{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CH 301 Exam 1 Help Sheet(2)

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

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

±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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern