The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. The wavelength of the second member of the Balmer series (in nm) is:

Solution

Physics sir · 4 steps

Do not find R at all: take the ratio of the two Balmer lines and the Rydberg constant cancels.

1. Identify the members. Balmer lines end on n = 2. The first member is the jump 3 → 2 and the second member is 4 → 2.
2. Write both lines. For 3 → 2: 1/λ₁ = R(1/4 - 1/9) = 5R/36. For 4 → 2: 1/λ₂ = R(1/4 - 1/16) = 3R/16.

   λ₂/λ₁ = (5/36)/(3/16) = 80/108 = 20/27

3. Put in numbers. λ₂ = 6561 × 20/27 Å = 243 × 20 Å = 4860 Å.
4. So. Since 1 nm = 10 Å, λ₂ = 486 nm, so the answer is 486.

The common slip: 4860. Leaving the answer in ångström gives 4860, not 486. Divide by 10 to convert Å to nm.

The idea to keep. When one line of a series is given and another is asked, divide the two Rydberg expressions. The larger energy jump always gives the shorter wavelength.

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