Question

Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of light coming out of your mouth as you chewed on the candy; and, without knowing it, you have experienced a physical phenomenon called triboluminescence. In this problem you will analyze some of the key elements of triboluminescence in wintergreen candies.
Part A
Imagine that an electron in an excited statein a nitrogen molecule decays to its ground state, emitting aphoton with a frequency of 8.88×1014 Hz. Whatis the change in energy, ΔE , that the electronundergoes to decay to its ground state?
Express your answer in electron voltsto three significant figures.
ΔE= eV
Part B
Image that the radiation emitted by thenitrogen at a frequency of 8.88×1014 Hz isabsorbed by an electron in a molecule of methyl salicylate. As aresult, the electron in the wintergreen oil molecule jumps to anexcited state. Before returning to its ground state, the electrondrops to an intermediate energy level, releasing two-thirds of theenergy previously absorbed and emitting a photon. What is thewavelength λw of the photon emitted by the wintergreen oilmolecule?
Express your answer in nanometers tothree significant figures.
λw= nm

Answer 1

Answer:

Part a)

$E = 3.66 eV$

Part b)

$\lambda = 508.5 nm$

Explanation:

Part a)

change in the energy due to decay of photon is given as

$E = h\nu$

here we know that

$\nu = 8.88 \times 10^{14} Hz$

now we have

$E = (6.6 \times 10^{-34})(8.88 \times 10^{14})$

$E = 5.86 \times 10^{-19} J$

$E = 3.66 eV$

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

$\Delta E = \frac{2}{3}(3.66 eV)$

$\Delta E = 2.44 eV$

now to find the wavelength we have

$\Delta E = \frac{hc}{\lambda}$

$2.44 = \frac{1242}{\lambda}$

$\lambda = 508.5 nm$