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ATOMIC PHYSICS1BE-PHYSICS-ATOMIC PHYSICS-2010-11MIT- MANIPALTOPICSText BookPHYSICS for Scientists and Engineers with Modern Physics (6th ed) By Serway & JewettAtomic spectra of gasesEarly models of the atomBohrs model of the hydrogen atomThe quantum model of the hydrogen atomThe wave functions for hydrogenPhysical interpretation of the quantum numbersThe X-ray spectrum of atomsX-rays and the numbering of the elementsLasers and laser lightTOPICSText BookPHYSICS, 5TH Edition Vol 2Halliday, Resnick, Krane

ATOMIC SPECTRA OF GASESEmission spectra: All objects emit thermal radiation characterized by a continuous distribution of wavelength (continuous spectrum). When a gas at low pressure is subjected to an electric discharge it emits radiations of discrete wavelengths (line spectrum).No two elements have the same line spectrum. This principle is used in identifying the element by analyzing its line spectrum.

HHg

The wavelengths of the Balmer series lines in the hydrogen spectrum are given by the equation n = 3, 4, 5, . . .

Rydberg constant RH= 1.097 x 107/mAbsorption spectra: An absorption spectrum is obtained by passing white light from a continuous source through a gas or a dilute solution of the element being analyzed. The absorption spectrum consists of a series of dark lines superimposed on the continuous spectrum of the light source.

SOLAR SPECTRUMFRAUNHOFER LINES

VISIBLE HYDROGEN SPECTRUM BALMER SERIES LINESH(656.3 nm) H(656.3 nm) H(656.3 nm) H(656.3 nm)

The wavelengths of the other series lines in the hydrogen spectrum are given by the equation LymanSeriesn = 2, 3, 4, . . .

PaschenSeriesn = 4, 5, 6, . . .

BrackettSeriesn = 5, 6, 7, . . .

Although no theoretical basis existed for these equations, they are in agreement with the experimental results.

EARLY MODELS OF THE ATOM

EARLY MODELS OF THE ATOM

EARLY MODELS OF THE ATOM

[1] (a) What value of n is associated with the 94.96-nm spectral line in the Lyman series of Hydrogen ? (b) Could this wavelength be associated with the Paschen or Balmer series ?SOLUTION:(a)Lyman Series

(b) Paschen Series

The shortest wavelength for this series corresponds to n = for ionization. For n = , gives = 820 nm. This is larger than 94.96 nm, so this wave length cannot be associated with the Paschen series Balmer Series

with n = for ionization, min = 365 nm. Once again the shorter given wavelength cannot be associated with the Balmer series

BOHRS MODEL OF THE HYDROGEN ATOM In his semi classical model of the H-atom Bohr postulated that-[1] The electron moves in circular orbits around the proton under the influence of the electric force of attraction as shown in the figure

v+emeer

[2] Only certain electron orbits are stable (stationary states). When in one of the stationary state, the atom does not radiate energy. Hence the total energy of the atom remains constant in a stationary state.

[3] When the atom makes a transition from higher energy state (Ei) to lower energy state (Ef) [ie, the electron makes a transition from a stable orbit of larger radius to that of smaller radius], radiation is emitted. The frequency (f) of this radiation (photon) is given by Ei Ef = h f .The frequency f of the photon emitted is independent of the frequency of electrons orbital motion.

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[4] The angular momentum of the electron in any stable orbit is quantized mev r = n n = 1, 2, 3, . . . me = mass of the electronv = speed of the electron in the orbitr = radius of the electrons orbit

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Electric potential energy of the H-atom is

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The total energy of the H-atom is Apply Newtons 2ND law to the electron, the electric force exerted on the electron must be equal to the product of mass and its centripetal acceleration (a=v2/r)

ke= Coulomb constant

The total energy of the H-atom is From Newtons 2ND law equation and orbit quantization equation

Thus the electron orbit radii are quantized rn= n2 aon = 1, 2, 3, . . .

Bohr radius

+ee4aoao9ao

Energy quantizationSubstitute rn= n2 ao in the total energy equation

n = 1, 2, 3, . . .

E1= 13.606 eV

Ionization energy = minimum energy required to ionize the atom in its ground state = 13.6 eV for H-atomFrom the equation Ei Ef = h fFrequency of the photon emitted during transition of the atom from state i to state f is

Use c = f

RH = 1.097 x 107 /m

Extension of Bohrs theory to other one-electron atoms - Nuclear charge = + Z eradius

Energy

Limitations of Bohrs theory:When spectroscopic techniques improved, it was found that many of the lines in the H-spectrum were not single lines but closely spaced groups of lines. The lines appear split when the H-vapour was kept in magnetic field.

Bohrs correspondence principle:Quantum physics agrees with classical physics when the difference between quantized levels becomes vanishingly small.

PROBLEMS

[1] Spectral lines from the star -Puppis: Some mysterious lines in 1896 in the emission spectrum of the star -Puppis fit the empirical equation

Show that these lines can be explained by the Bohrs theory as originating from He+.

SOLUTION: The ion He+ has z=2, Thus allowed energy levels are given by

[2] (A) The electron in a H-atom makes a transition from the n=2 energy level to the ground level (n=1). Find the wavelength and the frequency of the emitted photon. (B) In interstellar space highly excited hydrogen atoms called Rydberg atoms have been observed. Find the wavelength to which radioastronomers must tune to detect signals from electrons dropping from n=273 level to n=272.(C) What is the radius of the electron orbit for a Rydberg atom for which n=273 ? (D) How fast is the electron moving in a Rydberg atom for which n=273?(E) What is the the wavelength of the radiation from the Rydberg atom in part (B) if treated classically ?

SOLUTION(A)

SOLUTION(B)

SOLUTION(C)

rn= n2 ao= 2732 (0.0529nm)

r273=3.94m

SOLUTION(D)

SOLUTION(E)

We have speed v and radius r from (C) and (D)

[3] According to classical physics, a charge e moving with an acceleration a radiates at a rate

(a) Show that an electron in a classical hydrogen atom spirals into the nucleus at a rate

(b) Find the time interval over which the electron will reach r = 0, starting from ro = 2.00 x 1010 m

SOL:A The total energy is given by

The centripetal acceleration a is given by

SOL:B

[4] A hydrogen atom is in the first excited state (n = 2). Using the Bohr theory of the atom, calculate (a) the radius of the orbit (b) the linear momentum of the electron (c) the angular momentum of the electron (d) the kinetic energy of the electron the potential energy of the system and the total energy of the system.

[5] A photon is emitted as a hydrogen atom undergoes a transition from the n = 6 state to the n = 2 state. Calculate (a) the energy (b) the wavelength the frequency of the emitted photon.Solution b:

Solution a:

Solution c:

[6] (a) Construct an energy-level diagram for the He+ ion (Z = 2). (b) What is the ionization energy for He+ ?

Solution a: The energy levels of a hydrogen-like ion whose charge number is Z are given byThus for Helium (Z = 2), the energy levels are

(b) What is the ionization energy for He+ ?Solution b: For He+ , Z = 2 , so we see that the ionization energy (the energy required to take the electron from the n = 1 to the n = state) is

THE QUANTUM MODEL OF THE HYDROGEN ATOM 36The potential energy function for the H-atom is

ke = 8.99 x 109 N.m2/C2 is Coulomb constantr = radial distance of electron from proton [H-nucleus] (at r = 0)The time-independent schrodinger equation in 3-dimensional space is

Since U has sperical symmetry, it is easier to solve the schrodinger equation in spherical polar coordinates (r, , ):where is the angle between z-axis and

Pyxz

is the angle between the x-axis and the projection of onto the xy-plane. It is possible to separate the variables r, , as follows:(r, , ) = R(r) f() g()By solving the three separate ordinary differential equations for R(r), f(), g(), with conditions that the normalized and its first derivative are continuous and finite everywhere, one gets three different quantum numbers for each allowed state of the H-atom. The quantum numbers are integers and correspond to the three independent degrees of freedom.

Pyxz

The radial function R(r) of is associated with the principal quantum number n. From this theory the energies of the allowed states for the H-atom are

The polar function f() is associated with the orbital quantum number l. The azimuthal function g() is associated with the orbital magnetic quantum number ml. The application of boundary conditions on the three parts of leads to important relationships among the three quantum numbers: [1] n can range from 1 to . [2] l can range from 0 to n1 ; [n allowed values]. [3] ml can range from l to +l ; [(2l+1) allowed values].which is in agreement with Bohr theory.

All states having the same principal quantum are said to form a shell. All states having the same values of n and l are said to form a subshell:

n = 1 K shell l = 0 s subshelln = 2 L shell l = 1 p subshelln = 3 M shell l = 2 d subshelln = 4 N shell l = 3 f subshelln = 5 O shell l = 4 g subshelln = 6 P shell l = 5 h subshell. . . . . . . .. . . . . . . . . . . . . . . .. . . . . . . .

The potential energy for H-atom depends only on the radial distance r between nucleus and electron. some of the allowed states for the H-atom can be represented by wave functions that depend only on r (spherically symmetric function). The simplest wave function for H-atom is the 1s-state (ground state) wave function (n = 1, l = 0):

ao = Bohr radius.

|1s|2 is the probability density for H-atom in 1s-state.

