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okay so you have attempted part b)i) right?
SO POTASSIUM WILL EMIT ELECTRONS WHEN 350nm WAVE IS INCIDENT ON IT.
Basically, we were asked to find a pair of metal and wave in which the photon energy of the wave was greater than work function of the metal.
First of all, see what they are asking for: the force experienced by the wires.part C only please
View attachment 63236 plz complete it
thank you so muchI am not in A2 ...but still i read in the syllabus that for the examination in 2018 may june
Learning outcomes removed from the syllabus content
The following learning outcomes have been removed from the 2015 version of the syllabus. A Level learning outcomes are indicated in bold.
It is on page:74..part(r)
View attachment 63237
yh shes right ive left these quesI am not in A2 ...but still i read in the syllabus that for the examination in 2018 may june
Learning outcomes removed from the syllabus content
The following learning outcomes have been removed from the 2015 version of the syllabus. A Level learning outcomes are indicated in bold.
It is on page:74..part(r)
View attachment 63237
Lots of thanksI am not in A2 ...but still i read in the syllabus that for the examination in 2018 may june
Learning outcomes removed from the syllabus content
The following learning outcomes have been removed from the 2015 version of the syllabus. A Level learning outcomes are indicated in bold.
It is on page:74..part(r)
View attachment 63237
me
When nuclei are placed in magnetic field, they precess at Larmor frequency and emit r.f. pulses. This frequency depends on strength of magnetic field as w = Bγ. Different rf pulse frequency emitted allow nuclei to be located. Changing field also allows position of detection to be changed.Explain the function of the non-uniform magnetic field that is superimposed on a large uniform
magnetic field in diagnosis using nuclear magnetic resonance imaging (NMRI) (4marks) nov16/p42
View attachment 63277
I am having difficulty to write a model answer for this question.
For part 1, the photoelectric effect will not help because it supports the particulate nature of light.can someone help me for part 1
i dont understand it, i tried comparing the question to the observation of photoelectric effect but still got no luck :c
for the second part:
it says that wavelength is reduced so i thought that frequency is increased now since the wavelength already exceeds the threshold frequency it means that all it will do is allow electrons to have a greater maximum kinetic energy. But the markscheme says photon energy is larger so hence larger to kenietic energy is also larger what does that even mean!!!!!
and more maximum photoelectric current it says fewer photons per unit time so maximum current is smaller????
ohh achaaaa it makes sense now. thxSecond part
Remember this:
1- When wavelength decreases, frequency increases.
2- When frequency increases, photon energy increases.
3- When photon energy increases, Max k.e. increases.
(ONLY FOR PARTICULATE THEORY)
And vice versa.
learn the formula I=nhf where I is the current, n is the no. of photons and f is the frequency.ohh achaaaa it makes sense now. thx
but what would happen to the photoelectric current tho
It's because the ray passes twice through the layer of fat before it's intensity is detected.i understand how to do part one, its basically alpha=(Z2-Z1)^2/(Z2+Z1)^2 simple acoustic calculation
but part 2 I done understand how they did it,
i'm stuck what do to next?
this is what i did
0.012I=Ie^alphax
alpha is 48 from the table
but why did the marksheme say that alpha x is (48 *2x) and then they multiplied 0.018 with alpha when Ln (ing) bothsides ik that the 0.018 is the reflection coefficient at the boundary between fat and muscle.
but the part 2 only asks about the thickness of fat so surely i can just say the that alpha is 48 since its the absorbtion coefficient for fat???
It's because the ray passes twice through the layer of fat before it's intensity is detected.
So the expression is more like:
0.018 x (e^-48 x x*10-2)2
If you solve it this way you will get the same answer
Can you please explain why are you multiplying 0.018 by e^(-48*10-2)2 iIt's because the ray passes twice through the layer of fat before it's intensity is detected.
So the expression is more like:
0.018 x (e^-48 x x*10-2)2
If you solve it this way you will get the same answer
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