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AQA Physics Unit 1 PHYA1 20th May 2013

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Original post by BajoLily
This may seem silly but can anyone give me a definition to learn for emf?


It's a bit easier if you understand it in relation to potential difference.

The potential difference is the energy transferred per coulomb through any two points in a circuit whereas the electromotive force is the electrical energy per unit charge PRODUCED by the source.

Also the mark scheme definition is “work done per unit charge".
Original post by StalkeR47
just did this question yesterday. HERE.... sorry for the poor quality...


Thank you.
Original post by SpartASH
Thank you.


Your welcome m8. Hope you can read the answers :tongue:
Can anyone please explain how to get 6v, 4v and 2v on that resistor question in the Jan 13 paper? Im so confused :frown:

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Original post by BajoLily
This may seem silly but can anyone give me a definition to learn for emf?


Here is the smaller one to remember... work done per energy supplied per unit charge by the battery.:cool:
Anyone got any bets on what the long answer question will be on?
Original post by x-Sophie-x
Can anyone please explain how to get 6v, 4v and 2v on that resistor question in the Jan 13 paper? Im so confused :frown:

Posted from TSR Mobile


Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps... :colondollar:
can anyone explain q7b on jan 11. thanks
Original post by Qari
The question is, why the existence of threshold frequency supports the particle nature of electromagnetic waves


Hey sorry I misread the question. Here is the better explanation and the answer... The photoelectric effect is the emission of electrons from a metal surface due to incident photons of energy (h * f) greater than or equal to the work function (h * f(threshold)) of the metal, the minimum energy required to liberate an electron. Each electron can absorb only one photon. Emitted electrons have maximum kinetic energy equal to the difference between the photon energy and work function, and this varies up to a maximum due to electrons below the surface having to do work to get to the surface before being emitted. The photoelectric effect provides evidence for the particle nature of light, as if light were acting as waves, energy could be absorbed by the electrons over a period of time until reaching the required amount and being emitted. This is not the case - electrons are emitted immediately once f is greater than or equal to the threshold frequency. Furthermore, if light were acting as waves, the energy transferred by the light would be dependant on its intensity. This is not the case - if f < f(threshold), then increasing the intensity will have no effect; and if f is greater than or equal to f(threshold), then increasing the intensity will increase the rate of emission of electrons, indicating that there are more photons landing on the metal surface per second.:rolleyes:
Original post by rainerised
can anyone explain q7b on jan 11. thanks


Hello! This is the answer I wrote.... 1200ohm resistor will get more current as the overall parallel resistance is decreased and 1200ohm resistor stays same. Thus, the voltage will also increase at 1200ohm resistor R2.
Original post by StalkeR47
Hello! This is the answer I wrote.... 1200ohm resistor will get more current as the overall parallel resistance is decreased and 1200ohm resistor stays same. Thus, the voltage will also increase at 1200ohm resistor R2.


I thought the pd across the 1200ohm resistor decreases?

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Original post by StalkeR47
Hey sorry I misread the question. Here is the better explanation and the answer... The photoelectric effect is the emission of electrons from a metal surface due to incident photons of energy (h * f) greater than or equal to the work function (h * f(threshold)) of the metal, the minimum energy required to liberate an electron. Each electron can absorb only one photon. Emitted electrons have maximum kinetic energy equal to the difference between the photon energy and work function, and this varies up to a maximum due to electrons below the surface having to do work to get to the surface before being emitted. The photoelectric effect provides evidence for the particle nature of light, as if light were acting as waves, energy could be absorbed by the electrons over a period of time until reaching the required amount and being emitted. This is not the case - electrons are emitted immediately once f is greater than or equal to the threshold frequency. Furthermore, if light were acting as waves, the energy transferred by the light would be dependant on its intensity. This is not the case - if f < f(threshold), then increasing the intensity will have no effect; and if f is greater than or equal to f(threshold), then increasing the intensity will increase the rate of emission of electrons, indicating that there are more photons landing on the metal surface per second.:rolleyes:


When you have an incident electron colliding with an orbital electron does the orbital electron take all of the incident electrons energy or not, which could result in kinetic energy which is the excess energy from the collision. Also when you have an electron exiting due to the absorption of a photon, can you get an excess in energy or is it that the phon had to have the exact amount of energy in order to excite the orbital electron, any higher or lower value won't excite the electron?


