The wavelength of the radiation emitted is λ₀ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will 20/xλ₀. The value of x is _____________.

Solution

Physics sir · 4 steps

Turn each excited state into its n first. Then take the ratio of the two Rydberg brackets and R cancels out.

1. Name the levels. The ground state is n = 1, so the first excited state is n = 2, the second is n = 3 and the third is n = 4. The two jumps are 3 → 2, which gives λ₀, and 4 → 2, which gives λ.

   1/λ = R(1/n₁²-1/n₂²)

2. Both brackets. For 3 → 2: 1/4-1/9 = 5/36. For 4 → 2: 1/4-1/16 = 3/16.
3. Take the ratio. Wavelength is inversely proportional to the bracket, so λ/λ₀ = (5/36)/(3/16) = 80/108 = 20/27.
4. So. λ = 20/27λ₀, which gives x = 27.

The common slip: 20. Taking the third excited state as n = 3 makes the second jump 3 → 2 again, so λ = λ₀ and x comes out as 20.

The idea to keep. Excited-state number plus one gives n. For a ratio of two wavelengths, R cancels, and the bigger energy gap gives the shorter wavelength.

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