How do you simplify #(n!)/((n-2)!)#? - Socratic
2015年11月18日 · We can write it as: n ⋅ (n −1)(n − 2)(n − 3)! (n − 2)(n −3)! =. where you used the fact that n! = n(n − 1)! and so: n ⋅ (n −1)(n − 2) (n − 3)! (n − 2) (n − 3)! = n(n − 1) Check it with n = 4. n! (n − 2)! = 4! 2! = 12. n(n −1) = 4 ⋅ 3 = 12. Answer link.
How many electrons are in n = 2? What about n = 4, l = 3? What …
2015年11月17日 · Here's what I got. Since the question is a bit ambiguous, I will assume that you're dealing with three distinct sets of quantum numbers. In addition to this, I will also assume that you're fairly familiar with quantum numbers, so I won't go into too much details about what each represents. 1^"st" set -> n=2 The principal quantum number, n, tells you the energy level on which an electron ...
How do you simplify the factorial expression #((n+2)!)/(n!)
2017年2月8日 · ((n+2)!)/(n!) = (n+2)(n+1) Remember that: n! =n(n-1)(n-2)...1 And so (n+2)! =(n+2)(n+1)(n)(n-1) ... 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \=(n+2)(n+1)n!
How do you simplify #((n+2)! )/( n!)#? - Socratic
n^2+3n+2 We can rewrite the numerator as: ((n+2) * (n+2-1) * (n+2-2)!)/((n)!) =((n+2) * (n+1) * (n)!)/((n)!) We can cancel (n)! and (n)! out: =((n+2) * (n+1) * 1)/1 ...
How do you simplify ((2n)!)/(n!)? + Example - Socratic
2015年12月10日 · Explanation: While there isn't a simplification of (2n)! n!, there are other ways of expressing it. For example. (2n)! n! = n−1 ∏ k=0(2n −k) = (2n)(2n − 1)...(n +1) This follows directly from the definition of the factorial function and canceling common factors from the numerator and denominator. (2n)! n! = 2nn−1 ∏ k=0(2k +1) = 2n(1 ...
What electron could have quantum numbers n=2, l=1, m_l = 0
2016年6月26日 · The magnetic quantum number, ml, tells you the exact orbital in which the electron is located. The p-subshell contains a total of three orbitals, by convention assigned as. ml = − 0 → the 2pz orbital. ml = − 1 → the 2py orbital. ml = + 1 → the 2px orbital. In your case, ml = 0, which means that your electron is located in the 2pz orbital.
An electron changes from an n = 2 to an n = 6 energy state
2017年11月27日 · 4.85 * 10^(-19) "J" The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from n_i = 2 to n_f = 6. A good starting point here will be to calculate the energy of the photon emitted when the electron falls from n_i = 6 to n_f = 2 by using the Rydberg equation. 1/(lamda) = R * (1/n_f^2 - 1/n_i^2) Here lamda si the ...
What are the molecular orbital configurations for N_2^+, N_2
2015年11月2日 · If we build the MO diagram for "N"_2, it looks like this: First though, notice that the p orbitals are supposed to be degenerate. They weren't drawn that way on this diagram, but they should be. Anyways, for the electron configurations, you would use a notation like the above. g means "gerade", or even symmetry upon inversion, and u means "ungerade", or odd symmetry upon inversion. It's not ...
Quantum Numbers - Chemistry - Socratic
Quantum numbers can be used to describe the quantum state of an electron. There are four quantum numbers for atoms: n = 1,2,3,... - principal quantum number; describes the energy level. l = 0,1,2,...,n − 1 - angular momentum quantum number; describes the shape of the orbital. 0 ↔ s,1 ↔ p,2 ↔ d,3 ↔ f,..., etc. The ordering is s,p,d,f,g ...
How do you calculate the wavelength of the light emitted by
2016年12月16日 · How do you calculate the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level? Recall that for hydrogen #E_n = -2.18 xx 10^-18 J(1/n^2)#