A galaxy is moving away from the earth at a speed of 286 kms⁻¹. The shift in the wavelength of a redline at 630 nm is x × 10⁻¹⁰ m. The value of x, to the nearest integer, is ____________. [Take the value of speed of light c, as 3 × 10⁸ ms⁻¹]

Solution

Physics sir · 4 steps

A receding galaxy stretches its light, and the fractional stretch Δλ/λ equals v/c.

1. Doppler for light. When the source speed is much smaller than c, the shift in wavelength is set by the ratio v/c. A source moving away gives a longer wavelength, which is a red shift.

   Δλ = v/c λ

2. Use SI units. v = 286 km s⁻¹ = 2.86 × 10⁵ m s⁻¹ and λ = 630 nm = 630 × 10⁻⁹ m.
3. Put in numbers. Δλ = (2.86 × 10⁵)/(3 × 10⁸) × 630 × 10⁻⁹ = 6.006 × 10⁻¹⁰ m.
4. So. The shift is about 6 × 10⁻¹⁰ m, so x = 6.

The common slip: 0.6. Leaving the shift in nanometres gives 0.6, but the question wants it in units of 10⁻¹⁰ m, and 0.6 nm = 6 × 10⁻¹⁰ m.

The idea to keep. For light from a slow source, Δλ/λ = v/c. Convert every quantity to metres first, then read off the power of ten the question asks for.

Answer from the NTA final answer key. Solution by PrepWiser.