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(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \fr

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Violet

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(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.For the transition $n = 4 	o n = 2$,the energy emitted corresponds to blue light.For the transition $n = 5 	o n = 2$,the energy difference is $\Delta E_{5 	o 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{25} ight) = 13.6 \left( \frac{21}{100} ight) = 2.856 	ext{ eV}$.For the transition $n = 4 	o n = 2$,the energy difference is $\Delta E_{4 	o 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{16} ight) = 13.6 \left( \frac{3}{16} ight) = 2.55 	ext{ eV}$.Since $\Delta E_{5 	o 2} > \Delta E_{4 	o 2}$,the frequency of the emitted light will be higher,and the wavelength will be shorter.In the visible spectrum,violet light has a higher frequency and shorter wavelength than blue light. Therefore,the atom emits violet light.

$A$ hydrogen atom emits blue light when it changes from the $n = 4$ energy level to the $n = 2$ level. Which colour of light would the atom emit when it changes from the $n = 5$ level to the $n = 2$ level?

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