QM1 - Tutorial 1 The Bohr Atom and Mathematical Introduction

Size: px

Start display at page:

Download "QM1 - Tutorial 1 The Bohr Atom and Mathematical Introduction"

Augusta Chastity Logan
5 years ago
Views:

1 QM - Tutorial The Bohr Atom and Mathematical Introduction 26 October 207 Contents Bohr Atom. Energy of a Photon - The Photo Electric Eect The Discrete Energy Spectrum of the Hydrogen Atom ight Emitted by an Atom Fourier Transform 4 2. Orthogonality of Exponential Functions Fourier Series Fourier Transform Dirac Delta Function 8 3. Denition, Representations and Properties Fourier Transform of the Dirac Delta Function Bohr Atom. Energy of a Photon - The Photo Electric Eect Experimental Observation When light hits a metal it can free electrons from it. This is in itself surprising, but even more so is the fact that the eect depends on the frequency or wavelength) of the light - not on the intensity. Physical Interpretation Einstein 905) ight is not continuous. It consists of particles lets call them photons) that carry energy and momentum. The energy E ph of a microscopic) photon depends on the frequency ν of the macroscopic) light wave. E ph hν Consider a single photon with energy E ph hitting a metal. The photon will be absorbed by the metal i.e. it disappears) while its energy is transferred to an electron in the metal. The equation for conservation of energy is E elec E ph φ where φ is the work function of the metal. It is the minimal amount of energy needed to pull an electron out to the metal. This means that the energy of the photon must be larger than the work function for the photo electric eect to happen. E ph φ The additional energy that the photon carried becomes kinetic energy of the electron. It is called a function although it does not depend on the parameters of the problem. It depends on the type of metal.

2 .2 The Discrete Energy Spectrum of the Hydrogen Atom Incompatibility of Classical Mechanics According to classical physics the atom consists of a heavy positive nucleus with negative electrons circling it as in a typical Kepler set up. The problem is that such a circular motion of a charged particle causes electromagnetic radiation. Electromagnetic radiation carries energy away form the electron causing it to decay into the nucleus. This means that classical matter is unstable. But matter is stable! Niels Bohr invented a new model for the atom that does not have this problem. His model is still not correct but it is a good start and shows the idea of quantum mechanics. Figure : eft: The classical atom must decay due to radiation. Right: Bohr assumed discrete allowed radii. The electron can jump from one radius to another but while it is in one it does not radiate and is thus stable. The Hydrogen Atom We will derive this theory for the Hydrogen atom. Its nucleus consists of one proton charge +e, mass M) and there is one electron charge e, mass m) circling it. Our goal is to show that the energy of the Hydrogen atom can only take distinct discrete values. Having discrete energy values means that a continuous decay as in the classical atom) is not possible. Bohr's Quantization Now we start quantum mechanics. We quantize. What does this mean? It means that a certain quantity can only take on multiples of some number. Here this quantity is angular momentum. We postulate that the value of the angular momentum must be a multiple of the number 2. Why? No reason. It is a postulate. We know it works because the consequence of the postulate discrete energy levels) agrees with experiments. The postulate states that n, n N Discrete Radii The angular momentum can be written as r p mrv To nd the velocity v we assume classical trajectories: an electron circling a proton with radius r. The velocity v is then found by setting the Coulomb attraction equal to the centripetal acceleration a v 2 /r. m v2 r e2 4πɛ 0 r 2 v2 e2 4πɛ 0 m 2 The units of are the same as the units angular momentum. Maybe this makes the postulate a little easier to understand. In SI units the value is 34 kg m s r 2

3 The angular momentum squared can hence be written as Setting this equal to 2 n ) 2 we obtain We dene the Bohr radius with which we can right 2 m 2 r 2 v 2 m 2 r 2 e 2 4πɛ 0 m r e2 mr 4πɛ 0 e 2 4πɛ 0 mr n 2 2 r 2 4πɛ 0 e 2 m n2 a B 2 4πɛ 0 e 2 m 0.529Å r n a B n 2 We have found that only specic radii are allowed. We say: the radius is quantized. The little n used in the subscript of r n is used to show that the radius r is discrete. r a B, r 2 4a B, r 3 9a B,... Discrete Energy Spectrum Also here we start of classically and nd an expression for the energy 3. E T + V 2 mv2 e2 4πɛ 0 r e 2 2 4πɛ 0 r e2 4πɛ 0 r e2 4πɛ 0 2r Plugging in the quantized radii r n we get the quantized energies E n. E n e2 4πɛ 0 2a 0 n 2 3.6eV n 2 The quantized energies E n are called the energy levels of the states of the Hydrogen atom. We have shown that the electron cannot decay into the nucleus. The minimal radius is r a B not r 0). The energy of the atom in this minimal state is E 3.6eV. We call this the ground state of the Hydrogen atom..3 ight Emitted by an Atom Statement of the Problem Consider a Hydrogen atom with energy E 5. How many possible wavelengths can be emitted by the atom? Find two of them explicitly. Physical Principle A photon is emitted from an atom when it decays from a high energy level to a low energy level. The energy of the emitted light photon) is equal to the energy dierence of the two atomic states. E ph E init E final From the photoelectric eect we know that the energy of a photon is E ph hν and by using the relation λν c we get λ c ν hc E ph We conclude that when the atom decays from level n to level m a photon with wave length λ m n is emitted, where hc λ m n E m E n 3 It is not surprising that we nd T V /2. This is just the virial theorem. 3

