In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given hc =1245 eV nm, e=1.6 × 10⁻¹⁹ C).

Solution

Physics sir · 3 steps

The first dip at 10.2 V is the energy gap, so the photon coming back carries 10.2 eV.

1. Read the dip. The current first drops when the electrons gain just enough energy to lift a hydrogen atom from n = 1 to n = 2. An electron accelerated through 10.2 V carries 10.2 eV, so the gap E₂ - E₁ = 10.2 eV.
2. Photon energy. When the atom falls back from n = 2 to n = 1, it emits a photon carrying exactly this 10.2 eV. With hc given in eV nm, the wavelength comes out directly in nm.

   λ = hc/(Δ E) = (1245 eV nm)/(10.2 eV) = 122.06 nm

3. So. The emitted wavelength is about 122 nm, the Lyman-alpha line, so the answer is 122.

The common slip: 92. Using the ionisation energy 13.6 eV in place of the 10.2 eV gap gives 1245/13.6 ≈ 92 nm, not 122.

The idea to keep. In Franck-Hertz, the dip voltage in volts equals the excitation energy in eV. Then use λ = hc/E with hc in eV nm, and you never need the electron charge.

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