AAMC 3R Answers 11-26

[ocwaveoc]

Here's the link
http://www.aamc.org/students/mcat/pssampleitems.pdf

And does anyone have the explanation for questions 11, 15 and 16?

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[HippocratesX]

I was having trouble with the same passage. IN aamc 4R there is a similar passage that combines the electrochemistry with some type of cell having a cathode and anode and puts it inside a circuit, like the picture for this passage. Can anybody explain this conceptually, in terms of what is going on...how the voltage is related to the electrons flowing from anode to cathode?

You might want to post this question in the physics study thread too.

 

[killinsound]

in relation to electron flow refer to V = IR



As your e- concentration increases (I said concentration and not just the amount because if you just increase the amount of e-'s with the amount of other atoms, you won't gain voltage because the amount of e-'s relative to everything else did not increase) your current increases. Since the resistance of a resistor that you put in a circuit is constant barring temperature fluctuations, etc., your voltage increases.

e-'s ALWAYS move from anode to cathode.

 

[killinsound]

Now for linking Voltage to anode/cathode.

Think of voltage as the GH in PE = MGH.

In a Galvanic cell, the voltage is positive, GH is positive, and the electron is allowed to "fall". The possibility of falling is the electric potential. If you are already at the ground then your potential is 0. Therefore, if the e- is already at the cathode your Voltage is 0.

Finally, in an electrolytic cell, you have to use energy or put in that amount of "falling potential (gh or Voltage)" to get it back to where it was. You have to put energy into a mass in the amount of GH difference between the height and the ground state. That is why you have to supply Voltage that is at least as much as the opposite of the potential of a electrolytic cell.

I'm not looking through the kaplan 3R because I'm going to take the PS section tomorrow

 

[HippocratesX]

thanks, that helps ALOT! I was having trouble realizing the cathode is ground and thus potential of zero.

 

[killinsound]

Here's the link
http://www.aamc.org/students/mcat/pssampleitems.pdf

And does anyone have the explanation for questions 11, 15 and 16?

11 - Bremmstrahlung is continuous as said in the passage and p1 and p2 are clearly peaks in the graph synonomous with line spectra, and the graph shows intensity vs wavelength. You should know that when electrons fall from higher n's, they give off photons that equal the energy level between the n's. therefore C is your only answer

15 - It tells you in the passage that HV regulates e- acceleration. Which means that qV = MA of the electron, so you want to increase the V of the hV

16 - Intensity = how many x-ray's emitted. P2 has the greatest intensity. If more xrays are being emitted by p2, p2 is more probable

 

[ocwaveoc]

11 - Bremmstrahlung is continuous as said in the passage and p1 and p2 are clearly peaks in the graph synonomous with line spectra, and the graph shows intensity vs wavelength. You should know that when electrons fall from higher n's, they give off photons that equal the energy level between the n's. therefore C is your only answer

15 - It tells you in the passage that HV regulates e- acceleration. Which means that qV = MA of the electron, so you want to increase the V of the hV

16 - Intensity = how many x-ray's emitted. P2 has the greatest intensity. If more xrays are being emitted by p2, p2 is more probable

Thanks for the reply.
Regarding your response to #15, how is qV = MA? I'm thinking that qV you refer to is the potential energy = charge x voltage difference the charge goes through and what you refer to MA as F=MA. Am I missing something?

 

[1nicelady]

Thanks for the reply.
Regarding your response to #15, how is qV = MA? I'm thinking that qV you refer to is the potential energy = charge x voltage difference the charge goes through and what you refer to MA as F=MA. Am I missing something?

I got this question wrong too. I put by increasing LV because I thought increasing temperature would increase the kinetic energy. But what I forgot about was they were asking about the electrons. Although KE and temperature are related, increasing the acceleration of the electrons (HV) is the better answer. If you increase the acceleration then the velocity increases and KE=1/2mv^2