Deduce that the energy of a photon of wavelength 658 nm is 1.89 eV.

(i) On diagram 1, draw an arrow to show the electron transition between energy levels that gives rise to the emission of a photon of wavelength 658 nm. Label this arrow with the letter A.

(ii) On diagram 1, draw arrows to show the electron transitions between energy levels that give rise to the emission of photons of wavelengths 488 nm, 435 nm and 411 nm. Label these arrows with the letters B, C and D.

Explain why the lines in the emission spectrum of atomic hydrogen, shown in diagram 2, become closer together as the wavelength of the emitted photons decreases.

\(E = \frac{{hc}}{\lambda }\);
\( = \frac{{6.63 \times {{10}^{ - 34}} \times 3.00 \times {{10}^8}}}{{658 \times {{10}^{ - 9}}}} = 3.02 \times {10^{ - 19}}\);
=\(\frac{{3.02 \times {{10}^{ - 19}}}}{{1.60 \times {{10}^{ - 19}}}}\);
=1.89eV

or

the photon of wavelength 658nm is the longest (in the emission graph);
therefore it has the shortest frequency and lowest energy (from E=hf );
therefore it arises from the transition between the –1.51eV and the –3.40eV energy levels which have a difference of 1.89eV;

(i) see diagram below;

(ii) see diagram below;
All three must be correct for the mark.

at higher energy levels, energy levels become closer together;
the energy differences between higher energy levels and the lower level (n=2) become more equal;
hence the difference in wavelength of emitted photons decreases / OWTTE;