14. What is the probability of finding a n = 2, /= 1 electron 26 between 15 a For a hydrogen and 2ao? Orbital angular momentum Orbital magnetic quantum number |L|= √√(+1)h 7.2 Angular probability P(0, 4) = 7.5 (1= 0, 1, 2,...) density |0lm,(0)m, ($)|2 L₁ = m₁h 7.2 Orbital magnetic ₁ =-(e/2m)L 7.6 (m,=0,11,12,..., ±1) dipole moment Spatial quantization L₁₂ m₁ cos 0 = = 7.2 Spin magnetic dipole s = -(e/m)Š 7.6- |L| √(+1) moment Angular momentum uncertainty relationship Hydrogen quantum numbers Hydrogen energy levels Hydrogen wave functions ΔΙΑΦ ΣΑ n = 1,2,3,... = 0, 1, 2,...,n-1 m₁ =0,11,12,...,±l 7.2 Spin angular |S|= √√s(s+1)h= 7.6 momentum √√3/4h (for s = 1/2) Spin magnetic 7.3 S=mh (m,±2) 7.6 quantum number Spectroscopic notation En=- me 1 32л²² n² 7.3 7.3 Selection rules for photon emission Normal Zeeman effect s(10), p (1 = 1), d(l=2),f(l= 3),... ΔΙ = +1 Δ = 0,11 7.7 7.7, 7.8 2² R()(0) (0) Radial probability density P(r) = r²|R(r)|2 7.4 Fine-structure estimate Δλ= AE = mc2a/n = AE = BB 7.8 7.9 (a≈1/137)

14. What is the probability of finding a n = 2, /= 1 electron 26 between 15 a For a hydrogen and 2ao? Orbital angular momentum Orbital magnetic quantum number |L|= √√(+1)h 7.2 Angular probability P(0, 4) = 7.5 (1= 0, 1, 2,...) density |0lm,(0)m, ($)|2 L₁ = m₁h 7.2 Orbital magnetic ₁ =-(e/2m)L 7.6 (m,=0,11,12,..., ±1) dipole moment Spatial quantization L₁₂ m₁ cos 0 = = 7.2 Spin magnetic dipole s = -(e/m)Š 7.6- |L| √(+1) moment Angular momentum uncertainty relationship Hydrogen quantum numbers Hydrogen energy levels Hydrogen wave functions ΔΙΑΦ ΣΑ n = 1,2,3,... = 0, 1, 2,...,n-1 m₁ =0,11,12,...,±l 7.2 Spin angular |S|= √√s(s+1)h= 7.6 momentum √√3/4h (for s = 1/2) Spin magnetic 7.3 S=mh (m,±2) 7.6 quantum number Spectroscopic notation En=- me 1 32л²² n² 7.3 7.3 Selection rules for photon emission Normal Zeeman effect s(10), p (1 = 1), d(l=2),f(l= 3),... ΔΙ = +1 Δ = 0,11 7.7 7.7, 7.8 2² R()(0) (0) Radial probability density P(r) = r²|R(r)|2 7.4 Fine-structure estimate Δλ= AE = mc2a/n = AE = BB 7.8 7.9 (a≈1/137)

Similar questions

13
Apply the Pauli exclusion principle to determine the number of electrons that could occupy the quantum states described by the following. HINT (a) n = 4, { = 3, m2 : = -1 electrons (b) n = 3, { = 2 %3D electrons (C) n = 4 electrons
excited nydrogen atoms need less energy to oxidize than normal ones. It is in the ground condition when Can an electron in a hydrogen atom be ionized with less than 170.825 eV 07 energy? n=?
2. The following simplified mechanism captures some of the features of a 3-level laser: X+ hv -→ I I → A A → X+ hv’ r= kp f(hv)[X] r = k(IC1) [I] r=kf [A] r = kị f(hv') [A] pumping intersystem crossing spontaneous emission lasing hv' + A → X+ 2hv' f(hv) = flux of photons with energy = hv f(hv') = flux of photons with energy = hv' (usually a constant value) (a constant value different from f(hv)) Note that spontaneous emission (usually fluorescence) is detrimental to efficient lasing. a) Sketch a diagram of the system and identify all the states including metastable states also known as intermediates. b) Derive the steady-state concentrations for all metastable states. c) Show that population inversion ([A]ss/[X] constant is much larger than the sum of the rate constant of spontaneous emission plus the lasing rate constant. [A]ss = steady-state concentration of A. 1) is possible if the pumping rate >>
-7 For hydrogen atom electron, with Spin , in the state a -A(), Determine A, , of Sx, Sy,S2 13217水 and with spin State x く>,くら、?,くら,>, and
4. In describing the atom, we often refer to the "electron cloud." Does this mean that the electron is spread out in space in the form of a cloud? Just what is that cloud? In hat ratio would vou expet Ma and Clto combine? ? Why?
7. For the 3p and 3d states of hydrogen, show that (-) avg =1/1290. ao. n²