Vectorize output of for loop, but using specific vector values

I am sorry if this is already posted somewhere, but I can't find an example that uses a vector with specific values, not say, [0,1,2,3].

I need to use the following code:

k=1.38E-23;

E0= 2.72361635536024e-21

M=3;

myVector = zeros(M+1,1);

for T = [100,300,1000,3000]

ERatio0=exp(-E0/(k*T));

myVector(T) = ERatio0;

end

to give me a vector 4 values long, with the resulting ERatio0 values for the corresponding T values, should be only 4 of them. Matlab is calculating all 3000 values right now as if I wanted all the integer values. Probably simple, I'm just missing it somehow.

If someone really wants to be nice, can they explain how I could do this for multiple Eratio's, each using a different E0, E1, so on:

T=1000; %Kelvins

k=1.38E-23;

h=6.626e-34;

h_bar=6.626e-34/(2*pi);

c=2.998*10^8;

ccm=2.998*10^10; %speed of light in cm

w=1730 ; %in cm

xe=.00815; %in cm

%% a

E0=(h_bar*ccm*w*((0+.5)-xe*(0+.5)^2)); % Gives answer in J

E1=(h_bar*ccm*w*((1+.5)-xe*(1+.5)^2));

E2=(h_bar*ccm*w*((2+.5)-xe*(2+.5)^2));

E3=(h_bar*ccm*w*((3+.5)-xe*(3+.5)^2));

E4=(h_bar*ccm*w*((4+.5)-xe*(4+.5)^2));

E0_w=(h_bar*ccm*w*((0+.5)-xe*(0+.5)^2))/1.986e-23 % Gives answer in wavenumber (cm^-1)

E1_w=(h_bar*ccm*w*((1+.5)-xe*(1+.5)^2))/1.986e-23

E2_w=(h_bar*ccm*w*((2+.5)-xe*(2+.5)^2))/1.986e-23

E3_w=(h_bar*ccm*w*((3+.5)-xe*(3+.5)^2))/1.986e-23

E4_w=(h_bar*ccm*w*((4+.5)-xe*(4+.5)^2))/1.986e-23

ERatio0=(exp(-E0/(k*T)));

ERatio1=(exp(-E1/(k*T)));

ERatio2=(exp(-E2/(k*T)));

ERatio3=(exp(-E3/(k*T)));

ERatio4=(exp(-E4/(k*T)));

sumEratio=ERatio0+ERatio1+ERatio2+ERatio3+ERatio4;

nRatio0=ERatio0/sumEratio

nRatio1=ERatio1/sumEratio

nRatio2=ERatio2/sumEratio

nRatio3=ERatio3/sumEratio

nRatio4=ERatio4/sumEratio

nRatio0+nRatio1+nRatio2+nRatio3+nRatio4% summed, equals 1 as it should

p1=plot([0,1,2,3,4],[nRatio0,nRatio1,nRatio2,nRatio3,nRatio4]) %T=1000K

%%

%T=100K

T100=100;

ERatio0=(exp(-E0/(k*T100)));

ERatio1=(exp(-E1/(k*T100)));

ERatio2=(exp(-E2/(k*T100)));

ERatio3=(exp(-E3/(k*T100)));

ERatio4=(exp(-E4/(k*T100)));

sumEratio=ERatio0+ERatio1+ERatio2+ERatio3+ERatio4;

nRatio0=ERatio0/sumEratio

nRatio1=ERatio1/sumEratio

nRatio2=ERatio2/sumEratio

nRatio3=ERatio3/sumEratio

nRatio4=ERatio4/sumEratio

nRatio0+nRatio1+nRatio2+nRatio3+nRatio4% summed, equals 1 as it should

Again, I am missing some easy stuff here, but I bet this is simple for some of you.

Eventually I want to be able to make a plot of Eratios vs Temperature. There are 4 temperature values, and 5 Eratios, as you can see. Maybe output into vectors is the way?

Thanks.

Answer 1

[Voss]

k=1.38E-23;
E0= 2.72361635536024e-21
M=3;
myVector = zeros(M+1,1);
T = [100,300,1000,3000];
for ii = 1:numel(T)
   ERatio0=exp(-E0/(k*T(ii)));
   myVector(ii) = ERatio0;
end