If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Same series, same lower level: take the ratio of the two Rydberg brackets and R cancels out.

1. Identify the jumps. Every Paschen line ends at n₁ = 3. The first line is the smallest jump, 4 → 3, and the second line is 5 → 3.

   1/λ = R(1/3²-1/n₂²)

2. Work out brackets. For 4 → 3: 1/9-1/16 = 7/144. For 5 → 3: 1/9-1/25 = 16/225.
3. Take the ratio. Wavelength is inversely proportional to the bracket, so λ₂/λ₁ = (7/144)/(16/225) = 1575/2304 = 0.6836.
4. So. λ₂ = 720 × 0.6836 = 492.2 nm, which rounds to 492.

The common slip: 1053. Flipping the ratio, so that the wavelength is multiplied by 2304/1575, gives 1053. The second line comes from a bigger jump, so its wavelength must be shorter than 720 nm.

The idea to keep. For two lines of one series, never compute R. Divide the brackets, and remember the bigger jump gives the bigger bracket and so the shorter wavelength.

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