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This essay question is worth 2 marks, or 20% of your total grade for this lab.
There are a couple of real-life complications to the type of experiment we would conduct above. For each, you should (quickly) research a suitable explanation and write it below.
Real-Life Complication #1:
For a given material with a set work function , there may be an increase in photocurrent (i.e. rate of photoelectrons being ejected) with an increase in frequency of light. This is especially noticeable when the frequency of light is just above but very close to the threshold frequency of the material (i.e. photon energy is very close to the work function).
By thinking about the structure of electrons within a metal, explain why this is the case; why would (assuming no changes to the number of photons being sent at the cathode) an increase in photon frequency cause an increase in the number of photoelectrons being emitted?
Real-Life Complication #2:
This one is really pedantic, and comes down to the definition of the term 'Intensity'.
Here, let us take 'Intensity' as the amount of power radiated by the light source per area, in Watts per Metre Squared.
With this definition, we get a very curious phenomenon that really only shows up in (some) simulations; if you increase the frequency of the light, you get a DECREASE in total photocurrent. This means that there is a decreased number of electrons being ejected, when the frequency of the light is increased! Why could this be the case? You may wish to think about the number of photons in a beam of light of a given energy.
When the transmitter and receiver are set up as shown in this diagram with the transmitter and reciever slits parallel:
the intensity on the receiver reads 0.98 mA. A polarising sheet is now placed between the transmitter and the receiver such that the polariser makes an angle of 30o with the slit in the transmitter. This is shown in the diagram below.
What signal strength would you expect the detector to detect now?
mA
(Note: the receiver measures the intensity in mA)
You measure the capacitance of two capacitors with uncertainties, C1 = (1603 ± 30) μF and C2 = (737 ± 15) μF. You connect these capacitors in series.
What is the uncertainty in the value for ?
1/μF (3 sig. fig.)
Note: this is asking for just the uncertainty, not the value, only enter the uncertainty into the box
The graph of potential difference (in volts) versus time (in seconds) across a charging capacitor is plotted below. It is a straight line with the equation V = 17.03 t . The current is measured to be a steady 1.31 mA. What is the capacitance of the capacitor?
C = mF
Configuration 2 (all in parallel):
C = mF
Configuration 3 (capacitor 1 in parallel with capacitors 2 and 3 in series):
C = mF
You have three capacitors, capacitor 1 has a capacitance 2.91 mF, capacitor 2 has capacitance 4.20 mF and capacitor 3 has a capacitance of 6.11 mF. Calculate the capacitance of the circuit in the following configurations:
Configuration 1 (all in series):
C = mF
This charged capacitor is now unplugged from the power supply without being discharged and plugged across a second uncharged capacitor with a capacitance of 4.82 mF, so that the positive and negative plates are connected. If a voltmeter is now used to measure the voltage across the first capacitor what value will it measure?
V
A capacitor with capacitance 9.90 mF has a voltage of 6.65 V applied across it. What is the charge on the positive plate?
C
THIS WRITTEN QUESTION WILL BE WORTH 15% OF THE TOTAL MARK FOR THE LAB (I.E., 1.5/10).
SUBMIT YOUR ANSWER BELOW ON MOODLE AND IT WILL BE MARKED AS A SEPARATE GRADE ITEM IN YOUR GRADEBOOK.
If you are confused, check the Week 1 Lab Video; this explains the marking procedure.
Question:
Which of the methods (Part 2, Part 5, Part 6) do you feel was the most accurate, and why? You should write 1-2 paragraphs.
You will be judged on your evaluation AND your reasoning.
IN YOUR ANSWER, MAKE SURE YOU INCLUDE ANY RELEVANT NUMBERS / FORMULAE AS YOUR MARKER WILL ONLY BE ABLE TO SEE WHAT YOU WRITE IN THIS SECTION.