A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x × 10¹⁵ Hz. The value of x is ____________.
(Given h =4.25 × 10⁻¹⁵ eVs )

Solution

Physics sir · 4 steps

Six lines means the atom was pushed up to n = 4. Find that jump's energy and divide by h.

1. Count the lines. An atom sitting in level n can fall back down by every possible pair of levels. That gives n(n-1)/2 different wavelengths in total.

   n(n-1)/2 = 6 ⇒ n = 4

2. Energy absorbed. The light took the atom from n = 1 to n = 4, so the photon energy is E = 13.6(1 - 1/16) = 13.6 × 15/16 = 12.75 eV.
3. Find the frequency. ν = E/h = (12.75 eV)/(4.25 × 10⁻¹⁵ eV s) = 3 × 10¹⁵ Hz.
4. So. The incident frequency is 3 × 10¹⁵ Hz, so x = 3.

The common slip: 3.2. Using the full ionisation energy 13.6 eV instead of the 1 → 4 gap of 12.75 eV gives 3.2, not 3.

The idea to keep. Number of emission lines from level n is n(n-1)/2. Use it to find the top level, then the absorbed photon carries exactly the energy gap from the ground state to that level.

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