The Bohr model of the hydrogen atom can be represented by the diagram shown - Scottish Highers Physics - Question 10 - 2023

A line emission spectrum is produced when an electron in an atom absorbs energy and moves to a higher energy level. When the electron transitions back to a lower energy level, it emits energy in the form of photons. Each transition corresponds to a specific energy difference and therefore produces a distinct line in the spectrum.

Some lines appear brighter than others because they correspond to transitions that involve a greater number of electrons dropping to lower energy levels. The more transitions that occur from a specific level, the more photons are emitted, resulting in a brighter line.

There are 10 possible emission lines caused by the transitions between the five energy levels shown.

The frequency of the photon emitted can be calculated using the formula:

f = rac{ riangle E}{h}

where ( \triangle E = E_1 - E_3 = (-0.871 × 10^{-19}) - (-2.42 × 10^{-19}) = 1.55 × 10^{-19} J ) and ( h = 6.63 × 10^{-34} J·s ). Therefore,

f = rac{1.55 × 10^{-19}}{6.63 × 10^{-34}} = 2.34 × 10^{14} Hz.

The wavelength of the photons corresponding to the blue-green spectral line is 486 nm.

The observed wavelength can be determined using the Doppler effect formula for wavelengths:

rac{\lambda_o}{\lambda} = 1 + \frac{v}{c}

where ( \lambda_o = 486 × 10^{-9} ) m, ( v = 4.52 × 10^6 ) m/s, and ( c = 3.00 × 10^{8} ) m/s.

Calculating gives:\n$\lambda = \lambda_o \left( 1 + \frac{4.52 × 10^{6}}{3.00 × 10^{8}} \right) \approx 486 × 10^{-9}(1 + 0.01507) \approx 486 × 10^{-9} × 1.01507 \approx 493 × 10^{-9} m.$