OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

Wait i've just read that the you sometimes take the terminal pd to be the EMF of the supply because you are ignoring internal resistance so therefore the terminal pd must be the sum of the individual pd's i guess!!! :biggrin:

Woo, I'm right :woo:

Reply 101

JoshC.

2

What I'm surprised by is how low the grade boundaries are. In January, it was only 66/100 for an A! :confused:

I really hate this unit. In fact, I has OCR in general - it's just silly. However, I don't feel too bad in about it - just need to practice a couple more papers and I should be fine.

Reply 102

rafcan

Can someone please explain the difference between P.d & emf? thanks

Reply 103

Dale12

Original post by rafcan

Can someone please explain the difference between P.d & emf? thanks

e.m.f is the amount of energy gained per unit charge by the battery... the chemical reaction within the battery provides the charge and the energy... however p.d is the energy lost across a component by charges having to use energy to overcome resistance.. hope that helps

Reply 104

Dale12

Original post by A for Andromeda

Hi, if anyone could explain this one question to me, I will be very grateful :smile:

(Sorry if I explain it poorly, but I can't find the actually question paper online [Jan 2011 - Qu7 c.i.])

The light from a filament is veiwed through a diffraction grating, having 300 lines per mm. The continous first order spectrum appears between angles 7 and 12 (degrees) to the direction of the incident light;

State the colour of the light the is seen at an angle of;

7:
12:

The used: The angle = arcsin(wavelength/d)

And used typical values for red (~350nm) and violet (~700nm) and from this you get roughly 7 for red and 12 for violet, but in the mark scheme it has it the other way round :confused:

Probably very simple, but I really don't get how they got the two that was around :colondollar:

red has the 700nm wavelength, u got them the wrong way around

Reply 105

Dale12

Original post by Dale12

red has the 700nm wavelength, u got them the wrong way around

then that makes more sense because red will have the 12 degree angle and since it has the bigger wavelength it diffracts more

Reply 106

sweetascandy

18

Original post by rafcan

Can someone please explain the difference between P.d & emf? thanks

EMF= the energy transferred when other forms of energy is converted into electrical energy.
P.d. = energy transferred when electrical energy is turned into other forms of energy.

^Hope that makes it clear for you in some ways

Reply 107

sweetascandy

18

I have a question: when you're given the work function energy and the kinetic energy to work out the frequency of a radiation, how do you know which equation to use, out of 'f=(KE+work function)/h' or 'f=E/h'? Quote me please, thanks! Any help would be much appreciated :smile:

Reply 108

OllyHV

Say there are two bulbs connected in parallel, if one has 0 resistance will all the current flow through the 0 resistance one, meaning that neither will light up?

And if they are in series, and one bulb breaks, what happens?

Reply 109

the real viper

1

Original post by sweetascandy

I have a question: when you're given the work function energy and the kinetic energy to work out the frequency of a radiation, how do you know which equation to use, out of 'f=(KE+work function)/h' or 'f=E/h'? Quote me please, thanks! Any help would be much appreciated :smile:

Surely f=(KE+ work function)/h because you are given KE and work function and you know plancks constant. However you don't know the energy of the photon for E in E=hf.

Reply 110

The medjai

Original post by sweetascandy

I have a question: when you're given the work function energy and the kinetic energy to work out the frequency of a radiation, how do you know which equation to use, out of 'f=(KE+work function)/h' or 'f=E/h'? Quote me please, thanks! Any help would be much appreciated :smile:

I'd say the first because they have given work function and kinetic energy.

Reply 111

sohail.s

2

Original post by sweetascandy

I have a question: when you're given the work function energy and the kinetic energy to work out the frequency of a radiation, how do you know which equation to use, out of 'f=(KE+work function)/h' or 'f=E/h'? Quote me please, thanks! Any help would be much appreciated :smile:

they are both the same
Energy = hf = work function + KE(max)
So E=hf f=E/h

Reply 112

sweetascandy

18

Original post by the real viper

Surely f=(KE+ work function)/h because you are given KE and work function and you know plancks constant. However you don't know the energy of the photon for E in E=hf.

Original post by The medjai

I'd say the first because they have given work function and kinetic energy.

Maybe I shouldve mentioned it before, but I tried this equation and got the wrong answer. The mark scheme said use 'E=hf', hence I wanted to know if there was a way of working out which equation to use.

Reply 113

shuty

9

Helloo People.... Hows the revision going for everyone???

any thoughts on what might come up???

Reply 114

sweetascandy

18

Original post by shuty

Helloo People.... Hows the revision going for everyone???

any thoughts on what might come up???

Everything will come up in this exam. :scared:

Reply 115

shuty

9

Original post by sweetascandy

Everything will come up in this exam. :scared:

Reply 116

Abused Tampon

6

Original post by the real viper

Electricity

Radio: 10^-1 -------- 10^6
Microwaves: 10^-3 -------- 10^-1
Infrared: 7x10-7 -------- 10^-3
Ultraviolet:10^-8 ------- 4x10^-7
X-Rays: 10^-13 ------- 10^8
Gamma Rays: 10^-16 -------- 10^-1

do we need to know and memorise all the wavelength ranges?
if yes, is there any quick way of doing it?

Reply 117

Abused Tampon

6

does resistance in a component change current or voltage or both?

Reply 118

The medjai

Original post by Abused Tampon

does resistance in a component change current or voltage or both?

Both are affected, if you decrease Resistance then current increases hence less voltage needed.

(edited 12 years ago)

Reply 119

sweetascandy

18

Original post by Abused Tampon

do we need to know and memorise all the wavelength ranges?
if yes, is there any quick way of doing it?

Yeah we are, they do pretty much always ask them in past papers, so it would be a good idea. Just keep repeating them to yourself, and hopefully they'll sink in.

OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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