Recent content by elemis

Thanks! I just have one last question, if you imagine a porphyrin ring versus simple benzene. It is my belief that the upper curve (see above diagram) is displaced more to the right in benzene. This is because a single electronic excitation in a massive porphyrin ring (a multi electron system)...

Theory and my Understanding: So I understand how the frank condon principle let's us effect electronic transitions instantaneously, since the motion of nuclei (on the timescale of such electronic transitions) is quite slow. Consequently, when a photon of light is absorbed you can have an...

Hi Chet, So to be clear, the alveoli expansion simply results in an drop in the effective concentration (activity) of the surfactant and hence since their surface excess decreases we note an increase in surface tension ?

So, the way I understand this is as follows : The alveoli (pretend they're bubbles) have diameters of the order of microns implying a massive pressure required to inflate them by the Young-Laplace equation. p_{in}-p_{out}=\frac{2\gamma}{r} However, the presence of pulmonary surfactant...

Any ideas ?

My professor, in his handout (picture below), says the following about this diagram : I disagree with him partly. For \Delta_{mix} G/nRT<0 mixing is spontaneous and hence there solutions would be miscible. Hence at \beta=2.5 should we not expect the components to be fully miscible. ...

So starting from the time dependent Schrödinger equation I perform separation of variables and obtain a time and spatial part. The spatial part is in effect the time independent Schrödinger equation. Since we are dealing with a free particle I can take the time independent equation, set V = 0...

What is the correct form of the LHS of the equation ? I'm assuming the RHS is completely correct ? I realized that \hbar k = mv = p can be substituted into the last step of my workings to obtain j = A2v which is the correct answer. Where exactly ?

Homework Statement Calculate the probability current density vector \vec{j}for the wave function \psi = Ae^{-(wt-kx)}. Homework Equations From my very poor and beginner's understanding of probability density current it is : \frac{d(\psi...

Yes, I know that the absolute value of angular momentum would be equal for each astronaut, but my initial argument was that L will be equal in each astronaut's case but have opposite signs. The reason for the opposite signs (in my mind) is the opposite signs for velocity (moving in opoosite...

I assume you are referring to : lLl = lrllpl*sinθ Yes, what you say make sense. Hence, I agree that lLl is proportional to lpl in the scenario. However, wouldn't v have different signs for each astronaut since they are moving off in opposite directions ?

I comprehend everything in your answer except the bit in bold. Can you explain how they change ? I understand that lLl = lrllpl*sinθ... Does this have anything to do with your explanation ?

Lets imagine a binary system of two astronauts in space connected to one another via light rope. The rope is taut and they're spinning round and round with their axis of rotation being the the axis perpendicular to the their centre of mass. Now, my question is this. Let's say they each let go...

I've carried out the steps for the time average for cos2x for limits 0 to T. I've gotten : \frac{1}{T}[\frac{1}{2}T+\frac{1}{4}sin2T] What do I choose to be T in this case ? Infinity ?

Because the limits are 0 to infinity the cosine function goes zero and hence we can ignore it completely, correct ?