JEE Main 2022 · 26 June · evening · Single correct · +4 / −1
For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is :
- (a) 3 × 10⁸ ms⁻¹
- (b) 3 × 10¹⁰ ms⁻¹
- (c) 3.13 × 10⁵ ms⁻¹
- (d) 4.48 × 10⁵ ms⁻¹
Answer and solution
Answer (c) 3.13 × 10⁵ ms⁻¹
Solution
Physics sir · 4 steps
The wavelength stretched by 0.7 nm, so the galaxy moves away at c times that fractional shift.
- Red shift. The observed wavelength (670.7 nm) is longer than the emitted one (670 nm). This is a red shift, so the galaxy is moving away from us.
- Doppler for light. When v ≪ c, the fractional change in wavelength equals v/c.
Δλ/λ = v/c ⇒ v = c Δλ/λ
- Put in numbers. Δλ = 670.7 - 670 = 0.7 nm. So v = 3 × 10⁸ × 0.7/670 = 3 × 10⁸ × 1.045 × 10⁻³ ≈ 3.13 × 10⁵ m s⁻¹.
- So. The galaxy recedes at about 3.13 × 10⁵ m s⁻¹, which is option (c).
The idea to keep. For light from a moving source with v ≪ c, use Δλ/λ = v/c. A longer observed wavelength means red shift, so the source is moving away.
Answer from the NTA final answer key. Solution by PrepWiser.