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

Info icon This preview shows pages 4–7. Sign up to view the full content.

View Full Document Right Arrow Icon
(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.
Image of page 4

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 5
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?
Image of page 6

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern