Physics for Scientists and Engineers with Modern Physics
4th Edition
ISBN: 9780131495081
Author: Douglas C. Giancoli
Publisher: Addison-Wesley
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Chapter 39, Problem 19P
To determine
The proof that the mean value of
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Determine the integral
| P(r) dr
for the radial probability density for the ground state of the hydrogen atom
4
P(r) =
- r²e-2rla
a³
O 1
O-1
O 0.5
.55 The radial probability density for the ground state of the
hydrogen atom is a maximum whenr = a, where a is the Bohr ra-
dius. Show that the average value of r, defined as
P(r) r dr,
has the value 1.5a. In this expression for ravgs each value of P(r) is
weighted with the value of r at which it occurs. Note that the average
value of r is greater than the value of r for which P(r) is a maximum.
Consider an electron in the ground state of a Hydrogen atom:
a) Find (r) And (") in terms of the Bohr radius
b) Find () for the state n = 2, I = 1 and m = 1 considering that x =r cos 0 sin o.
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Chapter 39 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 39.2 - Prob. 1AECh. 39.2 - Prob. 1BECh. 39.3 - Prob. 1CECh. 39.4 - Prob. 1DECh. 39.4 - Prob. 1EECh. 39.5 - Prob. 1FECh. 39.7 - Prob. 1GECh. 39 - Prob. 1QCh. 39 - Prob. 2QCh. 39 - Prob. 3Q
Ch. 39 - Prob. 4QCh. 39 - Prob. 5QCh. 39 - Prob. 6QCh. 39 - Prob. 7QCh. 39 - Prob. 8QCh. 39 - Prob. 9QCh. 39 - Prob. 10QCh. 39 - Prob. 11QCh. 39 - On what factors does the periodicity of the...Ch. 39 - Prob. 13QCh. 39 - Prob. 14QCh. 39 - Prob. 15QCh. 39 - Prob. 16QCh. 39 - Prob. 17QCh. 39 - Prob. 18QCh. 39 - Prob. 19QCh. 39 - Prob. 20QCh. 39 - Prob. 21QCh. 39 - Prob. 22QCh. 39 - Prob. 23QCh. 39 - Prob. 24QCh. 39 - Prob. 25QCh. 39 - Prob. 26QCh. 39 - Prob. 27QCh. 39 - Prob. 28QCh. 39 - Prob. 29QCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3PCh. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 7PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 14PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 21PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 28PCh. 39 - Prob. 29PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 34PCh. 39 - Prob. 35PCh. 39 - Prob. 36PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40PCh. 39 - Prob. 41PCh. 39 - Prob. 42PCh. 39 - Prob. 43PCh. 39 - Prob. 44PCh. 39 - Prob. 45PCh. 39 - Prob. 46PCh. 39 - Prob. 47PCh. 39 - Prob. 48PCh. 39 - Prob. 49PCh. 39 - Prob. 50PCh. 39 - Prob. 51PCh. 39 - Prob. 52PCh. 39 - Prob. 53PCh. 39 - Prob. 54PCh. 39 - Prob. 55PCh. 39 - Prob. 56PCh. 39 - Prob. 57PCh. 39 - Prob. 58PCh. 39 - Prob. 59PCh. 39 - Prob. 60PCh. 39 - Prob. 61GPCh. 39 - Prob. 62GPCh. 39 - Prob. 63GPCh. 39 - Prob. 64GPCh. 39 - Prob. 65GPCh. 39 - Prob. 66GPCh. 39 - Prob. 67GPCh. 39 - Prob. 68GPCh. 39 - Prob. 69GPCh. 39 - Prob. 70GPCh. 39 - Prob. 71GPCh. 39 - Prob. 72GPCh. 39 - Prob. 73GPCh. 39 - Prob. 74GPCh. 39 - Prob. 75GPCh. 39 - Prob. 76GPCh. 39 - Prob. 77GP
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- Consider an electron in the ground state of a Hydrogen atom: a) Find () and (²) in terms of the Bohr radius a0. b) Find (-) for state n = 2, 1 = 1 and m = 1 considering that x = r cos 0 sin ø.arrow_forwardIf an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bhor radius is approximately 1. 0.68 2. 0.48 3. 0.28 4. 0.91arrow_forwardH-atom. The wave function of one of the electrons in the 2p orbital is given by (ignoring spin) r 2,1,0 (1,0,0)= - 7 exp(-270) c ao 1 |32πα cose Where do is the Bohr radius. In the Bohr model, the radius of the electron orbit is given by m=2 = n²ao = 4ao. The probability that the electron can be found at some radius between r and r + dr is given by 2π P(r) dr = √2 = √ ₁²ª d$ S ² What is the expectation value of the distance of the electron from the nucleus (r)? Clue: expected value is computed by (r) = forP(r) dr then do integration by parts do sin 0 de | Yn.l.m² (r, $,0)|²r² drarrow_forward
