Ex_2A_Fall 2015.pdf

# 7 pts a find the energy of the 1014 nm photon just

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(7 pts) (a). Find the energy of the 1014 nm photon. Just use E= hc/ λ . E = hc λ = (6.626 × 10 –34 Js)(2.9979 × 10 8 m/s) 1014 × 10 –9 m = 1.959 × 10 –19 J You could also find frequency first – should be 2.957 × 10 14 s -1 . (3 pts) (b). This process is summarized in the energy level diagram below, where the 3 arrows represent the transitions giving rise to the emission wavelengths listed above. Which arrow (A, B, or C) corresponds to the wavelength whose energy you found in part (a)? (You can assume the diagram is drawn at least approximately to scale.) A B C E A: longest wavelength = smallest energy I did not ask for a reason, so this is just right or wrong.

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A6 © 2015 L.S. Brown (8 pts) 5. A sample of N 2 gas occupies a volume of 1.25 L at 45.0°C and 0.975 atm. The temperature and the pressure of the gas are both changed, and as a result the gas expands to a final volume of 2.75 L. Find the density of the gas at the end of this process. We want density at the end. That’s mass/volume, and we know the volume. So we need to find the mass. Use the initial data to find moles, then mass. n = PV RT = (0.975 atm)(1.25 L) (0.08206 L atm mol K )(318.15 K) = 4.668 × 10 –2 mol N 2 4.668 × 10 –2 mol N 2 × 28.02 g 1 mol N 2 = 1.308 g 1.308 g 2.75 L = 0.476 g/L (8 pts) 6. Ultraviolet light with wavelength 363 nm is able to dissociate (or break) a carbon–chlorine bond. What is the C–Cl bond energy in kJ/mol? Start by converting the given wavelength into an energy: E = hc λ = (6.626 × 10 –34 Js)(2.9979 × 10 8 m/s) 363 × 10 –9 m = 5.472 × 10 –19 J That’s the bond energy for one bond, so we just need to convert that into kJ/mol. 5.472 × 10 –19 J × 1 kJ 1000 J × 6.022 × 10 23 1 mol = 330 kJ mol We split the points: half for getting the photon E and the other half for getting to kJ/mol.
NAME:_________________________________ © 2015 L.S. Brown A7 7. (8 pts) (a). A photoelectric effect experiment is carried out using ultraviolet light with a frequency of 1.370 × 10 15 s –1 and a silver (Ag) metal target. If electrons are emitted with a kinetic energy of 1.488 × 10 –19 J, what is the electron binding energy for silver?

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