INTRODUCTION
Here is an example of how to approach long 6 mark question in A-Level Physics. The example question is taken from AQA AS Physics Paper1 June 2022 Q4. The question is about the Photo Electric Effect. Understand this technique and you will be able to use it for any AS or A-level question.
- Reading the Question
- Think what is it Asking
- Remember what you Know
- Form the Answer
- Answer the Question
- Check it’s Correct
- Feedback
1. HOW TO READ THE QUESTION
Read the question part 1 then carry on to Read part 2 to glean all the information in the question.
Notice there are 6 marks for the long answer so with 90 minutes to score 70 marks you have 8 mins.
Notice the question is about photoelectric effect.
Reread part 1 and underline the key parts of the question.
Look for equations on the data sheet related to photons & the photoelectric effect.
From the formula sheet find the relavent equations E=hf and hf= φ + Ekmax.
We are not answering the second part of the question yet, but it is worth reading as it will usually have information useful to answering the first half of the question. That is the case here, we see that “charged particles are emitted from the metal plate” and the mention of the keyword “work function” They are answering the question for us!
The last source of information is the data sheet. They have already told us the question is about the photoelectric effect and photons, so we find the relevant section of the data sheet and see the formula we need.
2. THINK WHAT IS BEING ASKED
Think about what is going on:
A metal plate is given a negative charge ie eletrons are added. Electromagnetic radiation reduces that charge by the photoelectric effect.
So the rate of loss of charge must be related to the rate of loss of electrons. Which we are told is due to the PE effect.
3. REMEMBER WHAT YOU KNOW
Remember that the PE effect was explained by Einstein by assuming the light is composed of packets of energy called photons. The photons are packets of energy, the energy given by E=hf .That photon incident on a metal surface will transfer all of its energy to one electron in the surface of the metal. If the energy gained by the electron is above a threshold value φ the electron can escape from the metal.
4. FORM THE ANSWER
Collect the key words in your head, photoelectric effect, threshold frequency, photons and photon energy. Note the key formulae.
Note the three points that they ask for your answer to cover. 1) the plate loses charge,- photo electrons 2) The frequency must be above a certain value – threshold exceeded see equations 3) The rate of loss increases with intensity. – number of photons each only releasing one eletron.
Finally give the examiners back their words. (Underline quoted text)
5. PUTTING THAT TOGETHER THE ANSWER IS :
Answer to Question
The plate looses charge due to loss of electrons which are released from the metal surface by electromagnetic radiation due the photo-electric effect.
The photoelectric effect will happen when the frequency of the radiation is such that the energy of the photons given by E=hf is greater than φ the work function of the particular metal.
Once the incident radiation has photon energies above the cut off frequency then the rate of loss of charge will be directly proportional to the intensity, which is a measure of the rate at which photons are incident on the surface. Each photo is able to release only one electon.
Above the cut off frequency, the frequency of the radiation will not change the rare of release of electrons only their maximum kinetic energy Ekmax = hf – φ.
6. FINAL CHECK
FINALLY quickly read the question again to check that your answer covers all the points you underlined in the question. Then smile real big and move on to the next question knowing you have just gained another 6 marks which is another grade. WELL DONE!
6b. OH JUST A MINUTE THERE’S ANOTHER PART TO THIS QUESTION
Answer to Question part 2
And this is simple, you’ve already found Ekmax = hf – φ and E=hf
Given f = 1.2×10^15 Hz and Ekmax 1.1ev from data sheet h= 6.63 x 10^-34 J and 1eV = 1.6 x 10^-19
Φ=hf-Ekmax the equation works fine in eV instead of Joules
Work function = 6.63×10^-34 x 1.2 x 19^+15/1.6×10^-18 – 1.1. eV = 3.9 eV
7. FINAL FINAL CHECK
ANSWER CHECK does the number look right, YES expect a few eV
8. WHAT DID THE EXAMINER THINK?
If you are perparing for the exam you have the luxury of immediate feedback by checking the Marking Scheme, and possibly the Examiner’s Report. Both very useful sources of information.
9. SUMMARY
Answering the long 6 mark questions doesn’t have to be so daunting if follow the simple steps outlined above. The most important thing is not to shy away from them, but practice this method a lot. It will bring you a lot of marks on the day.
This is an example of the way I tutor. If you have found it useful please leave me a comment. Good luck with the exam!