00400 to obtain +1.20 × 10⁶ m s⁻¹ away from the observer (1).

Infer radial motion without a frequency-sign error

What a complete response should do

Your self-assessment