- (1) Find the average orbital radius for the electron in the 3p state of hydrogen. Compare your answer with the radius of the Bohr orbit for n=3. (2) What is the probability that this electron is outside the radius given by the Bohr model?arrow_forward2m || 1 2.2 The wave function of the ground state of hydrogen has the form. 1 100 2ma² e r To TT13 ㅠ 0 Find the probability of finding the electron in a volume dV around a given point.arrow_forward(b) A photon is emitted by a doubly ionised lithium atom (Li²+) when an electron makes a transition to the ground state. The wavelength of the photon is measured to be 10.83 nanometres. Determine the principal quantum number and the energy of the initial state The atomic number of lithium is Z = 3.arrow_forward
- The wavelengths of the Lyman series for hydrogen are given by: = RH(1-1), n = 2, 3, 4, ... For the second of this series; calculate the energy (in eV). Note: 1.60 x 10^-19 J = 1.0 eV O 4.10 x 10^3 eV 2.12 x 10^3 eV 3² O 1.21 x 10^3 eV 3.30 x 10^3 eVarrow_forward(a) What is the radius in nm of the electron orbit of a hydrogen atom for n = 4? (b) What is the energy in eV of the electron above?arrow_forwardConsidering the Bohr’s model, given that an electron is initially located at the ground state (n=1n=1) and it absorbs energy to jump to a particular energy level (n=nxn=nx). If the difference of the radius between the new energy level and the ground state is rnx−r1=5.247×10−9rnx−r1=5.247×10−9, determine nxnx and calculate how much energy is absorbed by the electron to jump to n=nxn=nx from n=1n=1. A. nx=9nx=9; absorbed energy is 13.4321 eV B. nx=10nx=10; absorbed energy is 13.464 eV C. nx=8nx=8; absorbed energy is 13.3875 eV D. nx=20nx=20; absorbed energy is 13.566 eV E. nx=6nx=6; absorbed energy is 13.22 eV F. nx=2nx=2; absorbed energy is 10.2 eV G. nx=12nx=12; absorbed energy is 13.506 eV H. nx=7nx=7; absorbed energy is 13.322 eVarrow_forward
- (i) Using Bohr model for atomic hydrogen, obtain energy levels for the 2s, 3s and 3p states in the actual number with the unit of [eV]. We consider a transition that electron in the 3p state emits a photon and make a transition to the 2s state. What is the frequency v of this photon ? (ii) Now we do not include electron spin angular momentum, and just estimate an effect of a magnetic field B on this transition (Normal Zeeman effect) with orbital angular momentum. How many lines of optical transition do we expect ? What is the interval of the frequency in the field B = 0.1 Tesla ? (iii) In this situation, we do not expect transition from 3s to 2s state if the electron is initially in the 3s state, Explain the reason. (iv) We now consider an effect of magnetic field B to a free electron spin (not in Hydrogen, but a free electron). The magnetic field of B = 1.0 Tesla will split the energy level into two (Zeeman) levels. Obtain the level difference in the unit of [eV] from the value of…arrow_forwardstate? (b) A hydrogen atom transitions from the n = frequency of the emitted photon is 1.574 x10" Hz, what is the value of n? 8 state to a final n state. If thearrow_forward(3) Calculate (x²) for in the quantum state {n = 2, l = 1, m = average can be expressed more simply as (2,1,1|x2|2,1,1) where the ket |n, l, m) defines the quantum state 1} for a hydrogen atom. Notice that this of a hydrogen atom, but, as done in class, the spin quantum number is not included.arrow_forward
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