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Photodissociation is the process in which a molecule fragments after absorbing a photon with sufficiAnalytical Chemistry Chemistry Question

Photodissociation of Cl2

Photodissociation is the process in which a molecule fragments after absorbing a photon with sufficient energy to break a chemical bond. The rupture of a chemical bond is one of the most fundamental chemical processes, and has been studied in great detail.

In a modified time-of-flight mass spectroscopy technique for studying Cl–Cl bond cleavage, a laser beam is crossed with a molecular beam of Cl2, and dissociation occurs at the crossing point. A second laser beam ionises the resulting Cl atoms (without affecting their velocities), so that a carefully tuned electric field may be used to guide them along a 40 cm flight path to a position sensitive detector. The image of the Cl fragments recorded at the detector is shown on the right. [VISUAL] Note that this represents a two-dimensional projection of the full three-dimensional velocity distribution.

15.1.

A potential of 3000 V is used to direct the ionised Cl atoms to the detector. What is their flight time? Take the molar mass of Cl to be 35 g mol–1.

Model Answer

The kinetic energy of the ions is eV = ½ m v2, so v = (2 eV / m)1/2 = 128 600 m s–1. The distance the ions fly is d = 0.4 m, so the flight time is t = d / v = 3.11 µs.

15.2.

The image appears as a single ring of Cl atoms as a result of conservation of energy and momentum. The outside diameter of the ring is 12.68 mm. What velocity did the Cl atoms acquire as a result of the photodissociation?

Model Answer

The atoms travel a radial distance of 6.34 mm in 3.11 µs, so their velocity is vCl = 2038 m s–1.

15.3.

The bond dissociation energy of Cl2 is 243 kJ mol–1. Use conservation of energy to determine the laser wavelength.

Model Answer

Conservation of energy requires that hν – D0 = 2 (½ mCl vCl 2).
From the data given, D0 = 4.035×10–19 J (2.519 eV), mCl = 5.812×10–26 kg, vCl = 2038 m s–1.
The photon energy is therefore hν = 6.449×10–19 J (4.026 eV), corresponding to a wavelength λ = hc / E of 308 nm.

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