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

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Can someone explain what the link between count rate and Activity is please.
Original post by moeeahmed
Can someone explain what the link between count rate and Activity is please.


Activity is the total number of emissions per second in all directions from the source.

Whereas if you use a Geiger tube, you will only measure a fraction of this value (count rate).

both are measured in Bq.
I've took the info out of two books and compiled the basic rules.!!!!!! Could someone just check I've got it all the right way round pleaseimage.jpg
Can someone explain to me electron diffraction in estimating the size of the nucleus thanks


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How would you answer part b(ii) of this question in the June 2012 paper3 (b) A γ ray detector with a cross-sectional area of 1.5 × 10–3 m2 when facing the source is placed 0.18 m from the source. A corrected count rate of 0.62 counts s–1 is recorded. 3 (b) (i) Assume the source emits γ rays uniformly in all directions. Show that the ratio number of γ photons incident on detector number of γ photons produced by source is about 4 × 10–3.
3 (b) (ii) The γ ray detector detects 1 in 400 of the γ photons incident on the facing surface of thedetector.Calculate the activity of the source. State an appropriate unit.
Original post by gcsestuff
Can someone explain to me electron diffraction in estimating the size of the nucleus thanks


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So you have a very thin material somewhere.
You give electrons lots of energy by passing high voltages through them. This gives the electrons a de broglie wavelength of slightly less than the diameter of a nucleus.

The electron beam is fired at the thin material and then a counter is used to plot a graph of counts against angle some distance behind the thin material. Also might want to mention the electrostatic force is attractive (protons and electrons)

The angle between the center and the first minimum (theta min) along with the de broglie wavelength of the electron beam is used to get a value of the size of a nucleus.
(edited 8 years ago)
Isn't it electrostatic attraction between the negatively charged electron and the positively charged nucleus?
Reply 1707
Anyone know of any disadvantages of a CCD?


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Original post by moeeahmed
Isn't it electrostatic attraction between the negatively charged electron and the positively charged nucleus?


Yep that's what I meant
Can someone give me a brief outline of how a Transmission Electron Microscope works?
image.jpgHow do you do these?
Reply 1711
Original post by Sbarron
image.jpgHow do you do these?


[br]p1V1=p2V2[br][br]p_1 V_1 = p_2 V_2[br]

[br](101×103)(10×104)=p2(8×104)[br][br]\therefore (101 \times 10^{3})(10 \times 10^{-4}) = p_2 (8 \times 10^{-4})[br]

[br]p2=126×103 Pa[br][br]p_2 = 126 \times 10^3\ \text{Pa}[br]


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(edited 8 years ago)
Original post by CD223
[br]p1V1=p2V2[br][br]p_1 V_1 = p_2 V_2[br]

[br](101×103)(10×104)=p2(8×104)[br][br]\therefore (101 \times 10^{3})(10 \times 10^{-4}) = p_2 (8 \times 10^{-4})[br]

[br]p2=126×103 Pa[br][br]p_2 = 126 \times 10^3\ \text{Pa}[br]


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Thanks I just worked it out another way that took ten times as long lol I started finding the ratio of moles I didn't think I could do it that way because the volume of the tank stays the same
Original post by Sbarron
Thanks I just worked it out another way that took ten times as long lol I started finding the ratio of moles I didn't think I could do it that way because the volume of the tank stays the same


I didn't think of adding the two volumes together for V1
Reply 1714
Original post by Sbarron
Thanks I just worked it out another way that took ten times as long lol I started finding the ratio of moles I didn't think I could do it that way because the volume of the tank stays the same


How do you know the volume remains unchanged? It only says the temperature does.


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Original post by CD223
[br]p1V1=p2V2[br][br]p_1 V_1 = p_2 V_2[br]

[br](101×103)(10×104)=p2(8×104)[br][br]\therefore (101 \times 10^{3})(10 \times 10^{-4}) = p_2 (8 \times 10^{-4})[br]

[br]p2=126×103 Pa[br][br]p_2 = 126 \times 10^3\ \text{Pa}[br]


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surely V1 is 8 not 10 if they say the container has volume 8?
Then V2 is 10
Reply 1716
Original post by betbi3etwerrd
surely V1 is 8 not 10 if they say the container has volume 8?
Then V2 is 10


The air initially occupies both the pump and container. After one pump it now occupies just the container.

Also, if you think of it logically, the bike pump compresses the air into the container so the tyres can be pumped up. The volume must therefore reduce in order for the pressure within the tyre to increase. Assuming the temperature is constant.


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Reply 1717
Do we assume r0=1.4×1015r_0 = 1.4 \times 10^{-15}? I've seen several accepted values stated.


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How do I know when to use I1/i2 = (X1/x2 )^2

And when to use i1/i2=(x2/X1) ^2

I'm so confused :frown:

ImageUploadedByStudent Room1434374015.895155.jpg

ImageUploadedByStudent Room1434374027.705602.jpg


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Original post by CD223
Do we assume r0=1.4×1015r_0 = 1.4 \times 10^{-15}? I've seen several accepted values stated.


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I would've thought that they provided us with r_0 value?

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