4 Figure 2: Emission of photons as the atom decays from level n 5 to the ground state n. This can happen by emitting one photon with wave length λ 5, by emitting two photons or even more. Dierent Decay Paths One way the atom can decay is directly from level n 5 to the ground state n. Using the formula derived above we nd that its wave length is λ 5 hc E 5 E m 95nm The decay from n 5 can also happen in two stages, e.g. through level n 3. The atom will start of in n 5 and decay to n 3 while emitting a photon with λ 5 3. It will stay there for a certain time we will not be concerned with how long) and then decay from n 3 to n while emitting a photon with λ 3. The formula gives λ 5 3 hc hc 282nm, λ 3 03nm E 5 E 3 E 3 E Notice that λ 5 λ λ 3. Do you know why? Think about it. Number of Dierent Photons Two photons are dierent from one another if they have dierent energy. This means that we are looking for all the dierent wave lengths that can be emitted by the atom. When the atom is in n 5 it can decay in four possible ways. It can go to n 4, 3, 2,. If the atom managed to reach the state n 4 it can decay from there in three possible ways and so on. In total, the number of dierent photons that can be emitted is Spectral ines If a gas of hydrogen atoms so a large number of atoms) is excited to the level n 5 all the atoms will decay back done to the ground state n within a short time. As this happens 0 dierent types of photons will come ying out which we can observe as light in dierent colors. These colors are discrete and are thus called spectral lines. Search hydrogen spectrum on google and click on images.) 2 Fourier Transform 2. Orthogonality of Exponential Functions Statement of the Problem Consider the function 4 φ n x) dened in the interval [ /2, /2] φ n x) exp i n ) x, n N Is f n x) orthogonal to f m x) for n m? What happens if n m? 4 This is actually a set of functions {φ n x)}, one for each n. 4

5 Figure 3: All spectral lines for a Hydrogen atom excited to the n 5 energy level. Denition of Orthogonality The two vectors a and b are orthogonal if their inner product vanishes. a b ) b a a 2 a 3 b 2 0 b 3 This is true for any dimension. In the same way the two functions fx) and gx) are orthogonal if f x)gx) 0 In other words a function is a vector with an innite amount of dimensions. It is customary to write the inner product as f g f x)gx) The orthogonality condition is thus f g 0 Solution We plug the two function φ n x) and φ m x) into the integral φ n φ m. The borders of integration are /2 /2 because this is the interval in which the functions are dened. φ n φ m φ nx)φ m x) /2 /2 /2 /2 exp i n exp i ) x exp m n) x ) i m x This integral is zero unless n m in which case it is equal to. To see this we can use Euler, namely expiθ) cosθ) + i sinθ) φ n φ m /2 [ ) )] m n) m n) cos x + i sin x /2 ) The integral of a sinusoidal function over a full cycle vanishes. Hence, we get 0 if n m. φ n φ m 0, n m 5

6 If n m the cosine is and the sine is 0. We thus have for n m φ n φ m /2 /2, n m In summary we have φ n φ m { 0 n m n m δ nm 2.2 Fourier Series Orthogonal Vectors are Basis Vectors A set of orthogonal vectors is very useful since they span a basis. Take for instance the Cartesian base vectors ˆx, ŷ and ẑ. They span the three dimensional space. I.e. any 3D vector a can be written as a linear combination of these three vectors. a a ˆx + a 2 ŷ + a 3 ẑ The same is true for a set of orthogonal functions. As an example lets consider the set of functions { ) ) } 2 n 2 n sin x, cos x, n N which are dened in the interval [ /2, /2]. These are the base functions of all periodic functions with periodicity. Any periodic function fx), which satises fx + ) fx), can be written as a linear combination of said base functions. We know this simply as the Fourier series. fx) n0 ) n A n sin x + n0 ) n B n cos x Complex Fourier Series Instead of using sinusoidal functions we can use exponential functions. We know from Euler that they are equivalent. For a periodic function we can thus write fx) n C n exp i n ) x Notice that the exponential is just φ n x) used above. By replacing the coecients C n by f n we have fx) n f n exp i n ) x or fx) n f n φ n x) Expansion Coecients f n of the Complex Fourier Series We wish to nd the coecients f n of the expansion. We start from the expansion fx) f n exp i n ) x and use the orthogonality proved above φ n φ m /2 /2 n exp i n ) x exp i m ) x δ nm in order to nd the coecients f n. We multiply both sides of the expansion by exp imx/) / and integrate over x. /2 fx) exp i m ) x /2 f n exp i n ) x exp i m ) x /2 /2 n On the right hand side we can do the integral before the sum and recognize the δ nm. /2 /2 n f n exp i n ) x exp 6 i m x ) n C n δ nm f m

