Commit 90794576 authored by Niklas Wahl's avatar Niklas Wahl
Browse files

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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