Show that the probability density for the ground-state solution of the one-dimensional Coulomb potential energy has its maximum at x = a.

icon
Related questions
Question
Show that the probability density for the ground-state
solution of the one-dimensional Coulomb potential
energy has its maximum at x = a.
Transcribed Image Text:Show that the probability density for the ground-state solution of the one-dimensional Coulomb potential energy has its maximum at x = a.
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, ($)|2
L₁ = m₁h
7.2
Orbital magnetic
₁ =-(e/2m)L
7.6
(m,=0,11,12,..., ±1)
dipole moment
Spatial quantization
L₁₂ m₁
cos 0 =
=
7.2
Spin magnetic dipole
s = -(e/m)Š
7.6-
|L|
√(+1)
moment
Angular momentum
uncertainty
relationship
Hydrogen quantum
numbers
Hydrogen energy
levels
Hydrogen wave
functions
ΔΙΑΦ ΣΑ
n = 1,2,3,...
= 0, 1, 2,...,n-1
m₁ =0,11,12,...,±l
7.2
Spin angular
|S|= √√s(s+1)h=
7.6
momentum
√√3/4h (for s = 1/2)
Spin magnetic
7.3
S=mh (m,±2)
7.6
quantum number
Spectroscopic notation
En=-
me 1
32л²² n²
7.3
7.3
Selection rules for
photon emission
Normal Zeeman effect
s(10), p (1 = 1),
d(l=2),f(l= 3),...
ΔΙ = +1 Δ = 0,11
7.7
7.7, 7.8
2²
R()(0) (0)
Radial probability
density
P(r) = r²|R(r)|2
7.4
Fine-structure estimate
Δλ=
AE = mc2a/n
= AE = BB
7.8
7.9
(a≈1/137)
Transcribed Image Text:Orbital angular momentum Orbital magnetic quantum number |L|= √√(+1)h 7.2 Angular probability P(0, 4) = 7.5 (1= 0, 1, 2,...) density |0lm,(0)m, ($)|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)
Expert Solution
steps

Step by step

Solved in 3 steps with 11 images

Blurred answer