THE WAVE FUNCTIONS FOR HYDROGEN

The radial probability density P(r) is the probability per unit radial length of finding the electron in a spherical shell of radius r and thickness dr.

P(r) dr is the probability of finding the electron in this shell.P(r) dr = ||2 dv = ||2 4r2 dr P(r) = 4r2 ||2 Radial probability density for H-atom in its ground state:

Plot of the probability of finding the electron as a function of distance from the nucleus for H-atom in the 1s (ground) state. P1s(r) is maximum when r = ao (Bohr radius). Cross-section of the spherical electronic charge distribution of H-atom in 1s-state

rMOST PROBABLE = aorAVERAGE = 3ao/2

rMOST PROBABLE = 5aoThe next simplest wave function for the H-atom is the 2s-state wave function (n = 2, l = 0):

2s is spherically symmetric (depends only on r). E2 = E1/4 = 3.401 eV (1ST excited state).

[1] - SJ-Example-42.3: For a H-atom, determine the number of allowed states corresponding to the principal quantum number n = 2, and calculate the energies of these states.Solution:When n= 2, l can have the values 0 and 1. If l=0, ml can only be 0.If l=1, ml can be -1, 0, or +1.Hence, we have a 2s state with quantum numbersn= 2, l=1, ml =0and three 2p states for which the quantum numbers aren= 2, l=1, ml =-1n= 2, l=1, ml =0n= 2, l=1, ml =+1All these states have the same principal quantum number, n=2, they also have the same energy, En =(-13.66eV) Z2 /n2 E2 =-(13.66eV)/22=-3.401eV

[2] SJ-Example-42.4.Calculate the most probable value of r (= distance from nucleus) for an electron in the ground state of the H-atom. Also calculate the average value r for the electron in the ground state.Solution: The most probable distance is the value of r that makes the radial probability P(r) a maximum. The slope here is zero, so the most probable value of r is obtained by setting dP/dr= 0 and solving for r.

The expression is satisfied ifThe most probabale value of r is the Bohr radius

The average value of r is the expectation value of r

The expectation value is given by

[3] SJ-Example-42.5 Probabilities for the electron in H-atom: Calculate the probability that the electron in the ground state of H-atom will be found outside the Bohr radius.Solution: The probability is found by integrating the radial probability density for this state, P1s(r), from the Bohr radius a0 to .

We can put the integral in dimensionless form by changingvariables from r to z = 2r/a0. Noting that z=2 when r=a0, and that dr=(a0/2)dz, we get

This is about 0.677, or 67.7%.

[5] SJ-Problem-42.16: A general expression for the energy levels of one-electron atoms and ions is

where ke is the the Coulomb constant, q1 and q2 are the charges of the electron and the nucleus, and is the reduced mass, given byThe wavelength for n = 3 to n = 2 Transition of the hydrogen atom is 656.3 nm (visible red light). What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron, and (b) singly ionized helium ?

Solution:The reduced mass of positronium is less than hydrogen, so the photon energy will be less for positronium than for hydrogen. This means that the wavelength of the emitted photon will be longer than 656.3 nm.

On the other hand, helium has about the same reduced mass but more charge than hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than 656.3 nm.

All the factors in the given equation are constant for this problem except for the reduced mass and the nuclear charge. Therefore, the wavelength corresponding to the energy difference for the transition can be found simply from the ratio of mass and charge variables.

so the energy of each level is one half as large as in hydrogen, which we could call protonium. The photon energy is inversely proportional to its wavelength , so forpositronium,so the transition energy is 22 = 4 times larger than hydrogen.

[6] SJ-Problem-42.17: An electron of momentum p is at a distance r from a stationary proton. The electron has a kinetic energy The atom has a potential energy and total energy E = K + U. If the electron is bound to the proton to form a H-atom, its average position is at the proton, but the uncertainty in its position is approximately equal to the radius r of its orbit. The electrons average vector momentum is zero, but its averaged squared momentum is equal to the squared uncertainty in its momentum, as given by the uncertainty principle.

An electron of momentum p is at a distance r from a stationary proton. Treating the atom as one-dimensional system, estimate the uncertainty in the electrons momentum in terms of r. Estimate the electrons kinetic, potential, and total energies in terms of r. The actual value of r is the one that minimizes the total energy, resulting in a stable atom. Find that value of r and the resulting total energy. Compare your answer with the predictions of the Bohr theory.

[5] SJ-Problem-42.21: For a spherically symmetric state of a H-atom the schrodinger equation in spherical coordinates is Show that the 1s wave function for an electron in H-atom satisfies the schrodinger equation.

Solution:

This is true , so the schrodinger equation is satisfied