Physics: Post your doubts here!

[Aishayasin]

Part bii in detail please

 

[anastasia grey113]

View attachment 63174


Part bii in detail please
View attachment 63175

okay so you have attempted part b)i) right?
that shows you know what work function is
you are half way through to the answer
the point that is to be noted is that it is the MINIMUM energy required
this will help us understand the table
on tungsten, if a wave having photons of energy lesser than 4.49 is incident, it will NOT emit any electrons
you can read off sodium and potassium's values off the table in the same way

we are given two waves of wavelengths 350nm and 700nm
v just have to figure out which metal will emit electrons when one of the waves is incident on it
for this we will have to find the energy of photons in each wave using
E = hc/ λ where h=Planck's constant, c= speed of light and λ=wavelength of wave
Now putting in the values for the first wave we have,
Wave (700nm)
E= (6.63x10-34 x 3x108)/700x10-9
the answer will be 2.84x10-19 Joules

Use the same formula for the 2nd wave
the answer will be 5.68x10-19 Joules

Now as you can see that the values we have calculated are in Joules but the work functions are in eV
so convert these values using
eV = E/1.6x10-19
This way the answers will be
first wave (700nm) = 1.78 eV
2nd wave (350nm) = 3.55 eV

Now it will be easier to analyse the values. We just have to find the metal which has the work function LESSER THAN the energy of photons in one of the waves.
The first wave is definitely not suitable for any of the three metals since all of them have work functions higher than its energy.
The 2nd wave?
Only Potassium seems to fit in because of its lower work function that the energy of 2nd wave.
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.

 

[Holmes]

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.

nicely explained Huh? [/QUOTE]
anastasia grey113

 

[GarryTheGhost]

part C only please

 

[anastasia grey113]

part C only please

First of all, see what they are asking for: the force experienced by the wires.
And F is directly proportional to B as B=F/IL
so in order to have a high magnitude of force between the two wires, a high value of B is required.
But as you can see, the constant in the formula given is very small (2x10-7)
so even if a force is produced under normal circumstances, it will be so small that it would be easily overlooked

so for a very noticeable value of B (let's suppose 1T), I/d would have to be 1/2 x10-7 which is a very small value
and for that the current has to be very large around 5000A and the distance has to be as small as 1mm
but this is practically impossible so you see this ratio can never be achieved

and even if it is, the force produced might not exactly produce the same effect because the wires also have weight which acts in a perpendicular direction.
so overall resultant will be downwards as weight of wire must be very large (let's say 0.2N) but the relative force must be very small with a usual 220V ac current and a distance of around minimum 1cm.

 

[Deeksha Tamang]

plz complete it

 

[Hamnah Zahoor]

View attachment 63236 plz complete it

I 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)

 

[Deeksha Tamang]

I 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

thank you so much

 

[anastasia grey113]

I 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 ques

 

[Hussein Yasir]

T

I 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 thanks

 

[Holmes]

Aishayasin , anastasia grey113 , Hamnah Zahoor
Holmes
who else is doing A Levels???

 

[Deeksha Tamang]

Aishayasin , anastasia grey113 , Hamnah Zahoor
Holmes
who else is doing A Levels???

me

 

[GarryTheGhost]

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????

 

[anastasia grey113]

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.

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.

 

[anastasia grey113]

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????

For part 1, the photoelectric effect will not help because it supports the particulate nature of light.
Look into the wave nature instead which is very different.
According to that theory, there is SUFFICIENT TIME DELAY between electron emmision and light incidence.
This is because electrons GAIN ENERGY FROM MORE THAN ONE PHOTONS and when they 'collect' all this energy and it's equal to work function, they escape. This obvio takes some time.
This also means, that if suppose a low frequency light is shone, the electrons will still be emitted, only more time will be taken because as low energy photons are present it will take more time for electrons to gather up energy equal to work function. So electrons will be EMITTED AT ALL FREQUENCIES.
But if intensity is increased, more photons given out per unit time so electrons will gain more energy and more quickly so their max. kinetic energy will also increase.
Write the points I wrote in capital.
And remember: THIS IS JUST THE WAVE THEORY OF LIGHT. THE PARTICULATE THEORY IS VERY DIFFERENT FROM THIS.

Second 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.

 

[GarryTheGhost]

Second 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.

ohh achaaaa it makes sense now. thx
but what would happen to the photoelectric current tho

 

[anastasia grey113]

ohh achaaaa it makes sense now. thx
but what would happen to the photoelectric current tho

learn the formula I=nhf where I is the current, n is the no. of photons and f is the frequency.
You can also write it as I=nh3x108/λ so that means
I is inversely proportional to the wavelength.
So if n (intensity or no.) is constant and wavelength is reduced, the current will increase.

 

[GarryTheGhost]

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???

 

[anastasia grey113]

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

 

[GarryTheGhost]

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

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 i
I understand e^ part but why is 0.018 being multiplied to it.
shouldn't it just be this:
0.012=e^((-48*10-2)2)
i just dont understand why you're using the coefficient alpha thats been calculated in the first part of the question
Because itn't the formula:
I=Io e^-αx
the Intensity cancel out, but what i dont understand is that why are we using the α and the µ at the same time in this question.
i'm so confused.