Physik IV - Lösungen - Aufgaben März 2010

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1 Physik IV - Lösungen - Aufgaben März

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8 4. (a) from the Bragg law 2d 220 sin θ = nλ it follows that d 220 = pm. In reality the lattice distance can be determined to an accuracy of tens of femtometers. (b) N A is determined by the ratio between the molar volume V mol and the atomic volume V atom, N A = V mol /V atom. The atomic volume can be determined from the lattice structure of a silicon crystal and is given by V atom = m 3. From the atomic mass of 28 Si the number of mole in 1kg silicon can be calculated, n Si = mol, and by multiplication with the molar volume V mol = N A V atom = cm 3 /mol, we obtain the volume of the sphere, V = n Si V mol = n Si N A V atom = cm 3. The radius r = 3 3V/4π = cm follows. (c) Gaussian error propagation: dn A = N A /10 6 = N A = V mol /V atom = V m 1 m mol V atom = 4πr3 m 1 3m mol V atom dn A = 4πr2 m 1 m mol V atom dr m = 1 kg, m/mol = kg/mol, r = cm dr < 7.8 nm (d) For natural composition of Si we have an atomic weight of 28.1 g/mol instead of 28 g/mol and the radius reduces to cm, 0.03 mm less than with a pure 28 Si crystal. Note, that this difference is huge compared to the required precision in r and therefore the purity is of highest importance as well. Note: the lattice distance d 220 obtained here is twice as small as in the paper referenced in the problem sheet. This is because a wavelength of 0.2 nm should be used to observe the first diffraction maximum at the indicated angle. 2

9 Übungsaufgabe 2 Init Laden von Paketen und definition von Konstanten Set Styles Aufgabe 1 a) Energiedichte gemäss Planck (in Frequenz) : In[63]:= u Ν_, T_ : 8 Π h c 3 Ν 3 Exp h Ν 1 kb T Wien' sches Verschiebungsgesetz aus du Ν dν 0 : In[17]:= solderiv Solve D u Ν, Ν 0, Ν InverseFunction::ifun : Inverse functions are being used. Values may be lost for multivalued inverses. Solve::verif : Potential solution Ν 0 possibly discarded by verifier should be checked by hand. May require use of limits. Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[17]= Ν kb T 3 ProductLog 3 3 h In[48]:= solderiv. kb BoltzmannConstant, h PlanckConstant N Out[48]= Ν T Kelvin Second ergibt das Wien' sche Verschiebungsgesetz : Ν max T Hz K 1 In[120]:= wiendispl T_ : Evaluate Ν. First solderiv. kb kboltz, h hplanck N wiendispl T Out[121]= T

10 2 solution_exm2.nb Aufgabe 1 b) Weg 1 Berechnung direkt aus der Frequenzdarstellung durch Ersetzen von Ν max e E max h: In[158]:= wiendisplen T_ : wiendispl T hplanck ec wiendisplen T Out[159]= T Berechnung der Temperatur für E ev : In[165]:= solt First Solve wiendisplen T 13.6, T Out[165]= T Weg 2 Energiedichte als Funktion der Photonenenergie in ev : Ν -> E/(q h) (q... Ladung Elektron) in Energie anstatt in Frequenz : P Ν dν P E dν de de P' E de mit Ν q h E bzw. dν q h de : In[136]:= uen En_, T_ : 8 Π h c 3 Ν 3 Exp h Ν 1 kb T q h. Ν q h En In[138]:= uen En, T Out[138]= c 3 8 En 3 Π q 4 En q 1 kb T h 3 In[139]:= solderiven Solve D uen En, T, En 0, En InverseFunction::ifun : Inverse functions are being used. Values may be lost for multivalued inverses. Solve::verif : Potential solution En 0 possibly discarded by verifier should be checked by hand. May require use of limits. Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[139]= En kb T 3 ProductLog 3 3 q

11 solution_exm2.nb Energie (in ev) bei Maximum des Spektrums : In[147]:= solderiven. kb BoltzmannConstant, h PlanckConstant, q ElectronCharge Out[147]= En Joule T Coulomb Kelvin In[148]:= wiendisplen T_ : Evaluate En. First solderiven. kb kboltz, h hplanck, q ec wiendisplen T Out[149]= T Berechnung der Temperatur für E ev : In[161]:= solt First Solve wiendisplen T 13.6, T Out[161]= T Spektrum In[164]:= Plot uen x, T. solt. kb kboltz, h hplanck, q ec, c clight, x, 0, , GridLines wiendisplen T. solt, None, Frame True, FrameLabel "E ev ", "u J m 3 ev 1 " Out[164]= u J m 3 ev Aufgabe 1 c) Integriere über den Teil des Spektrum mit E < 13.6 ev In[189]:= Ebelow Integrate Evaluate uen En, kb kboltz, h hplanck, q ec, c clight, En, 0, 13.6 Chop Out[189]= Gesamtenergie

12 4 solution_exm2.nb In[190]:= Entot Simplify Integrate uen En, T, En, 0,, Assumptions q 0, kb 0, T 0. kb kboltz, h hplanck, q ec, c clight, T 3000 Out[190]= Anteil überhalb 13.6 ev : In[191]:= Out[191]= Entot Ebelow Entot