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Original post by StalkeR47
Hello! This is the answer I wrote.... 1200ohm resistor will get more current as the overall parallel resistance is decreased and 1200ohm resistor stays same. Thus, the voltage will also increase at 1200ohm resistor R2.


It would decrease as the series resistor would have an increased pd, so the pd reading will decrease across the parallel combination of resistors. Or you could think about it as the total resistance decreases across the parallel combination, do less resisyance on the 1200 ohms resistor do less pd needed :smile:


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Original post by rainerised
can anyone explain q7b on jan 11. thanks


When the temperature increases:

The resistance of the thermistor and so the whole circuit will decrease.
More current will pass through the circuit.
The p.d. of the 540 ohm resistor will therefore increase.
Which will result in the p.d. across the parallel combination (and hence the 1200 ohm resistor) decreasing. :smile:
Original post by rainerised
can anyone explain q7b on jan 11. thanks


Lol I just copied what I wrote. Sophie is correct! the pd across 1200ohm increases because for the battery, 1200ohm will have bigger resistance.
Original post by Jimmy20002012
It would decrease as the series resistor would have an increased pd, so the pd reading will decrease across the parallel combination of resistors. Or you could think about it as the total resistance decreases across the parallel combination, do less resisyance on the 1200 ohms resistor do less pd needed :smile:


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I know lol I just copied what I had written. I did not read the question. HAHA
Original post by x-Sophie-x
Can anyone please explain how to get 6v, 4v and 2v on that resistor question in the Jan 13 paper? Im so confused :frown:

Posted from TSR Mobile

Ok, so you can work out the total resistance to be RT=(140×103+115×103)−1=120×10311 Ω⇒I=VR=1.1×10−3AR_{T} = \left( \dfrac{1}{40 \times 10^{3}} + \dfrac{1}{15 \times 10^{3}} \right)^{-1} = \dfrac{120 \times 10^{3}}{11} \ \Omega \Rightarrow I = \dfrac{V}{R} = 1.1 \times 10^{-3} A

Now, when the voltage branches off into the 2 parallel branches, the voltage stays the same (each coulomb of charge still has the same amount of energy) but the current splits as the charges split. Call the current across A-E I1I_{1} and the current across B-F I2I_{2}

From conservation of energy, you know that the potential difference (i.e. the voltage drop) across A-E has to be 12V as when the current branches from E and goes back to the cell, there should be a drop of 12J in every coulomb of charge, therefore you know that by Ohm's law 12=I1×(20×103+20×103)⇒I1=1240×103=3×10−4A12 = I_{1} \times (20 \times 10^{3} + 20 \times 10^{3} ) \Rightarrow I_{1} = \dfrac{12}{40 \times 10^{3}} = 3 \times 10^{-4} A

Thus, the potential difference across A-C is 3×10−4×20×103=6V3 \times 10^{-4} \times 20 \times 10^{3} = 6V

By similar arguments, you can calculate the potential difference across BD and the difference in C-D is just the difference in potential difference from AC and BD.
(edited 10 years ago)
Original post by Jimmy20002012
When you have an incident electron colliding with an orbital electron does the orbital electron take all of the incident electrons energy or not, which could result in kinetic energy which is the excess energy from the collision. Also when you have an electron exiting due to the absorption of a photon, can you get an excess in energy or is it that the phon had to have the exact amount of energy in order to excite the orbital electron, any higher or lower value won't excite the electron?


Posted from TSR Mobile


The orbital electron takes energy as much as it needs in order to ionise. If an incident electron has more energy than what's called the work function, the excess energy becomes ke as a result. When an electron absorbs a photon, it excites to higher energy level. However, if the energy absorbed from the photon is more than the work function, the electron will ionise. An electron absorbs all the energy from a single photon so there will be no excess energy.
Original post by StalkeR47
The orbital electron takes energy as much as it needs in order to ionise. If an incident electron has more energy than what's called the work function, the excess energy becomes ke as a result. When an electron absorbs a photon, it excites to higher energy level. However, if the energy absorbed from the photon is more than the work function, the electron will ionise. An electron absorbs all the energy from a single photon so there will be no excess energy.


I am not talking in the context of photoelectric effect, its more to do with excitation in atoms?


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Original post by Jimmy20002012
I am not talking in the context of photoelectric effect, its more to do with excitation in atoms?


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What do you want to know then?:confused:

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