- #1
quantumworld
- 36
- 0
Hello brainy people,
I did solve this problem, but my answer is double the correct one, I wonder what went wrong.
here it is:
The Lyman alpha spectral line of hydrogen (lambda = 122 nm) differs by 1.8*10^-12 m in spectra taken at opposite ends of the sun's equator. What is the speed of a particle on the equator due to the Sun's rotation, in kilometers per second?
I used the doppler formula, delta lambda/lambda = v/c
so I got c = 4.4 Km/s, but the correct answer is 2.2 km/s
thanks for your effort
I did solve this problem, but my answer is double the correct one, I wonder what went wrong.
here it is:
The Lyman alpha spectral line of hydrogen (lambda = 122 nm) differs by 1.8*10^-12 m in spectra taken at opposite ends of the sun's equator. What is the speed of a particle on the equator due to the Sun's rotation, in kilometers per second?
I used the doppler formula, delta lambda/lambda = v/c
so I got c = 4.4 Km/s, but the correct answer is 2.2 km/s
thanks for your effort