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# AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] watch

1. (Original post by 000alex)
How to do this one?

Firstly because only 1 out of 400 photons are detected multiply the corrected count rate by 400 to give you the count rate of the detector if it detected all photons incident on it. Then divide that by the area of detector because that will give you the intensity of the source at the distance it is from and because it is uniform in all direction the intensity is constant at this point therefore, you can use the inverse square law equation I=I0/4pi(r)2 and re arrange to make I0 the subject and 4pi(r)2 is the area of the sphere the gamma photons have spread out at that distance so I0 is activity of source because it gives the number photons emitted per second which is same as activity (number of disintegrations per second)

Hope that helps
2. How much do we need to know about quantum tunnelling?
3. (Original post by Klaxoii)
There's two types - constant height and constant current, either is applicable but I find constant height easier to understand.

Basically, the microscope has a really sharp tip, which is negatively charged, the surface of the sample is positively charged. This means since electrons exhibit wave like properties they propagate between the gap (only from - to +). This transfer of electrons means there's a current between the tip and the surface. The currents magnitude depends on the gap between tip and surface - bigger the gap the less electrons that transfer and hence a smaller current. So as the tip scans across the surface the current increases and decreases if there is a dip or peak, and hence you can build up an image of the surface.

Probably not very clear sorry
In mark schemes for questions I've seen something to do with limited probability, whats that? Apart from that though that does help
4. (Original post by AR_95)
How much do we need to know about quantum tunnelling?
Essentially nothing for the core unit. The alpha particle gains kinetic energy as a result of its gain of binding energy and has a greater probability of overcoming the Coulomb barrier keeping it within the nucleus.

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5. (Original post by Sam.B57)
In mark schemes for questions I've seen something to do with limited probability, whats that? Apart from that though that does help
It's to do with the further away the two charged plates the less chance there is of the electron 'jumping' across

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6. June 2011 Q2b

For the reaction to take place the a particle must come within a distance, d, from
the centre of the aluminium nucleus.

Calculate d if the nuclear reaction occurs when the a particle is given an initial
kinetic energy of at least 2.18 × 10–12 J.

The electrostatic potential energy between two point charges Q1 and Q2 is
equal to where r is the separation of the charges and ε0 is the
permittivity of free space.

In this question the Q1 is 13x1.6x10^-19 , I understand that because that's the number of electrons aluminium has,
but I don't understand why Q2 has a charge of 2x1.6x10^-19. Can anyone explain why?
7. Does anyone know why with electron diffraction when you calculate the de broglie wavelength of an electron you use λ=hc/E instead of λ=h/mv ? What does c have to do with the electron energy?
8. (Original post by gcsestuff)
It's to do with the further away the two charged plates the less chance there is of the electron 'jumping' across

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right... I think that makes sense, probably left it a bit late to try and learn that bit
9. (Original post by Me123456789)
June 2011 Q2b

For the reaction to take place the a particle must come within a distance, d, from
the centre of the aluminium nucleus.

Calculate d if the nuclear reaction occurs when the a particle is given an initial
kinetic energy of at least 2.18 × 10–12 J.

The electrostatic potential energy between two point charges Q1 and Q2 is
equal to where r is the separation of the charges and ε0 is the
permittivity of free space.

In this question the Q1 is 13x1.6x10^-19 , I understand that because that's the number of electrons aluminium has,
but I don't understand why Q2 has a charge of 2x1.6x10^-19. Can anyone explain why?
It's to do with proton number. The aluminium nucleus has a charge of +13e because that's how many protons the nucleus of Aluminium contains.

The alpha particle is equivalent to a Helium nucleus. In a similar way, it has a charge of +2e because it contains two protons.

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10. Can anyone help me on second part of this question doing turning points don't seem to know how to do it?

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Does anyone know why with electron diffraction when you calculate the de broglie wavelength of an electron you use λ=hc/E instead of λ=h/mv ? What does c have to do with the electron energy?
Both equations give the de Broglie wavelength but the former is easier to calculate from the conditions of the experiment. Electrons have a charge "e" and are accelerated by a potential difference "V". As such, their energy E is equal to eV.

From last year unit 1 you know that:

As ,

Subbing this into the first equation gives

12. (Original post by CD223)
It's to do with proton number. The aluminium nucleus has a charge of +13e because that's how many protons the nucleus of Aluminium contains.

The alpha particle is equivalent to a Helium nucleus. In a similar way, it has a charge of +2e because it contains two protons.

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ah yes, the proton number, I got a bit confused
Thank you! completely forgot about helium
13. (Original post by Me123456789)
ah yes, the proton number, I got a bit confused
Thank you! completely forgot about helium
No problem!

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14. (Original post by MSB47)
Firstly because only 1 out of 400 photons are detected multiply the corrected count rate by 400 to give you the count rate of the detector if it detected all photons incident on it. Then divide that by the area of detector because that will give you the intensity of the source at the distance it is from and because it is uniform in all direction the intensity is constant at this point therefore, you can use the inverse square law equation I=I0/4pi(r)2 and re arrange to make I0 the subject and 4pi(r)2 is the area of the sphere the gamma photons have spread out at that distance so I0 is activity of source because it gives the number photons emitted per second which is same as activity (number of disintegrations per second)

Hope that helps

So you can use I=Io/area^2 instead of I=Io/distance^2
15. (Original post by Sbarron)
So you can use I=Io/area^2 instead of I=Io/distance^2
Also I don't understand why there is a k constant if we never use it?
16. (Original post by AR_95)
How much do we need to know about quantum tunnelling?
For Turning points you just need to know as much detail as is necessary to explain STMs and how they work from a basic level
17. (Original post by Leonacatherine)
Also can anyone explain how electron scattering is used to determine nulcear radius
Here you go
18. Can someone please explain to me why in the June 2014 paper (question 1)b)ii.) you don't include the mass of the neutrons released in the fission reaction when calculating the energy released, but in the June 2013 and June 2011 papers (both question 1)c)i.) you do have to include the mass of electrons/neutrons released in the reactions when calculating the energy released in order to obtain the correct answer.
19. (Original post by CD223)
For those of you asking for it, here's what I'd write for the derivation of kinetic theory:

Attachment 430961

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this is great, thank you! also I love your handwriting :O
20. Can anyone doing medical physics explain to me when we should substitute negative values into the lens equation 1/f = 1/u + 1/v?

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