7 And thus Plugging this in we get f m /2 fx) 2.3 Fourier Transform Continuous imit this we dene and thus /2 n n We plug both k n and k in the expansion fx) exp i m x exp i n ) x ) or f m /2 /2 /2 φ n x) fx )φ m x ) /2 /2 /2 fx)φ m x) fx ) exp i n ) x ets continue our way towards the Fourier transform. The next step is going to continuity. For fx) n n k n n k k n+ k n exp i n ) x exp ik n x) k /2 /2 /2 /2 k fx ) exp i n ) x fx ) exp ik n x ) and take. Note that when doing this k n k and k dk. Also, the sum over n becomes an integral over k. We can now read of the expansion fx) with the coecients which are now a function) This is the Fourier transform. exp ikx) dk fx) fk) fx ) exp ikx ) dk e ikx fk) e ikx fx) Physical Meaning There are two dierent spaces, the x-space and the k-space. Any function can be represented in either one of those spaces. The two representations are equivalent because we know exactly how to go from one to another. The functions we are using for the expansion are plane waves. φ k x) e ikx Given a function in x-space we can expand it in plane waves to get the same function in k-space and vice versa. Example: Statement of the Problem Find the Fourier transform fk) of a Gaussian fx) e αx2 You may use e ay2 dy π a 7

8 Solution Plug in to the formula Switch variables y x + ik/2α fk) exp e ikx e αx2 [ exp α x + ik 2α ] [ k2 exp 4α fk) exp 2α exp [ k2 4α ] [ k2 4α ] ) ] 2 k2 4α [ α x + ik 2α dy exp [ αy 2] ) 2 ] Properties of the Fourier Transform inearity Shift Derivative Scaling 3 Dirac Delta Function A few properties of the Fourier transform fk) F [fx)] include F [afx) + bgx)] af [fx)] + bf [gx)] a fk) + b gk) F [fx + x 0 )] F [fx)] e ikx0 ikx0 fk)e [ ] d F fx) ikf [fx)] ik fk) [ ] d n F n fx) ik) n F [fx)] ik) n fk) F [fax)] a f 3. Denition, Representations and Properties Denition while ) k a This is the continuous version of δ nm. It is written as δx x 0 ). By denition b a b a n δ nm c m na a < m < b fx)δx x 0 ) fx 0 ) a < x 0 < b b a δx x 0 ) The Dirac delta δx x 0 ) is only dened inside the integral. It is meaningless without it. Representations It can be represented in dierent ways. A few examples are ɛ δx) lim ɛ 0 + π x 2 + ɛ 2 δx) lim e x/ɛ)2 ɛ 0 + πɛ sinx/ɛ) δx) lim ɛ 0 + π x 8

9 Example: Proof of a Representation function. Prove that the following limit is a valid representation of the Dirac delta δx) lim e x/ɛ)2 ɛ 0 + πɛ Solution. 2. In order to prove this we need to show three properties: δx 0) δx 0) 0 3. δx) ets do this one by one.. 2. δx 0) lim ɛ 0 + πɛ Ω δx 0) lim e x/ɛ)2 lim e Ωx)2 0 ɛ 0 + πɛ Ω π 3. δx) lim e x/ɛ)2 lim πɛ2 ɛ 0 + πɛ ɛ 0 + πɛ Figure 4: Graph of exp x/ɛ) 2 )/ πɛ with dierent values of ɛ. One can see that as ɛ approaches zero the function becomes more and more δ-like. Properties of the Dirac Delta Function Symmetric Scaling A few properties include can be proven under the integral) δx) δ x) δax) a δx) Delta of a function gx) with roots x i δgx)) i g x i ) δx x i) 9

10 3.2 Fourier Transform of the Dirac Delta Function Statement of the Problem Apply the Fourier transform formula to the Dirac delta function to nd its transform. Then inverse transform your result to prove the identity Solution Doing the inverse transform we get F [δx x 0 )] δx x 0 ) Remember this. It will turn out to be very useful! dk eikx x0) e ikx δx x 0 ) e ikx0 δx x 0 ) F [F [δx x 0 )]] [ ] F e ikx0 dk e ikx e ikx0 dk eikx x0) 0

OpenStax-CNX module: m The Bohr Model. OpenStax College. Abstract

OpenStax-CNX module: m The Bohr Model. OpenStax College. Abstract OpenStax-CNX module: m51039 1 The Bohr Model OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 By the end of this section, you will