Commit 90794576 by Niklas Wahl

### name changes

parent 5b5ee5a9
 ... ... @@ -71,17 +71,17 @@ dr = round(dr,9,'decimal') %% Ladungen %% q = 6*pi*eta * d * r/UC .* (v(:,1) + v(:,2)) ; q = round(q,22,'decimal') q = round(q,21,'decimal') tmp = 6*pi*eta*d / UC; dq = sqrt( (tmp * (v(:,1) + v(:,2)) .*dr).^2 + (tmp * r .* dv(:,1)).^2 + (tmp * r .* dv(:,2)).^2); dq = round(dq,22,'decimal') dq = round(dq,21,'decimal') qk = q ./ sqrt((1 + b ./ r).^3); qk = round(qk,22,'decimals') qk = round(qk,21,'decimals') dqk = sqrt( (3*b*q ./ (2*r.^2) .* (1 + b./r).^(-5/2) .* dr).^2 + ((1 + b./r).^(-3/2) .* dq).^2) dqk = round(dqk,22,'decimal') dqk = sqrt( (3*b*q ./ (2*r.^2) .* (1 + b./r).^(-5/2) .* dr).^2 + ((1 + b./r).^(-3/2) .* dq).^2); dqk = round(dqk,21,'decimal') %% Plot %% ... ... @@ -90,6 +90,8 @@ errorbar((1:nT)',qk,dqk,'vertical','o') %% eTmp = min(qk); en = round(qk / eTmp) den = dqk / eTmp; errorbar((1:nT)',en,den,'vertical','o'), grid on e = mean(qk ./ en); de = sqrt(sum((dqk ./ (en * nT)).^2)); e = round(e,4,'significant') ... ...
 %% Aufgabe 1 clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458 ... ...
 %% Aufgabe 4 clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458 ... ...
 %% Aufgabe 5 clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458 ... ...
 %% Aufgabe 5 clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; mp = 1.6727e-27 ... ... @@ -13,4 +13,4 @@ dv = 0.25 * v %% Ortsunsicherheit %% dp = mp * dv dx = h / (2*pi*dp) \ No newline at end of file dx = h / (4*pi*dp) \ No newline at end of file
 %% Aufgabe 7 clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; mp = 1.6727e-27 ... ... @@ -14,7 +14,7 @@ T = 4000 %K r = 1.75e10 %% Maximum %% lambdaMax = 2987.8e-6 / T lambdaMax = 2897.8e-6 / T %% Strahlungsleistung %% A = 4*pi*r^2 ... ...
 %% Wie rechne ich... die Austrittsarbeit beim Photoeffekt aus? clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; %% Gegeben: %% ... ...
 %% Wie rechne ich... die Energieniveaus in einem Potentialtopf aus? clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458 c = 299792458; %% Gegeben: %% b = 0.2e-9 ... ...
 %% Aufgabe 1) clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458; ... ...
 %% Wie rechne ich... Energiebergnge im Atom aus? clear all; e = 1.609e-19; e = 1.602e-19; h = 6.626e-34; me = 9.11e-31; c = 299792458; ... ...
 ... ... @@ -3,7 +3,7 @@ clear all; u = 1.6605e-27 c = 299792458 e = 1.609e-19 e = 1.602e-19 %% b) %% ... ...
 ... ... @@ -3,7 +3,7 @@ clear all; u = 1.6605e-27 c = 299792458 e = 1.609e-19 e = 1.602e-19 h = 6.626e-34; hQuer = h / (2*pi); me = 9.11e-31; ... ...
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