The ratio of wavelength of spectral lines H_α and H_β in the Balmer series is x/20. The value of x is _________.

Solution

Physics sir · 4 steps

Both lines end on n = 2: H-alpha starts from 3, H-beta from 4. Then wavelength is the reciprocal.

1. Identify the jumps. Every Balmer line ends on n = 2. H_α is the first line, from n = 3, and H_β is the second, from n = 4.

   1/λ = R(1/2²-1/n²)

2. Each line. For H_α: 1/λ_α = R(1/4-1/9) = 5R/36. For H_β: 1/λ_β = R(1/4-1/16) = 3R/16.
3. Take the ratio. Wavelength is the reciprocal of these, so λ_α/λ_β = (3R/16)/(5R/36) = (3 × 36)/(16 × 5) = 108/80 = 27/20.
4. So. The ratio is 27/20, so x = 27.

The idea to keep. The Rydberg formula gives 1/λ, not λ. When you compare wavelengths, flip the ratio. The smaller energy jump always gives the longer wavelength.

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