{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CH 301 Exam 1 Help Sheet(2)

CH 301 Exam 1 Help Sheet(2) - Name De Broglie wavelength...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
±²͵ͲͳǦͷͲͻ͹Ͳ ² ³ ´ ͓ͳ ³ ͻǡ ʹͲͳͳ µ ͳ Name Equation Explanation E = mc 2 Relationship between energy and mass c = ȜȞ Indirect proportionality of Ȝ and Ȟ E = h Ȟ Direct proportionality between E and Ȟ De Broglie wavelength Ȝ = h/mv (v = velocity) Moving body has wavelength Bohr Energy ǻ E = -2.178x10 -18 J (Z 2 /n 2 f - Z 2 /n 2 i ) Z=atomic no., n=principal quantum no. Photoelectric effect KE = ½ mv² = h Ȟ - ĭ ĭ = work function = h Ȟ 0 Schrödinger H Ȍ = E Ȍ H = Hamiltonian operator, E = total atom energy, Ȍ wave function Psi Ȍ n, Ɛ ,m Ɛ Representation of an orbital, origin of quantum numbers Psi squared Ȍ ² n, Ɛ ,m Ɛ Represents probability distribution of e- position in space Heisenberg uncertainty ʄ /2 = ǻ ǻ p = ǻ x·m ǻ v m = mass, x = position, p = momentum, v = velocity, ǻ = uncertainty Nodes in s orbital number = n-1 Quantum number Rules Principal quantum no. n = 1, 2, 3, 4, 5, Main electron energy level or shell Angular momentum Ɛ = 0, 1, 2, 3, n-1 Energy sublevel or subshell; all orbitals in subshell are degenerate Magnetic quantum no. m Ɛ = - Ɛ 0 Ɛ Specifies orbital, no. of values of m
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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